Q: 1. (b) Find the range, rank, kernel and nullity of the linear transformation T : R^4 - R^3 given by T(x, y, z, w) = (x - w, y + z, z - w). (10 Marks)

1. Conceptual Foundation & Definitions Let $T: V to W$ be a linear transformation from a vector space $V$ into a vector space $W$ over the field $mathbb{R}$. In this problem: $$V = mathbb{R}^4 quad text{and} quad W = mathbb{R}^3$$ The mapping is explicitly defined by: $$T(x, y, z, w) = (x - w, y + z, z - w)$$ We recall the key definitions: 1. Kernel (Null Space): The set of all vectors in the domain $V$ that map to the zero vector of the codomain $W$: $$ker(T) = left{ (x, y, z, w) in mathbb{R}^4 mid T(x, y, z, w) = (0, 0, 0) right}$$ 2. Nullity: The dimension of the kernel as a subspace of $V$: $$operatorname{nullity}(T) = dim(ker(T))$$ 3. Range (Image Space): The set of all vectors in the codomain $W$ that are images under $T$: $$operatorname{range}(T) = operatorname{Im}(T) = left{ T(x, y, z, w) mid (x, y, z, w) in mathbb{R}^4 right}$$ 4. Rank: The dimension of the range as a subspace of $W$: $$operatorname{rank}(T) = dim(operatorname{range}(T))$$ 5. The Dimension Theorem (Rank-Nullity Theorem): For any linear transformation from a finite-dimensional vector space $V$: $$operatorname{rank}(T) + operatorname{nullity}(T) = dim(V) = 4$$ --- 2. Determination of the Kernel and Nullity A vector $(x, y, z, w) in mathbb{R}^4$ belongs to $ker(T)$ if and only if: $$T(x, y, z, w) = (x - w, y + z, z - w) = (0, 0, 0)$$ Equating each component to zero yields a homogeneous system of three linear equations in four unknowns: $$begin{aligned} (1) & quad x - w = 0 implies x = w (2) & quad y + z = 0 implies y = -z (3) & quad z - w = 0 implies z = w end{aligned}$$ From equation (3), we have $z = w$. Substituting $z = w$ into equation (2): $$y = -w$$ And from equation (1): $$x = w$$ Here, $w$ is a free parameter ($w in mathbb{R}$). Any vector $(x, y, z, w) in ker(T)$ can therefore be parameterized in terms of $w$: $$(x, y, z, w) = (w, -w, w, w) = w(1, -1, 1, 1)$$ Thus, the kernel of $T$ is the one-dimensional subspace spanned by the single non-zero vector $(1, -1, 1, 1)$: $$ker(T) = oper

Q: 1. (a) Can the set {(0, 0, 0, 3), (1, 1, 0, 0), (0, 1, -1, 0)} be extended to form a basis of the vector space R^4 ? Justify your answer. (10 Marks)

1. Conceptual Foundation & Mathematical Framework In linear algebra, a basis of a finite-dimensional vector space $V$ over a field $mathbb{F}$ is defined as a subset $mathcal{B} subset V$ satisfying two fundamental axioms: 1. Linear Independence: No vector in $mathcal{B}$ can be expressed as a linear combination of the remaining vectors in $mathcal{B}$. 2. Spanning Property: The linear span of $mathcal{B}$ equals the entire space, i.e., $operatorname{span}(mathcal{B}) = V$. For the Euclidean space $mathbb{R}^4$, the dimension is known to be: $$dim(mathbb{R}^4) = 4$$ Consequently, by the fundamental dimension theorem, any basis of $mathbb{R}^4$ must consist of precisely 4 linearly independent vectors. The foundational Basis Extension Theorem (a direct corollary of the Steinitz Exchange Lemma) asserts that: If $V$ is a finite-dimensional vector space with $dim(V) = n$, and $S = {v1, v2, dots, vk}$ is an arbitrary linearly independent subset of $V$ where $k < n$, then $S$ can always be extended to a complete basis of $V$ by adjoining $(n - k)$ appropriately chosen vectors from any spanning set of $V$ (such as the standard canonical basis ${e1, e2, dots, en}$). Therefore, to answer the question affirmatively, we must rigorously prove that the given set of three vectors: $$S = {v1, v2, v3} = {(0, 0, 0, 3), (1, 1, 0, 0), (0, 1, -1, 0)}$$ is linearly independent in $mathbb{R}^4$. If it is linearly independent, the Basis Extension Theorem guarantees that it can be extended to form a basis of $mathbb{R}^4$. --- 2. Rigorous Test of Linear Independence Let $c1, c2, c3 in mathbb{R}$ be scalars such that the linear combination vanishes: $$c1 v1 + c2 v2 + c3 v3 = mathbf{0} = (0, 0, 0, 0)$$ Substituting the coordinates of $v1, v2, v3$: $$c1 (0, 0, 0, 3) + c2 (1, 1, 0, 0) + c3 (0, 1, -1, 0) = (0, 0, 0, 0)$$ Expanding into coordinate components: $$(c2, c2 + c3, -c3, 3c1) = (0, 0, 0, 0)$$ Equating respective components gives the homogeneous linear system: $$begin{aligned} text{Compon

Q: 1. (e) Find the equation of the cone whose vertex is the point (1, 1, 0) and whose guiding curve is y = 0, x^2 + z^2 = 4. (10 Marks)

1. Conceptual Foundation & Definition of a Cone In three-dimensional analytical geometry, a cone is a ruled surface generated by a variable straight line (called a generator) that passes through a fixed point $V(alpha, beta, gamma)$ (called the vertex) and continuously intersects a given fixed space curve (called the guiding curve or directrix). Here, we are provided with: - Vertex ($V$): $(alpha, beta, gamma) = (1, 1, 0)$ - Guiding Curve: Given by the intersection of the plane and the circular cylinder: $$begin{cases} y = 0 x^2 + z^2 = 4 end{cases}$$ This represents a circle of radius $R = 2$ centered at the origin lying in the $xz$-plane ($y = 0$). --- 2. Equations of Variable Generators Passing through the Vertex Any straight line passing through the fixed vertex $V(1, 1, 0)$ can be written in symmetric form with direction ratios $l, m, n$: $$frac{x - 1}{l} = frac{y - 1}{m} = frac{z - 0}{n} = r$$ where $r$ is a scalar parameter along the generator. Expressed in parametric form, the coordinates of an arbitrary point $P(x, y, z)$ on this generator are: $$x = 1 + lr, quad y = 1 + mr, quad z = nr$$ --- 3. Intersection of the Generator with the Guiding Curve If this generator intersects the guiding curve, there must exist a point on the generator lying in the plane $y = 0$. Setting the $y$-coordinate to zero: $$y = 1 + mr = 0 implies mr = -1 implies r = -frac{1}{m} quad (m eq 0)$$ Substituting this specific value of the parameter $r = -frac{1}{m}$ into the coordinates for $x$ and $z$, we obtain the point of intersection $Q$ of the generator with the plane $y = 0$: $$begin{aligned} xQ &= 1 + lleft(-frac{1}{m}right) = 1 - frac{l}{m} = frac{m - l}{m} yQ &= 0 zQ &= nleft(-frac{1}{m}right) = -frac{n}{m} end{aligned}$$ Since point $Q(xQ, yQ, zQ)$ lies on the guiding curve, its coordinates must satisfy the second equation defining the curve: $$x^2 + z^2 = 4$$ Substituting $xQ$ and $zQ$: $$left(frac{m - l}{m}right)^2 + left(-frac{n}{m}right)^2 = 4$$ Multiplying through by $m^

Q: 2. (a) Let T : R^3 - R^2 be a linear transformation such that T(1, 1, -1) = (1, 0), T(4, 1, 1) = (0, 1) and T(1, -1, 2) = (1, 1). Find T. (15 Marks)

1. Conceptual Framework & Existence-Uniqueness Theorem In linear algebra, a foundational theorem states that a linear transformation $T: V to W$ between finite-dimensional vector spaces is uniquely determined by its action on a basis of the domain $V$. Specifically: Let $mathcal{B} = {v1, v2, dots, vn}$ be a basis of a vector space $V$ over a field $mathbb{F}$, and let $w1, w2, dots, wn$ be arbitrary vectors in a vector space $W$. Then there exists a unique linear transformation $T: V to W$ such that $T(vi) = wi$ for each $i = 1, 2, dots, n$. For this problem: - Domain: $V = mathbb{R}^3$, where $dim(V) = 3$. - Codomain: $W = mathbb{R}^2$. - Given vectors in $mathbb{R}^3$: $$v1 = (1, 1, -1), quad v2 = (4, 1, 1), quad v3 = (1, -1, 2)$$ - Prescribed images in $mathbb{R}^2$: $$w1 = (1, 0), quad w2 = (0, 1), quad w3 = (1, 1)$$ Our goal is to find the explicit closed-form formula for $T(x, y, z)$ for any arbitrary vector $(x, y, z) in mathbb{R}^3$. --- 2. Verification of Basis Property of ${v1, v2, v3}$ Before computing $T$, we must verify that the set ${v1, v2, v3}$ is linearly independent, and hence forms a valid basis for $mathbb{R}^3$. Construct the $3 times 3$ matrix $P$ having $v1, v2, v3$ as its columns: $$P = begin{bmatrix} 1 & 4 & 1 1 & 1 & -1 -1 & 1 & 2 end{bmatrix}$$ Let us compute the determinant of $P$: $$begin{aligned} det(P) &= 1 cdot begin{vmatrix} 1 & -1 1 & 2 end{vmatrix} - 4 cdot begin{vmatrix} 1 & -1 -1 & 2 end{vmatrix} + 1 cdot begin{vmatrix} 1 & 1 -1 & 1 end{vmatrix} &= 1 cdot (2 - (-1)) - 4 cdot (2 - 1) + 1 cdot (1 - (-1)) &= 1 cdot 3 - 4 cdot 1 + 1 cdot 2 &= 3 - 4 + 2 = 1 eq 0 end{aligned}$$ Because $det(P) = 1 eq 0$, the vectors $v1, v2, v3$ are strictly linearly independent and form a basis of $mathbb{R}^3$. Consequently, the transformation $T$ exists and is uniquely defined. --- 3. Representation of an Arbitrary Vector $(x, y, z)$ in the Basis Let $(x, y, z)$ be an arbitrary vector in $mathbb{R}^3$. We express $(x, y, z)$ as a unique linear comb

Q: 2. (b) Using Mean Value Theorem, prove that pi/6 + sqrt(3)/15 < sin^(-1)(3/5) < pi/6 + 1/8. (15 Marks)

1. Conceptual Foundation & Lagrange's Mean Value Theorem Lagrange's Mean Value Theorem (LMVT) is a fundamental cornerstone of real analysis. It states: Let $f: [a, b] to mathbb{R}$ be a real-valued function satisfying: 1. $f$ is continuous on the closed bounded interval $[a, b]$; 2. $f$ is differentiable on the open interval $(a, b)$. Then there exists at least one point $c in (a, b)$ such that: $$frac{f(b) - f(a)}{b - a} = f'(c) iff f(b) - f(a) = (b - a)f'(c)$$ LMVT provides a rigorous analytical bridge connecting local properties (derivatives at an intermediate point $c$) to global properties (increments of the function over $[a, b]$). When $f'(x)$ is strictly monotonic on $[a, b]$, the bounds on $c in (a, b)$ yield strict inequalities bounding the value of $f(b)$. --- 2. Selection of Function and Interval We examine the expression to be bounded: $$sin^{-1}left(frac{3}{5}right)$$ Notice that: - $frac{3}{5} = 0.60$ - We know the exact value of the inverse sine at a nearby standard angle: $$sinleft(frac{pi}{6}right) = frac{1}{2} = 0.50 implies sin^{-1}left(frac{1}{2}right) = frac{pi}{6}$$ Since $0.50 0$. Therefore, for $frac{1}{2} < c < frac{3}{5}$: $$gleft(frac{1}{2}right) < g(c) < gleft(frac{3}{5}right)$$ $$frac{1}{sqrt{1 - (1/2)^2}} < frac{1}{sqrt{1 - c^2}} < frac{1}{sqrt{1 - (3/5)^2}}$$ Let us compute the boundary values: - At the lower bound $x = frac{1}{2}$: $$sqrt{1 - left(frac{1}{2}right)^2} = sqrt{1 - frac{1}{4}} = sqrt{frac{3}{4}} = frac{sqrt{3}}{2}$$ Therefore: $$gleft(frac{1}{2}right) = frac{1}{frac{sqrt{3}}{2}} = frac{2}{sqrt{3}} = frac{2sqrt{3}}{3}$$ - At the upper bound $x = frac{3}{5}$: $$sqrt{1 - left(frac{3}{5}right)^2} = sqrt{1 - frac{9}{25}} = sqrt{frac{16}{25}} = frac{4}{5}$$ Therefore: $$gleft(frac{3}{5}right) = frac{1}{frac{4}{5}} = frac{5}{4}$$ Substituting these evaluated extremes into the inequality: $$frac{2sqrt{3}}{3} < frac{1}{sqrt{1 - c^2}} < frac{5}{4}$$ --- 5. Multiplying by the Interval Width Multiply the entire inequality by the pos

Q: 2. (c) (i) Find the equation of the cylinder whose generators are parallel to the line x/1 = y/2 = z/3 and that passes through the curve x^2 + y^2 = 16, z = 0. (10 Marks)

1. Conceptual Framework & Definition of a Cylinder In three-dimensional analytical solid geometry, a cylinder is defined as a ruled surface generated by a variable straight line (called a generator) that moves such that it remains constantly parallel to a fixed straight line and continuously intersects a fixed plane or space curve (called the guiding curve). We are given: 1. Direction of Generators: Parallel to the straight line: $$frac{x}{1} = frac{y}{2} = frac{z}{3}$$ The direction ratios of the generators are therefore proportional to: $$(l, m, n) = (1, 2, 3)$$ 2. Guiding Curve: The circle in the $xy$-plane ($z = 0$) given by: $$begin{cases} x^2 + y^2 = 16 z = 0 end{cases}$$ --- 2. Equations of Variable Generators Let $P(x1, y1, z1)$ be any arbitrary point lying on the surface of the cylinder. Since the generator passing through $P(x1, y1, z1)$ is parallel to the given line $frac{x}{1} = frac{y}{2} = frac{z}{3}$, its direction ratios are $(1, 2, 3)$. The equations of this generator line in symmetric form are: $$frac{x - x1}{1} = frac{y - y1}{2} = frac{z - z1}{3} = t$$ where $t$ is a real scalar parameter along the generator. In parametric form, any point on this generator has coordinates: $$x = x1 + t, quad y = y1 + 2t, quad z = z1 + 3t$$ --- 3. Point of Intersection with the Guiding Curve Plane ($z = 0$) Since the generator must intersect the guiding curve, it must specifically pass through the plane $z = 0$. Setting the $z$-coordinate to zero: $$z = z1 + 3t = 0 implies 3t = -z1 implies t = -frac{z1}{3}$$ Substituting this parameter value $t = -frac{z1}{3}$ into the expressions for $x$ and $y$, we determine the coordinates of the point of intersection $Q$ of the generator with the plane $z = 0$: $$begin{aligned} xQ &= x1 + 1left(-frac{z1}{3}right) = x1 - frac{z1}{3} = frac{3x1 - z1}{3} yQ &= y1 + 2left(-frac{z1}{3}right) = y1 - frac{2z1}{3} = frac{3y1 - 2z1}{3} zQ &= 0 end{aligned}$$ --- 4. Enforcing the Guiding Curve Condition Since the point $Q(xQ, yQ, 0)$ lie

Q: 2. (c) (ii) Find the shortest distance between the straight lines (x - 3)/3 = (y - 8)/(-1) = (z - 3)/1 and (x + 3)/(-3) = (y + 7)/2 = (z - 6)/4. (10 Marks)

1. Conceptual Framework & Analytical Formula for Shortest Distance Two straight lines in three-dimensional space that are neither parallel nor intersecting are called skew lines. They lie in distinct parallel planes. The shortest distance ($d$) between two skew lines is measured along the unique common perpendicular segment that intersects both lines orthogonally. Let the two lines be given in symmetric canonical form: - Line 1 ($L1$): Passes through point $A(x1, y1, z1)$ with direction vector $vec{b}1 = (l1, m1, n1)$: $$frac{x - x1}{l1} = frac{y - y1}{m1} = frac{z - z1}{n1}$$ - Line 2 ($L2$): Passes through point $B(x2, y2, z2)$ with direction vector $vec{b}2 = (l2, m2, n2)$: $$frac{x - x2}{l2} = frac{y - y2}{m2} = frac{z - z2}{n2}$$ The shortest distance $d$ is given analytically by the projection of the displacement vector $vec{AB} = (x2 - x1, y2 - y1, z2 - z1)$ onto the unit common normal vector $hat{n} = frac{vec{b}1 times vec{b}2}{|vec{b}1 times vec{b}2|}$: $$d = frac{left| begin{vmatrix} x2 - x1 & y2 - y1 & z2 - z1 l1 & m1 & n1 l2 & m2 & n2 end{vmatrix} right|}{sqrt{(m1 n2 - m2 n1)^2 + (n1 l2 - n2 l1)^2 + (l1 m2 - l2 m1)^2}}$$ --- 2. Extracting Line Parameters From the given equations: - Line 1 ($L1$): $$frac{x - 3}{3} = frac{y - 8}{-1} = frac{z - 3}{1}$$ Point on $L1$: $A(x1, y1, z1) = (3, 8, 3)$ Direction vector: $vec{b}1 = (l1, m1, n1) = (3, -1, 1)$ - Line 2 ($L2$): $$frac{x + 3}{-3} = frac{y + 7}{2} = frac{z - 6}{4} iff frac{x - (-3)}{-3} = frac{y - (-7)}{2} = frac{z - 6}{4}$$ Point on $L2$: $B(x2, y2, z2) = (-3, -7, 6)$ Direction vector: $vec{b}2 = (l2, m2, n2) = (-3, 2, 4)$ The displacement vector connecting $A$ and $B$: $$vec{AB} = (x2 - x1, y2 - y1, z2 - z1) = (-3 - 3, ; -7 - 8, ; 6 - 3) = (-6, -15, 3)$$ --- 3. Computation of the Cross Product $vec{b}1 times vec{b}2$ The vector normal to both lines is: $$begin{aligned} vec{n} = vec{b}1 times vec{b}2 &= begin{vmatrix} hat{i} & hat{j} & hat{k} 3 & -1 & 1 -3 & 2 & 4 end{vmatrix} &= hat{i}[(-1)(4) - (1)(2

Q: 3. (c) (ii) If u(x, y) = x f(y/x) + g(y/x), where f and g are arbitrary functions, then show that: I. x (du/dx) + y (du/dy) = x f(y/x), II. x^2 (d^2u/dx^2) + 2xy (d^2u/dxdy) + y^2 (d^2u/dy^2) = 0. (10 Marks)

1. Conceptual Framework: Euler's Theorem on Homogeneous Functions A function $v(x, y)$ is defined to be homogeneous of degree $n$ in $x$ and $y$ if for any scalar $t 0$: $$v(tx, ty) = t^n v(x, y)$$ Euler's Theorem on Homogeneous Functions states: If $v(x, y)$ is a continuously differentiable homogeneous function of degree $n$, then: 1. First-order Euler identity: $$x frac{partial v}{partial x} + y frac{partial v}{partial y} = n v$$ 2. Second-order Euler identity: $$x^2 frac{partial^2 v}{partial x^2} + 2xy frac{partial^2 v}{partial x partial y} + y^2 frac{partial^2 v}{partial y^2} = n(n - 1)v$$ Notice that the differential operator: $$mathcal{D} = x frac{partial}{partial x} + y frac{partial}{partial y}$$ acts on homogeneous functions of degree $n$ as an eigenvalue operator with eigenvalue $n$: $$mathcal{D}(v) = n v$$ Furthermore, the second-order operator satisfies the operational identity: $$x^2 frac{partial^2}{partial x^2} + 2xy frac{partial^2}{partial x partial y} + y^2 frac{partial^2}{partial y^2} = mathcal{D}(mathcal{D} - 1)$$ --- 2. Decomposition into Homogeneous Components The given function is: $$u(x, y) = x fleft(frac{y}{x}right) + gleft(frac{y}{x}right)$$ Let us decompose $u(x, y)$ into two distinct constituent parts: $$u(x, y) = v1(x, y) + v2(x, y)$$ where: - $v1(x, y) = x fleft(frac{y}{x}right)$ - $v2(x, y) = gleft(frac{y}{x}right)$ Let us check the degree of homogeneity for each part: 1. For $v1(x, y)$: $$v1(tx, ty) = (tx) fleft(frac{ty}{tx}right) = t left[ x fleft(frac{y}{x}right) right] = t^1 v1(x, y)$$ Thus, $v1(x, y)$ is a homogeneous function of degree $n1 = 1$. 2. For $v2(x, y)$: $$v2(tx, ty) = gleft(frac{ty}{tx}right) = gleft(frac{y}{x}right) = t^0 left[ gleft(frac{y}{x}right) right] = t^0 v2(x, y)$$ Thus, $v2(x, y)$ is a homogeneous function of degree $n2 = 0$. --- 3. Proof of Part I: $x frac{partial u}{partial x} + y frac{partial u}{partial y} = x fleft(frac{y}{x}right)$ Method A: Direct Application of Euler's Theorem Applying the first-order Eu

Question 1

1. (d) Given that f(x + y) = f(x) f(y) for all real x, y, f(x) != 0 for any real x and f'(0) = 2. Show that for all real x, f'(x) = 2f(x). Hence find f(x). (10 Marks)

Accepted Answer

1. Conceptual Foundation & Functional Equation Theory The given equation: $$f(x + y) = f(x) f(y) quad forall x, y in mathbb{R}$$ is the celebrated Cauchy's Exponential Functional Equation. It governs phenomena where translation in the domain transforms into scaling (multiplication) in the codomain. We are supplied with three critical conditions: 1. Functional Relation: $f(x + y) = f(x) f(y)$ for all $x, y in mathbb{R}$. 2. Non-vanishing Property: $f(x) eq 0$ for every $x in mathbb{R}$. 3. Differentiability at the Origin: $f'(0) = 2$. Our objective is twofold: - Part I: Establish rigorously from first principles that $f$ is differentiable everywhere on $mathbb{R}$ and satisfies the first-order differential equation $f'(x) = 2f(x)$. - Part II: Integrate this differential equation using appropriate initial values to uniquely determine $f(x)$. --- 2. Determination of the Initial Value $f(0)$ To work with derivatives at the origin, we first find the value of the function at $x = 0$. Substitute $x = 0$ and $y = 0$ into the functional relation: $$f(0 + 0) = f(0) f(0) implies f(0) = [f(0)]^2$$ This algebraic equation can be factored as: $$[f(0)]^2 - f(0) = 0 implies f(0)[f(0) - 1] = 0$$ This yields two algebraic possibilities: $f(0) = 0$ or $f(0) = 1$. However, the problem explicitly hypotheses that $f(x) eq 0$ for any real $x$. Specifically: $$f(0) eq 0$$ Hence, we must reject the root $f(0) = 0$. Therefore, we have uniquely: $$f(0) = 1$$ (Note on positivity: For any $x in mathbb{R}$, $f(x) = fleft(frac{x}{2} + frac{x}{2}right) = left[fleft(frac{x}{2}right)right]^2$. Since $f(x) eq 0$, the square of a real non-zero number is strictly positive, establishing that $f(x) 0$ for all $x in mathbb{R}$.) --- 3. Establishing $f'(x) = 2f(x)$ from First Principles By the definition of the derivative from first principles (the Newton quotient): $$f'(x) = lim{h to 0} frac{f(x + h) - f(x)}{h}$$ Using the functional relation $f(x + h) = f(x) f(h)$: $$f'(x) = lim{h to 0} frac{f(x)f(h) - f

Question 2

1. (c) A rectangular sheet of metal of length 6 meters and width 2 meters is given. Four equal squares are removed from the four corners. The sides of this sheet are now folded up to form an open rectangular box. Find approximately the height of the box, such that the volume of the box is maximum. (

Accepted Answer

1. Mathematical Modeling & Geometric Setup We are given a flat rectangular metal sheet with the following initial dimensions: - Initial Length: $L0 = 6text{ meters}$ - Initial Width: $W0 = 2text{ meters}$ Four congruent squares, each having side length $x$ (in meters), are cut out from each of the four corners of the sheet. The four flaps along the edges are then folded upwards along the cut lines at right angles to form an open-top rectangular box. Let us determine the physical dimensions of the resulting open box in terms of $x$: 1. Height of the box ($h$): When the flaps are folded vertically upward, the height of the box is precisely the side length of the removed square: $$h(x) = x$$ 2. Width of the base ($w$): Since a square of length $x$ is removed from both ends of the width: $$w(x) = 2 - 2x = 2(1 - x)$$ 3. Length of the base ($l$): Similarly, a square of length $x$ is removed from both ends of the length: $$l(x) = 6 - 2x = 2(3 - x)$$ Physical Domain & Feasibility Constraints: For the physical box to have non-degenerate, strictly positive real dimensions: - $h 0 implies x 0$ - $w 0 implies 2 - 2x 0 implies x 0 implies 6 - 2x 0 implies x 1$, this root lies outside our physically admissible interval $(0, 1)$. (At $x = 2.215$, the width $w = 2 - 2(2.215) = -2.43text{ m} 0$, we have: $$left. frac{d^2V}{dx^2} right|{x2} = -8sqrt{7} approx -21.166 < 0$$ Because the second derivative is strictly negative, $V(x)$ attains a strict local maximum at $x = frac{4 - sqrt{7}}{3}$. Since it is the unique critical point on the open interval $(0, 1)$ and $V(x) = 0$ on the boundaries, this point represents the unique global maximum. --- 5. Maximum Volume Calculation For completeness, substituting $x = 0.45142text{ m}$ into the dimensions: - Height: $h approx 0.4514text{ m}$ - Width: $w = 2 - 2(0.45142) = 1.0972text{ m}$ - Length: $l = 6 - 2(0.45142) = 5.0972text{ m}$ - Maximum Volume: $$V{max} = (0.45142) times (1.09716) times (5.09716) approx 2.5245text{ m}^3$$ --- 6. Final A

Question 3

1. (b) Find the range, rank, kernel and nullity of the linear transformation T : R^4 - R^3 given by T(x, y, z, w) = (x - w, y + z, z - w). (10 Marks)

Accepted Answer

1. Conceptual Foundation & Definitions Let $T: V to W$ be a linear transformation from a vector space $V$ into a vector space $W$ over the field $mathbb{R}$. In this problem: $$V = mathbb{R}^4 quad text{and} quad W = mathbb{R}^3$$ The mapping is explicitly defined by: $$T(x, y, z, w) = (x - w, y + z, z - w)$$ We recall the key definitions: 1. Kernel (Null Space): The set of all vectors in the domain $V$ that map to the zero vector of the codomain $W$: $$ker(T) = left{ (x, y, z, w) in mathbb{R}^4 mid T(x, y, z, w) = (0, 0, 0) right}$$ 2. Nullity: The dimension of the kernel as a subspace of $V$: $$operatorname{nullity}(T) = dim(ker(T))$$ 3. Range (Image Space): The set of all vectors in the codomain $W$ that are images under $T$: $$operatorname{range}(T) = operatorname{Im}(T) = left{ T(x, y, z, w) mid (x, y, z, w) in mathbb{R}^4 right}$$ 4. Rank: The dimension of the range as a subspace of $W$: $$operatorname{rank}(T) = dim(operatorname{range}(T))$$ 5. The Dimension Theorem (Rank-Nullity Theorem): For any linear transformation from a finite-dimensional vector space $V$: $$operatorname{rank}(T) + operatorname{nullity}(T) = dim(V) = 4$$ --- 2. Determination of the Kernel and Nullity A vector $(x, y, z, w) in mathbb{R}^4$ belongs to $ker(T)$ if and only if: $$T(x, y, z, w) = (x - w, y + z, z - w) = (0, 0, 0)$$ Equating each component to zero yields a homogeneous system of three linear equations in four unknowns: $$begin{aligned} (1) & quad x - w = 0 implies x = w (2) & quad y + z = 0 implies y = -z (3) & quad z - w = 0 implies z = w end{aligned}$$ From equation (3), we have $z = w$. Substituting $z = w$ into equation (2): $$y = -w$$ And from equation (1): $$x = w$$ Here, $w$ is a free parameter ($w in mathbb{R}$). Any vector $(x, y, z, w) in ker(T)$ can therefore be parameterized in terms of $w$: $$(x, y, z, w) = (w, -w, w, w) = w(1, -1, 1, 1)$$ Thus, the kernel of $T$ is the one-dimensional subspace spanned by the single non-zero vector $(1, -1, 1, 1)$: $$ker(T) = oper

Question 4

1. (a) Can the set {(0, 0, 0, 3), (1, 1, 0, 0), (0, 1, -1, 0)} be extended to form a basis of the vector space R^4 ? Justify your answer. (10 Marks)

Accepted Answer

1. Conceptual Foundation & Mathematical Framework In linear algebra, a basis of a finite-dimensional vector space $V$ over a field $mathbb{F}$ is defined as a subset $mathcal{B} subset V$ satisfying two fundamental axioms: 1. Linear Independence: No vector in $mathcal{B}$ can be expressed as a linear combination of the remaining vectors in $mathcal{B}$. 2. Spanning Property: The linear span of $mathcal{B}$ equals the entire space, i.e., $operatorname{span}(mathcal{B}) = V$. For the Euclidean space $mathbb{R}^4$, the dimension is known to be: $$dim(mathbb{R}^4) = 4$$ Consequently, by the fundamental dimension theorem, any basis of $mathbb{R}^4$ must consist of precisely 4 linearly independent vectors. The foundational Basis Extension Theorem (a direct corollary of the Steinitz Exchange Lemma) asserts that: If $V$ is a finite-dimensional vector space with $dim(V) = n$, and $S = {v1, v2, dots, vk}$ is an arbitrary linearly independent subset of $V$ where $k < n$, then $S$ can always be extended to a complete basis of $V$ by adjoining $(n - k)$ appropriately chosen vectors from any spanning set of $V$ (such as the standard canonical basis ${e1, e2, dots, en}$). Therefore, to answer the question affirmatively, we must rigorously prove that the given set of three vectors: $$S = {v1, v2, v3} = {(0, 0, 0, 3), (1, 1, 0, 0), (0, 1, -1, 0)}$$ is linearly independent in $mathbb{R}^4$. If it is linearly independent, the Basis Extension Theorem guarantees that it can be extended to form a basis of $mathbb{R}^4$. --- 2. Rigorous Test of Linear Independence Let $c1, c2, c3 in mathbb{R}$ be scalars such that the linear combination vanishes: $$c1 v1 + c2 v2 + c3 v3 = mathbf{0} = (0, 0, 0, 0)$$ Substituting the coordinates of $v1, v2, v3$: $$c1 (0, 0, 0, 3) + c2 (1, 1, 0, 0) + c3 (0, 1, -1, 0) = (0, 0, 0, 0)$$ Expanding into coordinate components: $$(c2, c2 + c3, -c3, 3c1) = (0, 0, 0, 0)$$ Equating respective components gives the homogeneous linear system: $$begin{aligned} text{Compon

Question 5

1. (e) Find the equation of the cone whose vertex is the point (1, 1, 0) and whose guiding curve is y = 0, x^2 + z^2 = 4. (10 Marks)

Accepted Answer

1. Conceptual Foundation & Definition of a Cone In three-dimensional analytical geometry, a cone is a ruled surface generated by a variable straight line (called a generator) that passes through a fixed point $V(alpha, beta, gamma)$ (called the vertex) and continuously intersects a given fixed space curve (called the guiding curve or directrix). Here, we are provided with: - Vertex ($V$): $(alpha, beta, gamma) = (1, 1, 0)$ - Guiding Curve: Given by the intersection of the plane and the circular cylinder: $$begin{cases} y = 0 x^2 + z^2 = 4 end{cases}$$ This represents a circle of radius $R = 2$ centered at the origin lying in the $xz$-plane ($y = 0$). --- 2. Equations of Variable Generators Passing through the Vertex Any straight line passing through the fixed vertex $V(1, 1, 0)$ can be written in symmetric form with direction ratios $l, m, n$: $$frac{x - 1}{l} = frac{y - 1}{m} = frac{z - 0}{n} = r$$ where $r$ is a scalar parameter along the generator. Expressed in parametric form, the coordinates of an arbitrary point $P(x, y, z)$ on this generator are: $$x = 1 + lr, quad y = 1 + mr, quad z = nr$$ --- 3. Intersection of the Generator with the Guiding Curve If this generator intersects the guiding curve, there must exist a point on the generator lying in the plane $y = 0$. Setting the $y$-coordinate to zero: $$y = 1 + mr = 0 implies mr = -1 implies r = -frac{1}{m} quad (m eq 0)$$ Substituting this specific value of the parameter $r = -frac{1}{m}$ into the coordinates for $x$ and $z$, we obtain the point of intersection $Q$ of the generator with the plane $y = 0$: $$begin{aligned} xQ &= 1 + lleft(-frac{1}{m}right) = 1 - frac{l}{m} = frac{m - l}{m} yQ &= 0 zQ &= nleft(-frac{1}{m}right) = -frac{n}{m} end{aligned}$$ Since point $Q(xQ, yQ, zQ)$ lies on the guiding curve, its coordinates must satisfy the second equation defining the curve: $$x^2 + z^2 = 4$$ Substituting $xQ$ and $zQ$: $$left(frac{m - l}{m}right)^2 + left(-frac{n}{m}right)^2 = 4$$ Multiplying through by $m^

Question 6

2. (a) Let T : R^3 - R^2 be a linear transformation such that T(1, 1, -1) = (1, 0), T(4, 1, 1) = (0, 1) and T(1, -1, 2) = (1, 1). Find T. (15 Marks)

Accepted Answer

1. Conceptual Framework & Existence-Uniqueness Theorem In linear algebra, a foundational theorem states that a linear transformation $T: V to W$ between finite-dimensional vector spaces is uniquely determined by its action on a basis of the domain $V$. Specifically: Let $mathcal{B} = {v1, v2, dots, vn}$ be a basis of a vector space $V$ over a field $mathbb{F}$, and let $w1, w2, dots, wn$ be arbitrary vectors in a vector space $W$. Then there exists a unique linear transformation $T: V to W$ such that $T(vi) = wi$ for each $i = 1, 2, dots, n$. For this problem: - Domain: $V = mathbb{R}^3$, where $dim(V) = 3$. - Codomain: $W = mathbb{R}^2$. - Given vectors in $mathbb{R}^3$: $$v1 = (1, 1, -1), quad v2 = (4, 1, 1), quad v3 = (1, -1, 2)$$ - Prescribed images in $mathbb{R}^2$: $$w1 = (1, 0), quad w2 = (0, 1), quad w3 = (1, 1)$$ Our goal is to find the explicit closed-form formula for $T(x, y, z)$ for any arbitrary vector $(x, y, z) in mathbb{R}^3$. --- 2. Verification of Basis Property of ${v1, v2, v3}$ Before computing $T$, we must verify that the set ${v1, v2, v3}$ is linearly independent, and hence forms a valid basis for $mathbb{R}^3$. Construct the $3 times 3$ matrix $P$ having $v1, v2, v3$ as its columns: $$P = begin{bmatrix} 1 & 4 & 1 1 & 1 & -1 -1 & 1 & 2 end{bmatrix}$$ Let us compute the determinant of $P$: $$begin{aligned} det(P) &= 1 cdot begin{vmatrix} 1 & -1 1 & 2 end{vmatrix} - 4 cdot begin{vmatrix} 1 & -1 -1 & 2 end{vmatrix} + 1 cdot begin{vmatrix} 1 & 1 -1 & 1 end{vmatrix} &= 1 cdot (2 - (-1)) - 4 cdot (2 - 1) + 1 cdot (1 - (-1)) &= 1 cdot 3 - 4 cdot 1 + 1 cdot 2 &= 3 - 4 + 2 = 1 eq 0 end{aligned}$$ Because $det(P) = 1 eq 0$, the vectors $v1, v2, v3$ are strictly linearly independent and form a basis of $mathbb{R}^3$. Consequently, the transformation $T$ exists and is uniquely defined. --- 3. Representation of an Arbitrary Vector $(x, y, z)$ in the Basis Let $(x, y, z)$ be an arbitrary vector in $mathbb{R}^3$. We express $(x, y, z)$ as a unique linear comb

Question 7

2. (b) Using Mean Value Theorem, prove that pi/6 + sqrt(3)/15 < sin^(-1)(3/5) < pi/6 + 1/8. (15 Marks)

Accepted Answer

1. Conceptual Foundation & Lagrange's Mean Value Theorem Lagrange's Mean Value Theorem (LMVT) is a fundamental cornerstone of real analysis. It states: Let $f: [a, b] to mathbb{R}$ be a real-valued function satisfying: 1. $f$ is continuous on the closed bounded interval $[a, b]$; 2. $f$ is differentiable on the open interval $(a, b)$. Then there exists at least one point $c in (a, b)$ such that: $$frac{f(b) - f(a)}{b - a} = f'(c) iff f(b) - f(a) = (b - a)f'(c)$$ LMVT provides a rigorous analytical bridge connecting local properties (derivatives at an intermediate point $c$) to global properties (increments of the function over $[a, b]$). When $f'(x)$ is strictly monotonic on $[a, b]$, the bounds on $c in (a, b)$ yield strict inequalities bounding the value of $f(b)$. --- 2. Selection of Function and Interval We examine the expression to be bounded: $$sin^{-1}left(frac{3}{5}right)$$ Notice that: - $frac{3}{5} = 0.60$ - We know the exact value of the inverse sine at a nearby standard angle: $$sinleft(frac{pi}{6}right) = frac{1}{2} = 0.50 implies sin^{-1}left(frac{1}{2}right) = frac{pi}{6}$$ Since $0.50 0$. Therefore, for $frac{1}{2} < c < frac{3}{5}$: $$gleft(frac{1}{2}right) < g(c) < gleft(frac{3}{5}right)$$ $$frac{1}{sqrt{1 - (1/2)^2}} < frac{1}{sqrt{1 - c^2}} < frac{1}{sqrt{1 - (3/5)^2}}$$ Let us compute the boundary values: - At the lower bound $x = frac{1}{2}$: $$sqrt{1 - left(frac{1}{2}right)^2} = sqrt{1 - frac{1}{4}} = sqrt{frac{3}{4}} = frac{sqrt{3}}{2}$$ Therefore: $$gleft(frac{1}{2}right) = frac{1}{frac{sqrt{3}}{2}} = frac{2}{sqrt{3}} = frac{2sqrt{3}}{3}$$ - At the upper bound $x = frac{3}{5}$: $$sqrt{1 - left(frac{3}{5}right)^2} = sqrt{1 - frac{9}{25}} = sqrt{frac{16}{25}} = frac{4}{5}$$ Therefore: $$gleft(frac{3}{5}right) = frac{1}{frac{4}{5}} = frac{5}{4}$$ Substituting these evaluated extremes into the inequality: $$frac{2sqrt{3}}{3} < frac{1}{sqrt{1 - c^2}} < frac{5}{4}$$ --- 5. Multiplying by the Interval Width Multiply the entire inequality by the pos

Question 8

2. (c) (i) Find the equation of the cylinder whose generators are parallel to the line x/1 = y/2 = z/3 and that passes through the curve x^2 + y^2 = 16, z = 0. (10 Marks)

Accepted Answer

1. Conceptual Framework & Definition of a Cylinder In three-dimensional analytical solid geometry, a cylinder is defined as a ruled surface generated by a variable straight line (called a generator) that moves such that it remains constantly parallel to a fixed straight line and continuously intersects a fixed plane or space curve (called the guiding curve). We are given: 1. Direction of Generators: Parallel to the straight line: $$frac{x}{1} = frac{y}{2} = frac{z}{3}$$ The direction ratios of the generators are therefore proportional to: $$(l, m, n) = (1, 2, 3)$$ 2. Guiding Curve: The circle in the $xy$-plane ($z = 0$) given by: $$begin{cases} x^2 + y^2 = 16 z = 0 end{cases}$$ --- 2. Equations of Variable Generators Let $P(x1, y1, z1)$ be any arbitrary point lying on the surface of the cylinder. Since the generator passing through $P(x1, y1, z1)$ is parallel to the given line $frac{x}{1} = frac{y}{2} = frac{z}{3}$, its direction ratios are $(1, 2, 3)$. The equations of this generator line in symmetric form are: $$frac{x - x1}{1} = frac{y - y1}{2} = frac{z - z1}{3} = t$$ where $t$ is a real scalar parameter along the generator. In parametric form, any point on this generator has coordinates: $$x = x1 + t, quad y = y1 + 2t, quad z = z1 + 3t$$ --- 3. Point of Intersection with the Guiding Curve Plane ($z = 0$) Since the generator must intersect the guiding curve, it must specifically pass through the plane $z = 0$. Setting the $z$-coordinate to zero: $$z = z1 + 3t = 0 implies 3t = -z1 implies t = -frac{z1}{3}$$ Substituting this parameter value $t = -frac{z1}{3}$ into the expressions for $x$ and $y$, we determine the coordinates of the point of intersection $Q$ of the generator with the plane $z = 0$: $$begin{aligned} xQ &= x1 + 1left(-frac{z1}{3}right) = x1 - frac{z1}{3} = frac{3x1 - z1}{3} yQ &= y1 + 2left(-frac{z1}{3}right) = y1 - frac{2z1}{3} = frac{3y1 - 2z1}{3} zQ &= 0 end{aligned}$$ --- 4. Enforcing the Guiding Curve Condition Since the point $Q(xQ, yQ, 0)$ lie

Question 9

2. (c) (ii) Find the shortest distance between the straight lines (x - 3)/3 = (y - 8)/(-1) = (z - 3)/1 and (x + 3)/(-3) = (y + 7)/2 = (z - 6)/4. (10 Marks)

Accepted Answer

1. Conceptual Framework & Analytical Formula for Shortest Distance Two straight lines in three-dimensional space that are neither parallel nor intersecting are called skew lines. They lie in distinct parallel planes. The shortest distance ($d$) between two skew lines is measured along the unique common perpendicular segment that intersects both lines orthogonally. Let the two lines be given in symmetric canonical form: - Line 1 ($L1$): Passes through point $A(x1, y1, z1)$ with direction vector $vec{b}1 = (l1, m1, n1)$: $$frac{x - x1}{l1} = frac{y - y1}{m1} = frac{z - z1}{n1}$$ - Line 2 ($L2$): Passes through point $B(x2, y2, z2)$ with direction vector $vec{b}2 = (l2, m2, n2)$: $$frac{x - x2}{l2} = frac{y - y2}{m2} = frac{z - z2}{n2}$$ The shortest distance $d$ is given analytically by the projection of the displacement vector $vec{AB} = (x2 - x1, y2 - y1, z2 - z1)$ onto the unit common normal vector $hat{n} = frac{vec{b}1 times vec{b}2}{|vec{b}1 times vec{b}2|}$: $$d = frac{left| begin{vmatrix} x2 - x1 & y2 - y1 & z2 - z1 l1 & m1 & n1 l2 & m2 & n2 end{vmatrix} right|}{sqrt{(m1 n2 - m2 n1)^2 + (n1 l2 - n2 l1)^2 + (l1 m2 - l2 m1)^2}}$$ --- 2. Extracting Line Parameters From the given equations: - Line 1 ($L1$): $$frac{x - 3}{3} = frac{y - 8}{-1} = frac{z - 3}{1}$$ Point on $L1$: $A(x1, y1, z1) = (3, 8, 3)$ Direction vector: $vec{b}1 = (l1, m1, n1) = (3, -1, 1)$ - Line 2 ($L2$): $$frac{x + 3}{-3} = frac{y + 7}{2} = frac{z - 6}{4} iff frac{x - (-3)}{-3} = frac{y - (-7)}{2} = frac{z - 6}{4}$$ Point on $L2$: $B(x2, y2, z2) = (-3, -7, 6)$ Direction vector: $vec{b}2 = (l2, m2, n2) = (-3, 2, 4)$ The displacement vector connecting $A$ and $B$: $$vec{AB} = (x2 - x1, y2 - y1, z2 - z1) = (-3 - 3, ; -7 - 8, ; 6 - 3) = (-6, -15, 3)$$ --- 3. Computation of the Cross Product $vec{b}1 times vec{b}2$ The vector normal to both lines is: $$begin{aligned} vec{n} = vec{b}1 times vec{b}2 &= begin{vmatrix} hat{i} & hat{j} & hat{k} 3 & -1 & 1 -3 & 2 & 4 end{vmatrix} &= hat{i}[(-1)(4) - (1)(2

Question 10

3. (c) (ii) If u(x, y) = x f(y/x) + g(y/x), where f and g are arbitrary functions, then show that: I. x (du/dx) + y (du/dy) = x f(y/x), II. x^2 (d^2u/dx^2) + 2xy (d^2u/dxdy) + y^2 (d^2u/dy^2) = 0. (10 Marks)

Accepted Answer

1. Conceptual Framework: Euler's Theorem on Homogeneous Functions A function $v(x, y)$ is defined to be homogeneous of degree $n$ in $x$ and $y$ if for any scalar $t 0$: $$v(tx, ty) = t^n v(x, y)$$ Euler's Theorem on Homogeneous Functions states: If $v(x, y)$ is a continuously differentiable homogeneous function of degree $n$, then: 1. First-order Euler identity: $$x frac{partial v}{partial x} + y frac{partial v}{partial y} = n v$$ 2. Second-order Euler identity: $$x^2 frac{partial^2 v}{partial x^2} + 2xy frac{partial^2 v}{partial x partial y} + y^2 frac{partial^2 v}{partial y^2} = n(n - 1)v$$ Notice that the differential operator: $$mathcal{D} = x frac{partial}{partial x} + y frac{partial}{partial y}$$ acts on homogeneous functions of degree $n$ as an eigenvalue operator with eigenvalue $n$: $$mathcal{D}(v) = n v$$ Furthermore, the second-order operator satisfies the operational identity: $$x^2 frac{partial^2}{partial x^2} + 2xy frac{partial^2}{partial x partial y} + y^2 frac{partial^2}{partial y^2} = mathcal{D}(mathcal{D} - 1)$$ --- 2. Decomposition into Homogeneous Components The given function is: $$u(x, y) = x fleft(frac{y}{x}right) + gleft(frac{y}{x}right)$$ Let us decompose $u(x, y)$ into two distinct constituent parts: $$u(x, y) = v1(x, y) + v2(x, y)$$ where: - $v1(x, y) = x fleft(frac{y}{x}right)$ - $v2(x, y) = gleft(frac{y}{x}right)$ Let us check the degree of homogeneity for each part: 1. For $v1(x, y)$: $$v1(tx, ty) = (tx) fleft(frac{ty}{tx}right) = t left[ x fleft(frac{y}{x}right) right] = t^1 v1(x, y)$$ Thus, $v1(x, y)$ is a homogeneous function of degree $n1 = 1$. 2. For $v2(x, y)$: $$v2(tx, ty) = gleft(frac{ty}{tx}right) = gleft(frac{y}{x}right) = t^0 left[ gleft(frac{y}{x}right) right] = t^0 v2(x, y)$$ Thus, $v2(x, y)$ is a homogeneous function of degree $n2 = 0$. --- 3. Proof of Part I: $x frac{partial u}{partial x} + y frac{partial u}{partial y} = x fleft(frac{y}{x}right)$ Method A: Direct Application of Euler's Theorem Applying the first-order Eu

Question 11

3. (c) (i) Evaluate iintR y dx dy, where R is the region bounded by y = x and y = 4x - x^2. (10 Marks)

Accepted Answer

1. Conceptual Framework & Double Integration Setup In multivariable integral calculus, a double integral of a scalar function $f(x, y)$ over a planar region $R subset mathbb{R}^2$: $$I = iintR f(x, y) , dA = iintR y , dx , dy$$ represents the integral of the integrand $f(x, y) = y$ over the area domain $R$. To compute the integral using Fubini's Theorem, we must: 1. Sketch and analyze the geometry of the bounding curves. 2. Determine the exact points of intersection of the boundaries. 3. Express $R$ as a vertically simple (Type I) or horizontally simple (Type II) domain. 4. Set up the iterated integral with proper limits and integrate sequentially. --- 2. Geometry of the Bounding Curves & Points of Intersection The region $R$ is bounded by: 1. The straight line passing through the origin at a $45^circ$ angle: $$y = x$$ 2. The downward-opening parabola: $$y = 4x - x^2 = -(x^2 - 4x) = -(x - 2)^2 + 4$$ The vertex of this parabola is at $(2, 4)$, and its $x$-intercepts are at $x = 0$ and $x = 4$. Finding the Points of Intersection: Equating the two expressions for $y$: $$x = 4x - x^2$$ $$x^2 - 3x = 0 implies x(x - 3) = 0$$ This yields two solutions for $x$: - $x = 0 implies y = 0$. Point of intersection: $O(0, 0)$. - $x = 3 implies y = 3$. Point of intersection: $P(3, 3)$. Relative Position of the Curves on $[0, 3]$: For any test point in the interior of the interval, say $x = 1$: - Parabola: $y{text{upper}} = 4(1) - (1)^2 = 3$ - Line: $y{text{lower}} = 1$ Since $3 1$, the parabola lies strictly above the straight line over the open interval $(0, 3)$: $$y{text{lower}} = x le y le 4x - x^2 = y{text{upper}} quad forall x in [0, 3]$$ --- 3. Formulation of the Iterated Integral (Type I Domain) Describing the region $R$ as a vertically simple (Type I) region: $$R = left{ (x, y) in mathbb{R}^2 mid 0 le x le 3, quad x le y le 4x - x^2 right}$$ Applying Fubini's theorem: $$I = int{x = 0}^{3} left[ int{y = x}^{4x - x^2} y , dy right] dx$$ --- 4. Inner Integration with respect to

Question 12

3. (b) Find the equations of the spheres which pass through the circle x^2 + y^2 + z^2 - 2x + 2y + 4z - 3 = 0, 2x + y + z = 4 and touch the plane 3x + 4y = 14. (15 Marks)

Accepted Answer

1. Conceptual Framework & Family of Spheres through a Circle In three-dimensional analytical solid geometry, a circle in space is uniquely defined by the intersection of a sphere $S = 0$ and a secant plane $P = 0$: $$begin{cases} S(x, y, z) = x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0 P(x, y, z) = Ax + By + Cz + D = 0 end{cases}$$ The general equation of any sphere passing through this circle of intersection belongs to the coaxial pencil (or family) given by: $$S + lambda P = 0$$ where $lambda in mathbb{R}$ is an arbitrary real parameter. We are given: - Sphere: $S equiv x^2 + y^2 + z^2 - 2x + 2y + 4z - 3 = 0$ - Secant Plane: $P equiv 2x + y + z - 4 = 0$ - Target Tangent Plane: $Pi equiv 3x + 4y - 14 = 0$ Condition of Tangency: A sphere with center $C(x0, y0, z0)$ and radius $R$ touches (is tangent to) a plane $Pi equiv ax + by + cz + d = 0$ if and only if the perpendicular distance from the center $C$ to the plane $Pi$ equals the radius $R$: $$operatorname{dist}(C, Pi) = frac{|a x0 + b y0 + c z0 + d|}{sqrt{a^2 + b^2 + c^2}} = R$$ Equivalently: $$[operatorname{dist}(C, Pi)]^2 = R^2$$ --- 2. Formulating the Family of Spheres Substituting $S$ and $P$ into the family equation: $$(x^2 + y^2 + z^2 - 2x + 2y + 4z - 3) + lambda (2x + y + z - 4) = 0$$ Grouping terms by powers of $x, y, z$: $$x^2 + y^2 + z^2 + (2lambda - 2)x + (lambda + 2)y + (lambda + 4)z - (4lambda + 3) = 0$$ $$x^2 + y^2 + z^2 + 2(lambda - 1)x + 2left(frac{lambda + 2}{2}right)y + 2left(frac{lambda + 4}{2}right)z - (4lambda + 3) = 0$$ Comparing this with the general sphere equation $x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0$: $$u = lambda - 1, quad v = frac{lambda + 2}{2}, quad w = frac{lambda + 4}{2}, quad d = -(4lambda + 3)$$ --- 3. Center and Radius of the Variable Sphere 1. Center $C$: $$C(-u, -v, -w) = left( 1 - lambda, ; -frac{lambda + 2}{2}, ; -frac{lambda + 4}{2} right)$$ 2. Radius squared ($R^2$): $$begin{aligned} R^2 &= u^2 + v^2 + w^2 - d &= (lambda - 1)^2 + left(frac{lambda + 2}{2}right)^2 + left

Question 13

3. (a) Reduce the following matrix to echelon form: A = [[2, -2, 2, 1], [-3, 6, 0, -1], [1, -7, 10, 2]]. (15 Marks)

Accepted Answer

1. Conceptual Framework & Definition of Row Echelon Form In linear algebra, a matrix $E$ is said to be in Row Echelon Form (REF) if it satisfies three strict canonical criteria: 1. Zero Rows at Bottom: All zero rows (rows consisting entirely of zeros), if any, are situated at the very bottom of the matrix. 2. Leading Entry (Pivot): The first non-zero entry in each non-zero row, reading from left to right, is called the leading coefficient or pivot. (While some texts require the leading coefficient to be normalized to $1$, standard echelon form permits any non-zero real number, though normalizing to $1$ is standard practice). 3. Staircase Property: The leading entry of any non-zero row is positioned strictly to the right of the leading entry of the row immediately preceding it. Any $m times n$ matrix $A$ can be systematically reduced to row echelon form using a sequence of Elementary Row Operations (EROs): - Type I ($Ri leftrightarrow Rj$): Interchanging two rows. - Type II ($Ri to k Ri, ; k eq 0$): Multiplying a row by a non-zero scalar. - Type III ($Ri to Ri + c Rj$): Adding a scalar multiple of one row to another row. Elementary row operations preserve the row space and rank of the matrix: $operatorname{rank}(A) = operatorname{rank}(E)$, where the rank is the number of non-zero rows in the echelon form. --- 2. Given Matrix Specification The given matrix $A$ is of dimension $3 times 4$ (3 rows and 4 columns): $$A = begin{bmatrix} 2 & -2 & 2 & 1 -3 & 6 & 0 & -1 1 & -7 & 10 & 2 end{bmatrix}$$ --- 3. Step-by-Step Elementary Row Reduction Step 1: Establish a Convenient Pivot in Column 1 To avoid fractions early in computation, we take advantage of the entry $1$ in row 3, column 1. We perform the Type I row interchange: $$R1 longleftrightarrow R3$$ The matrix becomes: $$A1 = begin{bmatrix} 1 & -7 & 10 & 2 -3 & 6 & 0 & -1 2 & -2 & 2 & 1 end{bmatrix}$$ Here, the leading entry of the first row is $a{11} = 1$. --- Step 2: Eliminate Entries in Column 1 below the Pivot We now

Question 14

4. (a) Show that there is no tangent plane to the sphere x^2 + y^2 + z^2 - 4x + 2y - 4z + 4 = 0 that can be passed through the straight line (x + 6)/2 = y + 3 = z + 1. (15 Marks)

Accepted Answer

1. Conceptual Framework & Geometric Criteria In three-dimensional solid geometry, a straight line $L$ in relation to a sphere $S$ of radius $R$ centered at $C$ can occupy one of three mutually exclusive geometric positions: 1. Secant Line: The line intersects the sphere in two distinct real points. In this case, the perpendicular distance $d$ from the sphere's center $C$ to the line $L$ is strictly less than the radius ($d 0$ rather than being tangent at a single point. 2. Tangent Line: The line touches the sphere at a single point ($d = R$). In this case, there exists exactly one tangent plane containing the line. 3. Exterior Line: The line passes entirely outside the sphere ($d R$). In this case, there exist two distinct real tangent planes passing through the line. Therefore, to show that no tangent plane can pass through the given line, we can establish either of two equivalent rigorous proofs: - Approach 1 (Distance Analysis): Prove that the perpendicular distance from the center of the sphere to the given line is strictly less than the radius of the sphere ($d 0$. No plane through $L$ can touch the sphere at a single point. Hence, no real tangent plane can pass through the line. --- 5. Method 2: Pencil of Planes Formulation To provide an alternative algebraic proof, express the line as the intersection of two planes: From $frac{x + 6}{2} = frac{y + 3}{1}$: $$x + 6 = 2y + 6 implies x - 2y = 0 quad (P1)$$ From $frac{y + 3}{1} = frac{z + 1}{1}$: $$y - z + 2 = 0 quad (P2)$$ The general equation of any plane passing through the line is the pencil $P1 + lambda P2 = 0$: $$(x - 2y) + lambda (y - z + 2) = 0$$ $$x + (lambda - 2)y - lambda z + 2lambda = 0$$ For this plane to be tangent to the sphere, the perpendicular distance from center $C(2, -1, 2)$ to the plane must equal the radius $R = sqrt{5}$: $$frac{| 1(2) + (lambda - 2)(-1) - lambda(2) + 2lambda |}{sqrt{1^2 + (lambda - 2)^2 + (-lambda)^2}} = sqrt{5}$$ Simplifying the numerator: $$| 2 - lambda + 2 - 2lambda + 2l

Question 15

4. (c) (i) Find the eigenvalues and the corresponding eigenvectors of the matrix A = [[1, 2, 0], [2, 1, -6], [2, -2, 3]]. (12 Marks)

Accepted Answer

1. Conceptual Framework & Definitions Let $A$ be a real square matrix of order $n times n$. 1. A scalar $lambda in mathbb{R}$ (or $mathbb{C}$) is called an eigenvalue (or characteristic value) of $A$ if there exists a non-zero column vector $X eq mathbf{0}$ such that: $$A X = lambda X iff (A - lambda I) X = mathbf{0}$$ 2. The non-zero vector $X$ is called an eigenvector (or characteristic vector) of $A$ corresponding to the eigenvalue $lambda$. 3. For a non-trivial solution $X eq mathbf{0}$ to exist, the coefficient matrix $(A - lambda I)$ must be singular, giving the characteristic equation: $$det(A - lambda I) = 0$$ 4. The set of all eigenvectors corresponding to an eigenvalue $lambda$, together with the zero vector, forms the null space $ker(A - lambda I)$, known as the eigenspace $Elambda$. We are given the $3 times 3$ matrix: $$A = begin{bmatrix} 1 & 2 & 0 2 & 1 & -6 2 & -2 & 3 end{bmatrix}$$ --- 2. Derivation of the Characteristic Equation The characteristic matrix is: $$A - lambda I = begin{bmatrix} 1 - lambda & 2 & 0 2 & 1 - lambda & -6 2 & -2 & 3 - lambda end{bmatrix}$$ Setting the determinant to zero: $$det(A - lambda I) = begin{vmatrix} 1 - lambda & 2 & 0 2 & 1 - lambda & -6 2 & -2 & 3 - lambda end{vmatrix} = 0$$ Expanding along the first row (since entry $a{13} = 0$ simplifies calculation): $$begin{aligned} det(A - lambda I) &= (1 - lambda) begin{vmatrix} 1 - lambda & -6 -2 & 3 - lambda end{vmatrix} - 2 begin{vmatrix} 2 & -6 2 & 3 - lambda end{vmatrix} + 0 end{aligned}$$ Evaluate the two $2 times 2$ determinants: 1. First determinant: $$begin{aligned} (1 - lambda)(3 - lambda) - (-6)(-2) &= (3 - 4lambda + lambda^2) - 12 &= lambda^2 - 4lambda - 9 end{aligned}$$ 2. Second determinant: $$2(3 - lambda) - (-6)(2) = 6 - 2lambda + 12 = 18 - 2lambda$$ Multiplying by their respective row coefficients: $$begin{aligned} det(A - lambda I) &= (1 - lambda)(lambda^2 - 4lambda - 9) - 2(18 - 2lambda) &= [1(lambda^2 - 4lambda - 9) - lambda(lambda^2 - 4lambda - 9)] - (36 -

Question 16

4. (b) If f(x, y) = xy (x^2 - y^2)/(x^2 + y^2) when (x, y) != (0, 0) and 0 when (x, y) = (0, 0), then find fxy(0, 0) and fyx(0, 0). (15 Marks)

Accepted Answer

1. Conceptual Framework & Schwarz's Theorem In multivariable calculus, Schwarz's Theorem (also known as Clairaut's Theorem on Equality of Mixed Partials) states: If a function $f(x, y)$ has partial derivatives $fx, fy, f{xy},$ and $f{yx}$ in a neighborhood of a point $(a, b)$, and if the mixed partial derivatives $f{xy}$ and $f{yx}$ are continuous at $(a, b)$, then: $$f{xy}(a, b) = f{yx}(a, b)$$ When the mixed partial derivatives are discontinuous at the origin $(0, 0)$, the equality can fail. The function presented in this problem: $$f(x, y) = begin{cases} xy frac{x^2 - y^2}{x^2 + y^2}, & (x, y) eq (0, 0) 0, & (x, y) = (0, 0) end{cases}$$ is the classic historical counterexample discovered by Hermann Schwarz (1873) and Giuseppe Peano demonstrating that continuity of the second-order mixed derivatives is an essential, indispensable hypothesis for their symmetry. --- 2. Definitions of Mixed Partial Derivatives from First Principles Recall the precise limit definitions for the second-order mixed partial derivatives evaluated at the origin $(0, 0)$: 1. Definition of $f{xy}(0, 0)$ (derivative with respect to $x$ first, then $y$, commonly denoted $frac{partial^2 f}{partial y partial x}(0, 0)$): $$f{xy}(0, 0) = lim{k to 0} frac{fx(0, k) - fx(0, 0)}{k}$$ 2. Definition of $f{yx}(0, 0)$ (derivative with respect to $y$ first, then $x$, commonly denoted $frac{partial^2 f}{partial x partial y}(0, 0)$): $$f{yx}(0, 0) = lim{h to 0} frac{fy(h, 0) - fy(0, 0)}{h}$$ To evaluate these limits, we must first compute: - $fx(0, 0)$ and $fy(0, 0)$ - $fx(0, k)$ for any $k eq 0$ - $fy(h, 0)$ for any $h eq 0$ --- 3. Computation of First-Order Partial Derivatives at the Origin By definition of the partial derivative with respect to $x$ at $(0, 0)$: $$fx(0, 0) = lim{h to 0} frac{f(0 + h, 0) - f(0, 0)}{h} = lim{h to 0} frac{f(h, 0) - 0}{h}$$ For any $h eq 0$: $$f(h, 0) = (h)(0) frac{h^2 - 0^2}{h^2 + 0^2} = 0$$ Therefore: $$fx(0, 0) = lim{h to 0} frac{0 - 0}{h} = 0$$ Similarly, for the partial de

Question 17

4. (c) (ii) Let Pn denote the vector space of all polynomials of degree <= n over R. Verify that dim(P4 / P2) = dim P4 - dim P2. (8 Marks)

Accepted Answer

1. Conceptual Framework & Definitions In abstract linear algebra: 1. The Vector Space of Polynomials $Pn$: $Pn$ denotes the real vector space consisting of all polynomials in an indeterminate $x$ with real coefficients having degree at most $n$: $$Pn = left{ p(x) = a0 + a1 x + a2 x^2 + dots + an x^n mid a0, a1, dots, an in mathbb{R} right}$$ The standard monomial basis for $Pn$ is: $$mathcal{B}n = {1, x, x^2, dots, x^n}$$ Since there are $n + 1$ linearly independent basis elements in $mathcal{B}n$, the dimension of $Pn$ is: $$dim(Pn) = n + 1$$ 2. Subspace Property: For any non-negative integers $m le n$, any polynomial of degree $le m$ is automatically a polynomial of degree $le n$. Hence, $Pm$ is a legitimate vector subspace of $Pn$: $$Pm subseteq Pn$$ 3. Quotient Vector Space $V / W$: Let $V$ be a vector space over $mathbb{R}$ and let $W$ be a subspace of $V$. The quotient space $V / W$ is the set of all cosets (affine equivalence classes) of $W$ in $V$: $$V / W = { v + W mid v in V }$$ endowed with the standard coset addition and scalar multiplication: $$(v1 + W) + (v2 + W) = (v1 + v2) + W, quad c(v + W) = (cv) + W$$ 4. Quotient Space Dimension Theorem: If $V$ is a finite-dimensional vector space and $W$ is a subspace of $V$, then: $$dim(V / W) = dim(V) - dim(W)$$ --- 2. Determination of $dim(P4)$ and $dim(P2)$ - For $P4$: Any polynomial $p(x) in P4$ has degree $le 4$: $$p(x) = a0 + a1 x + a2 x^2 + a3 x^3 + a4 x^4 quad (ai in mathbb{R})$$ The canonical basis of $P4$ is: $$mathcal{B}4 = {1, x, x^2, x^3, x^4}$$ The number of basis elements in $mathcal{B}4$ is $4 + 1 = 5$. Therefore: $$dim(P4) = 5$$ - For $P2$: Any polynomial $q(x) in P2$ has degree $le 2$: $$q(x) = b0 + b1 x + b2 x^2 quad (bi in mathbb{R})$$ The canonical basis of $P2$ is: $$mathcal{B}2 = {1, x, x^2}$$ The number of basis elements in $mathcal{B}2$ is $2 + 1 = 3$. Therefore: $$dim(P2) = 3$$ Notice that $mathcal{B}2 subset mathcal{B}4$, confirming that $P2$ is a 3-dimensional subspace of the 5-dimens

Question 18

5. (b) Form the differential equation of all ellipses whose axes coincide with coordinate axes. (10 Marks)

Accepted Answer

1. Conceptual Framework & Formation of Differential Equations In the theory of ordinary differential equations: 1. Geometric Family of Curves: A family of plane curves involving $n$ essential, independent arbitrary constants (parameters) requires an ordinary differential equation of order $n$ for its complete elimination. 2. Procedure: - Write the algebraic equation of the given geometric family. - Differentiate the equation with respect to $x$ successively $n$ times. - Eliminate the $n$ arbitrary constants between the original equation and its derived equations. - The resulting relation between $x, y, y', y'', dots, y^{(n)}$ is the required differential equation. --- 2. Equation of the Family of Ellipses The standard equation of any ellipse whose center is at the origin $(0, 0)$ and whose axes of symmetry coincide with the Cartesian coordinate axes ($x$-axis and $y$-axis) is given by: $$frac{x^2}{a^2} + frac{y^2}{b^2} = 1$$ where $a$ and $b$ are the lengths of the semi-axes along the $x$ and $y$ directions, respectively ($a 0, b 0$). To simplify the elimination, we can express the two arbitrary constants as: $$A = frac{1}{a^2} 0 quad text{and} quad B = frac{1}{b^2} 0$$ The equation becomes: $$A x^2 + B y^2 = 1 quad text{--- (Eq. 1)}$$ Here, there are two arbitrary constants ($A$ and $B$). Therefore, the resulting differential equation must be of the second order ($n = 2$). --- 3. First Differentiation Differentiating (Eq. 1) implicitly with respect to $x$: $$frac{d}{dx}(A x^2 + B y^2) = frac{d}{dx}(1)$$ $$2A x + 2B y frac{dy}{dx} = 0$$ Dividing by 2 and writing $y' = frac{dy}{dx}$: $$A x + B y y' = 0 quad text{--- (Eq. 2)}$$ Notice that we can immediately eliminate one constant by isolating the ratio $frac{A}{B}$: $$frac{A}{B} = -frac{y y'}{x} quad text{--- (Eq. 3)}$$ --- 4. Second Differentiation & Elimination of Constants Method 1: Differentiating the Isolated Ratio (Direct & Elegant) Since $A$ and $B$ are constants, their ratio $frac{A}{B}$ is also a constant. T

Question 19

5. (c) Prove that the time taken by the Earth to travel over half of its orbit, which is separated by the minor axis and is remote from the Sun, when the Sun is at the focus of the elliptic orbit, is two days more than half of the year. The eccentricity of the orbit is taken as 1/60. (10 Marks)

Accepted Answer

1. Conceptual Framework & Kepler's Laws of Planetary Motion In celestial mechanics and central orbit dynamics: 1. Kepler's First Law: The Earth moves in an elliptical orbit with the Sun located at one of the two foci $S$. 2. Kepler's Second Law (Areal Velocity Theorem): The radius vector joining the Sun to the Earth sweeps out equal areas in equal intervals of time: $$frac{dA}{dt} = frac{h}{2} = text{constant}$$ where $h$ is the angular momentum per unit mass. 3. Consequence of Kepler's Second Law: The time $t$ taken by the planet to sweep out any portion of its orbital path having focal area $A$ is strictly proportional to that area: $$frac{t}{T} = frac{text{Area swept by the radius vector from the Sun}}{text{Total area of the ellipse}}$$ where $T = 365text{ days}$ is the period of one complete revolution (one year). --- 2. Geometry of the Elliptic Orbit Let the equation of the Earth's elliptical orbit referred to its center $C(0, 0)$ as origin be: $$frac{x^2}{a^2} + frac{y^2}{b^2} = 1$$ where: - $a$ is the semi-major axis. - $b$ is the semi-minor axis ($b = asqrt{1 - e^2}$). - $e = frac{1}{60}$ is the orbital eccentricity. - The total area of the entire ellipse is: $$A{text{total}} = pi a b$$ Let the major axis lie along the $x$-axis and the minor axis lie along the $y$-axis: - Foci of the ellipse: $S(ae, 0)$ and $S'(-ae, 0)$. - Without loss of generality, let the Sun be located at the focus $S(ae, 0)$ on the positive $x$-axis. - The minor axis is the line segment joining $(0, b)$ and $(0, -b)$, dividing the ellipse into two halves: - The half closer to the Sun: $x ge 0$ (contains perihelion). - The half remote from the Sun: $x le 0$ (contains aphelion). --- 3. Focal Area Swept During the Journey Over the Remote Half The problem asks for the time $t1$ taken by the Earth to describe the half of the orbit remote from the Sun, separated by the minor axis. - Path of Earth: Moves along the elliptic arc from the positive end of the minor axis $B(0, b)$ through the aphel

Question 20

5. (d) Given that A and B are two points in the same horizontal line distant 2a apart. AO and BO are two equal heavy strings tied together at O and carrying their weight at O. If l is length of each string and d is depth of O below AB, then show that the parameter c of this catenary, in which the st

Accepted Answer

1. Conceptual Framework & Equations of the Common Catenary A uniform, flexible, inextensible heavy string suspended under the action of gravity hangs in the curve of a common catenary. Referred to the lowest point (vertex) as having vertical coordinate $y = c$ above the directrix (which is taken as the $x$-axis): 1. Cartesian Equation: $$y = c coshleft(frac{x}{c}right)$$ 2. Intrinsic Equation relating arc length $s$ from the vertex to coordinate $x$: $$s = c sinhleft(frac{x}{c}right)$$ 3. Fundamental Relation between $y, s,$ and $c$: $$y^2 = c^2 + s^2$$ 4. Hyperbolic Identity: $$cosh(2theta) - 1 = 2sinh^2(theta) iff cosh(u) - 1 = 2sinh^2left(frac{u}{2}right)$$ --- 2. Physical Setup & Geometric Configuration - $A$ and $B$ are two fixed suspension points situated in the same horizontal line at distance $2a$ apart. - Because of lateral symmetry, the vertical line bisecting $AB$ is the axis of symmetry of the system. - Let the vertical line through $O$ be the $y$-axis ($x = 0$), so: - $x$-coordinate of $A$: $x = -a$ - $x$-coordinate of $B$: $x = a$ - $AO$ and $BO$ are two equal strings of length $l$. - Since $AO$ and $BO$ are identical and carry their weight at $O$, by symmetry, each string forms a branch of the same catenary. - Let: - $O$ be the lowest joint at $x = 0$, having depth $d$ below the horizontal line $AB$. - Arc length of string $OB$: $s = l$. - Horizontal span from $O$ to $B$: $xB = a$. --- 3. Establishing Coordinate Relations Let the horizontal directrix of the catenary be at a distance $yO$ below $O$. Since $O$ lies on the $y$-axis ($x = 0$): $$yO = c coshleft(frac{0}{c}right) = c cosh(0) = c$$ Now, consider the suspension point $B(a, yB)$: - The depth of $O$ below $AB$ is $d$. - Since $A$ and $B$ are on the horizontal line, the vertical coordinate of $B$ is: $$yB = yO + d = c + d$$ On the other hand, since $B(a, yB)$ lies on the catenary: $$yB = c coshleft(frac{a}{c}right)$$ Therefore: $$c + d = c coshleft(frac{a}{c}right)$$ $$d = c left[ coshleft(frac{

Question 21

5. (e) If u = x + y + z, v = x^2 + y^2 + z^2 and w = xy + yz + zx, then show that grad u, grad v and grad w are coplanar. (10 Marks)

Accepted Answer

1. Conceptual Framework & Coplanarity Condition for Vectors In vector calculus: 1. Gradient of a Scalar Field: The gradient of a continuously differentiable scalar field $phi(x, y, z)$ is a vector field defined by: $$operatorname{grad} phi = abla phi = frac{partial phi}{partial x} hat{i} + frac{partial phi}{partial y} hat{j} + frac{partial phi}{partial z} hat{k}$$ 2. Coplanarity of Three Vectors: Three vectors $vec{A}, vec{B}, vec{C}$ in three-dimensional Euclidean space are coplanar if and only if their scalar triple product vanishes identically: $$[vec{A}, vec{B}, vec{C}] = vec{A} cdot (vec{B} times vec{C}) = begin{vmatrix} Ax & Ay & Az Bx & By & Bz Cx & Cy & Cz end{vmatrix} = 0$$ 3. Functional Dependence (Jacobian Criterion): By the Jacobian theorem on functional dependence, three functions $u, v, w$ of three variables $x, y, z$ are functionally dependent (i.e., there exists a non-trivial relation $F(u, v, w) = 0$) if and only if their Jacobian matrix has determinant zero: $$frac{partial(u, v, w)}{partial(x, y, z)} = [ abla u, abla v, abla w] = 0$$ This directly implies that the gradient vectors $ abla u, abla v, abla w$ are linearly dependent, and therefore coplanar at every point in space. We demonstrate both the direct vector determinant method and the functional dependence method. --- 2. Computation of Gradient Vectors We are given the three scalar functions: $$u = x + y + z$$ $$v = x^2 + y^2 + z^2$$ $$w = xy + yz + zx$$ Let us compute the gradient of each function: 1. Gradient of $u$: $$ abla u = frac{partial u}{partial x} hat{i} + frac{partial u}{partial y} hat{j} + frac{partial u}{partial z} hat{k} = 1hat{i} + 1hat{j} + 1hat{k} = (1, 1, 1)$$ 2. Gradient of $v$: $$ abla v = frac{partial v}{partial x} hat{i} + frac{partial v}{partial y} hat{j} + frac{partial v}{partial z} hat{k} = 2xhat{i} + 2yhat{j} + 2zhat{k} = (2x, 2y, 2z)$$ 3. Gradient of $w$: - $frac{partial w}{partial x} = y + z$ - $frac{partial w}{partial y} = x + z$ - $frac{partial w}{partial z} = x

Question 22

5. (a) Solve: (1 - y^2 + y^4/x^2)(dy/dx)^2 - 2(y/x)(dy/dx) + y^2/x^2 = 0. (10 Marks)

Accepted Answer

1. Conceptual Framework & Classification of First-Order Equations The given differential equation is of first order but of second degree in the derivative $p = frac{dy}{dx}$: $$left(1 - y^2 + frac{y^4}{x^2}right) p^2 - 2left(frac{y}{x}right) p + frac{y^2}{x^2} = 0$$ Such equations belong to the classical class of first-order differential equations of higher degree, generally represented as: $$A(x, y) p^2 + B(x, y) p + C(x, y) = 0$$ To solve this equation, we can pursue two standard analytical avenues: 1. Solvable for $p$: Treating the equation as an algebraic quadratic equation in $p = frac{dy}{dx}$ and solving for $p$ using the quadratic formula. 2. Change of Variables (Substitution): Noticing the homogeneous-like combination $frac{y}{x}$ and related algebraic structures to decouple variables into separable or Clairaut-type forms. We solve for $p = frac{dy}{dx}$ directly. --- 2. Solving Algebraically for $p = frac{dy}{dx}$ Let: $$A = 1 - y^2 + frac{y^4}{x^2}, quad B = -frac{2y}{x}, quad C = frac{y^2}{x^2}$$ Applying the quadratic formula $p = frac{-B pm sqrt{B^2 - 4AC}}{2A}$: Let us compute the discriminant $Delta = B^2 - 4AC$: $$begin{aligned} Delta &= left(-frac{2y}{x}right)^2 - 4 left(1 - y^2 + frac{y^4}{x^2}right) left(frac{y^2}{x^2}right) &= frac{4y^2}{x^2} - 4left(frac{y^2}{x^2}right) left(1 - y^2 + frac{y^4}{x^2}right) &= frac{4y^2}{x^2} left[ 1 - left(1 - y^2 + frac{y^4}{x^2}right) right] &= frac{4y^2}{x^2} left[ 1 - 1 + y^2 - frac{y^4}{x^2} right] &= frac{4y^2}{x^2} left( y^2 - frac{y^4}{x^2} right) &= frac{4y^4}{x^2} left( 1 - frac{y^2}{x^2} right) = frac{4y^4}{x^4} (x^2 - y^2) end{aligned}$$ Hence: $$sqrt{Delta} = sqrt{frac{4y^4}{x^2} left(y^2 - frac{y^4}{x^2}right)} = frac{2y^2}{x} sqrt{1 - y^2 + dots}$$ Let us alternatively expand the original equation to find an even simpler factorized form. --- 3. Alternative Insight: Grouping by Perfect Squares Let us rewrite the given equation: $$left(1 - y^2 + frac{y^4}{x^2}right) p^2 - 2frac{y}{x} p + frac{y^2}{x

Question 23

6. (c) (i) Find the absolute value of the directional derivative of phi(x, y, z) = x^2 y^2 z^2 at the point (1, 1, -1) in the direction of the tangent to the curve x = e^t, y = 2 sin t + 1, z = t - cos t, at t = 0. (10 Marks)

Accepted Answer

1. Conceptual Framework & Definition of Directional Derivative In vector differential calculus: 1. Directional Derivative: The directional derivative of a scalar field $phi(x, y, z)$ at a point $P$ in the direction of a unit vector $hat{u}$ is defined by: $$D{hat{u}} phi = abla phi cdot hat{u}$$ where $ abla phi = operatorname{grad} phi$ is the gradient vector of $phi$. 2. Tangent Vector to a Space Curve: For a space curve parameterized by $vec{r}(t) = x(t)hat{i} + y(t)hat{j} + z(t)hat{k}$, the tangent vector at any parameter value $t$ is given by: $$vec{v} = frac{dvec{r}}{dt} = frac{dx}{dt} hat{i} + frac{dy}{dt} hat{j} + frac{dz}{dt} hat{k}$$ 3. Unit Direction Vector $hat{u}$: The unit vector in the direction of the tangent is: $$hat{u} = frac{vec{v}}{|vec{v}|}$$ --- 2. Computation of Gradient $ abla phi$ at $(1, 1, -1)$ The given scalar field is: $$phi(x, y, z) = x^2 y^2 z^2$$ Computing the partial derivatives: - $frac{partial phi}{partial x} = 2x y^2 z^2$ - $frac{partial phi}{partial y} = 2x^2 y z^2$ - $frac{partial phi}{partial z} = 2x^2 y^2 z$ Evaluating these partial derivatives at the given point $P(1, 1, -1)$: - $left. frac{partial phi}{partial x} right|{(1, 1, -1)} = 2(1)(1)^2 (-1)^2 = 2$ - $left. frac{partial phi}{partial y} right|{(1, 1, -1)} = 2(1)^2 (1) (-1)^2 = 2$ - $left. frac{partial phi}{partial z} right|{(1, 1, -1)} = 2(1)^2 (1)^2 (-1) = -2$ Therefore, the gradient vector at $(1, 1, -1)$ is: $$ abla phi(1, 1, -1) = 2hat{i} + 2hat{j} - 2hat{k} = (2, 2, -2)$$ --- 3. Computation of the Tangent Vector $vec{v}$ at $t = 0$ The space curve is given parametrically by: $$x(t) = e^t$$ $$y(t) = 2sin t + 1$$ $$z(t) = t - cos t$$ Let us check the point corresponding to $t = 0$: - $x(0) = e^0 = 1$ - $y(0) = 2sin(0) + 1 = 1$ - $z(0) = 0 - cos(0) = -1$ This perfectly matches the given point $P(1, 1, -1)$. Now, compute the derivatives with respect to $t$: - $frac{dx}{dt} = frac{d}{dt}(e^t) = e^t$ - $frac{dy}{dt} = frac{d}{dt}(2sin t + 1) = 2cos t$ - $frac{dz}{dt} =

Question 24

6. (c) (ii) If del . E = 0, del . H = 0, del x E = -dH/dt and del x H = dE/dt, then show that del^2 H = d^2H/dt^2 and del^2 E = d^2E/dt^2. (10 Marks)

Accepted Answer

1. Conceptual Framework & Maxwell's Wave Equations The given equations are the dimensionless form of Maxwell's Equations in Free Space (in a source-free, non-conducting dielectric medium with zero charge density $rho = 0$ and zero current density $vec{J} = mathbf{0}$, where the wave speed is normalized to $c = 1$): 1. Gauss's Law for Electricity: $$ abla cdot vec{E} = 0 quad text{--- (Eq. 1)}$$ 2. Gauss's Law for Magnetism: $$ abla cdot vec{H} = 0 quad text{--- (Eq. 2)}$$ 3. Faraday's Law of Electromagnetic Induction: $$ abla times vec{E} = -frac{partial vec{H}}{partial t} quad text{--- (Eq. 3)}$$ 4. Ampère-Maxwell Law: $$ abla times vec{H} = frac{partial vec{E}}{partial t} quad text{--- (Eq. 4)}$$ Essential Vector Identity: To decouple these first-order coupled partial differential equations into second-order wave equations, we invoke the fundamental vector identity for the curl of the curl of any twice continuously differentiable vector field $vec{V}$: $$ abla times ( abla times vec{V}) = abla( abla cdot vec{V}) - abla^2 vec{V}$$ where $ abla^2 vec{V}$ is the vector Laplacian: $$ abla^2 vec{V} = abla^2 Vx hat{i} + abla^2 Vy hat{j} + abla^2 Vz hat{k}$$ --- 2. Derivation of the Wave Equation for the Magnetic Field $vec{H}$ To obtain the wave equation for $vec{H}$, we take the curl of both sides of the Ampère-Maxwell Law (Eq. 4): $$ abla times ( abla times vec{H}) = abla times left( frac{partial vec{E}}{partial t} right)$$ Using the vector identity on the left-hand side: $$ abla times ( abla times vec{H}) = abla( abla cdot vec{H}) - abla^2 vec{H}$$ Since spatial partial derivatives commute with temporal partial derivatives (by Clairaut's theorem for smooth fields), the right-hand side can be rewritten as: $$ abla times left( frac{partial vec{E}}{partial t} right) = frac{partial}{partial t}( abla times vec{E})$$ Equating the two expressions: $$ abla( abla cdot vec{H}) - abla^2 vec{H} = frac{partial}{partial t}( abla times vec{E})$$ Now, substitute the known relations:

Question 25

6. (a) If F(s) and G(s) are Laplace transforms of f(t) and g(t) respectively, then prove that L[int0^t f(x) g(t - x) dx] = F(s) G(s). Using this result, solve the equation y(t) = t + int0^t y(x) sin(t - x) dx. (15 Marks)

Accepted Answer

1. Conceptual Framework & Convolution Theorem In the theory of integral transforms: 1. Laplace Transform Definition: The Laplace transform of a function $f(t)$ defined for $t ge 0$ is: $$mathcal{L}{f(t)} = F(s) = int0^infty e^{-st} f(t) , dt$$ 2. Convolution Product: The convolution of two piecewise continuous functions $f(t)$ and $g(t)$ on $[0, infty)$ is defined by: $$(f g)(t) = int0^t f(x) g(t - x) , dx$$ 3. The Convolution Theorem: States that the Laplace transform of the convolution product of two functions is simply the algebraic product of their individual Laplace transforms: $$mathcal{L}{(f g)(t)} = mathcal{L}{f(t)} cdot mathcal{L}{g(t)} = F(s) G(s)$$ 4. Volterra Integral Equation of the Second Kind: An equation of the form: $$y(t) = f(t) + int0^t K(t, x) y(x) , dx$$ where the kernel is of convolution type $K(t, x) = k(t - x)$, can be transformed into an algebraic equation using the Convolution Theorem and solved analytically. --- 2. Proof of the Convolution Theorem By definition of the Laplace transform: $$mathcal{L}{(f g)(t)} = int{t = 0}^{infty} e^{-st} left[ int{x = 0}^{t} f(x) g(t - x) , dx right] dt$$ This expression is a double integral over the two-dimensional region $R$ in the $(x, t)$-plane defined by: $$R = { (x, t) mid 0 le t < infty, quad 0 le x le t }$$ Geometry of the Region of Integration: - In the $(x, t)$-plane, $x$ ranges from $0$ to $t$, meaning the region is bounded by the line $x = 0$ ($t$-axis) and the line $x = t$ extending into the first quadrant. - We reverse the order of integration using Fubini's Theorem: - The variable $x$ varies over the interval: $0 le x < infty$. - For each fixed $x$, $t$ varies from the boundary line $t = x$ to $infty$: $x le t < infty$. Rewriting the double integral with the reversed order of integration: $$mathcal{L}{(f g)(t)} = int{x = 0}^{infty} left[ int{t = x}^{infty} e^{-st} f(x) g(t - x) , dt right] dx$$ Since $f(x)$ does not depend on the inner variable $t$, we factor it out of the inner integral: $$

Question 26

6. (b) One end of an elastic string, having natural length a, is fixed at some point O and a heavy particle is attached to the other end of the string. The string is drawn vertically downward till it is four times its natural length at the point C and then released. If the modulus of elasticity of t

Accepted Answer

1. Conceptual Framework & Dynamical Principles This problem models the vertical motion of a particle attached to an elastic string under the combined action of gravity and Hooke's Law of Elasticity. Fundamental Physical Principles: 1. Hooke's Law: The tension $T$ in an elastic string of natural length $a$, stretched to length $a + x$ (extension $x$), is proportional to the extension: $$T = lambda frac{x}{a}$$ where $lambda$ is the modulus of elasticity. We are given that $lambda$ equals the weight of the particle: $$lambda = mg$$ Therefore: $$T = mg frac{x}{a}$$ 2. Equilibrium Position ($B$): In static equilibrium, the downward gravitational force equals the upward tension: $$mg = Te = mg frac{e}{a} implies e = a$$ Thus, in the equilibrium position $B$, the extension is $e = a$, meaning $B$ is at a distance $a + a = 2a$ below the fixed point $O$. 3. Two Distinct Regimes of Motion: Because a flexible string can support tension but cannot support thrust/compression: - Stretched Regime (Below natural length, distance $ a$ from $O$): The string is taut, experiencing both gravity and tension. The particle executes Simple Harmonic Motion (SHM) centered at the equilibrium position $B$. - Slack Regime (Above natural length, distance $< a$ from $O$): When the particle rises above the unstretched length $a$, the string becomes slack ($T = 0$). The particle moves freely under gravity alone as a projectile moving vertically upwards. --- 2. Motion in the Stretched State (Simple Harmonic Motion) Let $x$ be the downward displacement of the particle measured from the equilibrium point $B$ (located at depth $2a$ below $O$). - Depth below $O$ is $2a + x$. - Total extension of the string beyond its natural length $a$ is $(2a + x) - a = a + x$. - Tension in the string: $$T = mg frac{a + x}{a} = mg + frac{mg}{a} x$$ The equation of motion of the particle of mass $m$ is: $$m frac{d^2x}{dt^2} = mg - T = mg - left(mg + frac{mg}{a} xright) = -frac{mg}{a} x$$ $$frac{d^2x}{dt^2} = -frac{g}{a}

Question 27

7. (b) Verify Green's theorem in the plane for ointC [(xy + y^2) dx + x^2 dy], where C is the boundary of the region bounded by the curves y = x and y = x^2. (15 Marks)

Accepted Answer

1. Conceptual Framework & Green's Theorem in the Plane Green's Theorem in the Plane is a fundamental theorem of vector calculus that connects a line integral around a simple, closed, positively oriented piecewise-smooth curve $C$ to a double integral over the plane region $R$ bounded by $C$. Statement of Green's Theorem: Let $R$ be a regular, closed, bounded region in the $xy$-plane whose boundary $C = partial R$ is a positively oriented (counter-clockwise), piecewise-smooth, simple closed curve. If $M(x, y)$ and $N(x, y)$ are continuous functions having continuous first-order partial derivatives on an open region containing $R$, then: $$ointC (M , dx + N , dy) = iintR left( frac{partial N}{partial x} - frac{partial M}{partial y} right) dx , dy$$ To verify Green's Theorem, we must compute: 1. The line integral $ointC (M , dx + N , dy)$ along the boundary paths directly. 2. The double integral $iintR left( frac{partial N}{partial x} - frac{partial M}{partial y} right) dx , dy$ over the enclosed region $R$. 3. Show that both independent calculations yield the exact same numerical result. Here: $$M(x, y) = xy + y^2, quad N(x, y) = x^2$$ --- 2. Geometry of the Region $R$ and Boundary $C$ The region $R$ is bounded by: - The parabola: $y = x^2$ - The straight line: $y = x$ Points of Intersection: Equating $y = x^2$ and $y = x$: $$x^2 - x = 0 implies x(x - 1) = 0 implies x = 0 quad text{or} quad x = 1$$ - For $x = 0 implies y = 0$. Point: $O(0, 0)$. - For $x = 1 implies y = 1$. Point: $A(1, 1)$. On the interval $(0, 1)$, $x^2 < x$. Hence, the line $y = x$ lies above the parabola $y = x^2$. The boundary $C$ traversed in the positive (counter-clockwise) direction consists of two smooth curves: 1. Path $C1$: Along the parabola $y = x^2$ from $O(0, 0)$ to $A(1, 1)$. 2. Path $C2$: Along the straight line $y = x$ from $A(1, 1)$ back to $O(0, 0)$. Therefore: $$ointC (M , dx + N , dy) = int{C1} (M , dx + N , dy) + int{C2} (M , dx + N , dy)$$ --- 3. Evaluation of the Line Integral

Question 28

7. (c) (ii) Find the complete solution of x^3 (d^3y/dx^3) + 3x^2 (d^2y/dx^2) + x (dy/dx) + y = x log x. (10 Marks)

Accepted Answer

1. Conceptual Framework: Cauchy-Euler Equidimensional Differential Equations The given equation: $$x^3 frac{d^3y}{dx^3} + 3x^2 frac{d^2y}{dx^2} + x frac{dy}{dx} + y = x log x$$ is a classical Cauchy-Euler (or Euler-Cauchy) linear differential equation of order 3 with variable coefficients, where the coefficient of the $k$-th derivative $frac{d^k y}{dx^k}$ is proportional to $x^k$. Standard Transformation to Constant Coefficients: We introduce the logarithmic substitution for $x 0$: $$x = e^z iff z = log x$$ Let $D = frac{d}{dz}$. Using the chain rule, the differential operators transform as: - $x frac{dy}{dx} = D y$ - $x^2 frac{d^2y}{dx^2} = D(D - 1)y$ - $x^3 frac{d^3y}{dx^3} = D(D - 1)(D - 2)y$ The complete solution consists of: $$y(x) = yc(x) + yp(x)$$ where $yc$ is the Complementary Function (CF) solving the homogeneous equation, and $yp$ is the Particular Integral (PI) corresponding to the non-homogeneous source term. --- 2. Transformation of the Differential Equation Substitute the differential operator expressions into the left-hand side: $$begin{aligned} x^3 frac{d^3y}{dx^3} + 3x^2 frac{d^2y}{dx^2} + x frac{dy}{dx} + y &= left[ D(D - 1)(D - 2) + 3D(D - 1) + D + 1 right] y end{aligned}$$ Let us expand the polynomial operator in $D$: $$begin{aligned} D(D - 1)(D - 2) &= D(D^2 - 3D + 2) = D^3 - 3D^2 + 2D 3D(D - 1) &= 3D^2 - 3D end{aligned}$$ Adding the terms: $$begin{aligned} (D^3 - 3D^2 + 2D) + (3D^2 - 3D) + D + 1 &= D^3 + (-3 + 3)D^2 + (2 - 3 + 1)D + 1 &= D^3 + 0 D^2 + 0 D + 1 &= D^3 + 1 end{aligned}$$ The right-hand side in terms of $z$ is: $$x log x = e^z cdot z = z e^z$$ Thus, the differential equation transforms into the linear constant-coefficient ODE: $$(D^3 + 1)y = z e^z$$ --- 3. Finding the Complementary Function ($yc$) The auxiliary equation (characteristic equation) is: $$m^3 + 1 = 0$$ Factoring using the sum of cubes formula $m^3 + 1 = (m + 1)(m^2 - m + 1) = 0$: 1. Real root: $$m1 = -1$$ 2. Complex conjugate roots from $m^2 - m + 1 = 0$: $$m = frac{-

Question 29

7. (a) A solid sphere rests inside a fixed rough and hemispherical bowl of twice its radius. If a large amount of weight, whatsoever, is attached to the highest point of the sphere, then show that the equilibrium is stable. (15 Marks)

Accepted Answer

1. Conceptual Framework & Stability of Equilibrium of Rolling Bodies In analytical statics, when a body of one curved surface rests in equilibrium on another fixed curved surface (assumed sufficiently rough to prevent slipping): 1. The Fundamental Stability Criterion for Rolling Bodies: If a body whose surface has radius of curvature $rho1$ at the point of contact rests on a fixed surface of radius of curvature $rho2$ (where $rho2$ is positive if concave upwards), and $h$ is the height of the center of gravity ($G$) of the composite resting body above the point of contact $A$, then: - The equilibrium is stable if: $$frac{1}{h} frac{1}{rho1} + frac{1}{rho2} iff h frac{1}{r} - frac{1}{R}$$ $$h 0$$ Furthermore, the attached weight $W1$ is positive ($W1 0$): $$frac{W r}{W + W1} 0$$ Therefore, the term $frac{W r}{W + W1}$ is strictly positive for any finite weight $W1$, no matter how large $W1$ is! Consequently: $$h = 2r - frac{W r}{W + W1} 0$$ for all $h < 2r$. Since $h = 2r - frac{Wr}{W + W1} < 2r$ is always satisfied, the potential energy is at a strict local minimum. --- 6. Conclusion Since $h < 2r$ is guaranteed for any weight $W1$ whatsoever, the stability criterion is universally satisfied. Hence, the equilibrium is unconditionally stable.

Question 30

7. (c) (i) Find the general solution and singular solution of the differential equation (1 + dy/dx)^3 = (27/8a) (x + y) (1 - dy/dx)^3. (10 Marks)

Accepted Answer

1. Conceptual Framework & Classification The given differential equation is: $$left(1 + frac{dy}{dx}right)^3 = frac{27}{8a} (x + y) left(1 - frac{dy}{dx}right)^3$$ Writing $p = frac{dy}{dx}$, the equation becomes: $$(1 + p)^3 = frac{27}{8a} (x + y) (1 - p)^3$$ This is a non-linear first-order differential equation of degree 3. Notice the characteristic linear combination: - The base term on the left is $1 + p = 1 + frac{dy}{dx} = frac{d(x + y)}{dx}$. - The base term on the right is $1 - p = 1 - frac{dy}{dx} = frac{d(x - y)}{dx}$. - The spatial variable appearing is $(x + y)$. This striking structure strongly suggests a change of coordinate axes by a $45^circ$ rotation, or standard linear transformation of variables: $$u = x + y quad text{and} quad v = x - y$$ --- 2. Transformation of Variables Let: $$u = x + y quad implies quad du = dx + dy$$ $$v = x - y quad implies quad dv = dx - dy$$ Dividing the two differential expressions: $$frac{du}{dv} = frac{dx + dy}{dx - dy} = frac{1 + frac{dy}{dx}}{1 - frac{dy}{dx}} = frac{1 + p}{1 - p}$$ Let us define: $$P = frac{du}{dv} = frac{1 + p}{1 - p}$$ Now, rearrange the original differential equation by dividing by $(1 - p)^3$: $$left(frac{1 + p}{1 - p}right)^3 = frac{27}{8a} (x + y)$$ Substituting our new variables $u = x + y$ and $P = frac{du}{dv}$: $$P^3 = frac{27}{8a} u$$ $$u = frac{8a}{27} P^3$$ --- 3. Solving for the General Solution Notice that $v$ does not appear explicitly in the transformed equation $u = frac{8a}{27} P^3$. This is an equation solvable for $u$ in terms of $P = frac{du}{dv}$. Taking the reciprocal derivative: $$frac{dv}{du} = frac{1}{P}$$ $$dv = frac{1}{P} , du$$ Differentiating $u = frac{8a}{27} P^3$ with respect to $P$: $$du = frac{8a}{27} (3P^2 , dP) = frac{8a}{9} P^2 , dP$$ Substituting this expression for $du$ into the differential for $v$: $$begin{aligned} dv &= frac{1}{P} left( frac{8a}{9} P^2 , dP right) &= frac{8a}{9} P , dP end{aligned}$$ Integrating both sides directly: $$v = int frac{8a}{9} P

Question 31

8. (c) A particle is projected inside a fixed smooth cylinder with circular cross-section in a vertical plane from the lowest point with initial horizontal velocity u. Show that for: (i) u^2 = 5ag; the particle executes complete circular motion, and (iii) 2ag < u^2 < 5ag; the particle will leave the

Accepted Answer

1. Conceptual Framework & Dynamical Formulation We consider the planar vertical circular motion of a particle of mass $m$ constrained to move along the inner surface of a fixed, smooth cylinder of radius $a$. Let: - $O$ be the center of the circular cross-section of radius $a$. - $A$ be the lowest point of the circle (taken as reference for potential energy). - $u$ be the initial horizontal projection speed at $A$. - $P$ be the position of the particle at any time $t$, such that the radius vector $OP$ makes an angle $theta$ with the downward vertical $OA$. - $v$ be the speed of the particle at $P$. - $R$ be the normal reaction exerted by the smooth inner cylindrical wall on the particle, directed radially inward toward the center $O$. --- 2. Derivation of Velocity and Normal Reaction A. Principle of Conservation of Mechanical Energy: The height of point $P$ above the lowest point $A$ is: $$h = a - acostheta = a(1 - costheta)$$ By conservation of mechanical energy (kinetic + potential): $$frac{1}{2} m v^2 + mgh = frac{1}{2} m u^2 + 0$$ $$v^2 = u^2 - 2gh = u^2 - 2ga(1 - costheta) quad text{--- (Eq. 1)}$$ B. Equation of Motion in the Radial Direction: The forces acting on the particle at $P$ along the inward normal (toward the center $O$) are: - The inward normal reaction $R$. - The radial component of gravity, which acts outward as $mg costheta$. Applying Newton's second law along the inward normal: $$frac{m v^2}{a} = R - mg costheta$$ $$R = frac{m v^2}{a} + mg costheta quad text{--- (Eq. 2)}$$ Substituting $v^2$ from (Eq. 1) into (Eq. 2): $$begin{aligned} R &= frac{m}{a} [u^2 - 2ga(1 - costheta)] + mg costheta &= frac{m u^2}{a} - 2mg + 2mg costheta + mg costheta &= frac{m}{a} (u^2 - 2ag) + 3mg costheta quad text{--- (Eq. 3)} end{aligned}$$ --- 3. Critical Conditions for Motion The nature of the motion depends on which vanishes first: - Velocity vanishes ($v = 0$): The particle comes momentarily to rest. From (Eq. 1): $$0 = u^2 - 2ag(1 - costheta) implies costhetav =

Question 32

8. (a) Solve the differential equation (x + 2) (d^2y/dx^2) - (2x + 5) (dy/dx) + 2y = (1 + x) e^x by the method of variation of parameters. (15 Marks)

Accepted Answer

1. Conceptual Framework & The Method of Variation of Parameters In ordinary differential equations, the Method of Variation of Parameters (developed by Joseph-Louis Lagrange) is a powerful, general technique for finding the particular integral of a linear non-homogeneous second-order differential equation: $$y'' + P(x) y' + Q(x) y = R(x)$$ Theoretical Framework: 1. Complementary Function: First, determine two linearly independent solutions $y1(x)$ and $y2(x)$ of the associated homogeneous equation: $$y'' + P(x) y' + Q(x) y = 0$$ The complementary function is $yc(x) = c1 y1(x) + c2 y2(x)$. 2. Wronskian: The Wronskian $W(y1, y2)$ is defined by the determinant: $$W(x) = begin{vmatrix} y1 & y2 y1' & y2' end{vmatrix} = y1 y2' - y2 y1' eq 0$$ 3. Particular Integral via Variation of Parameters: We replace the arbitrary constants $c1, c2$ by variable functions $u(x), v(x)$: $$yp(x) = u(x) y1(x) + v(x) y2(x)$$ where $u(x)$ and $v(x)$ are determined by the formulas: $$u(x) = -int frac{y2(x) R(x)}{W(x)} , dx, quad v(x) = int frac{y1(x) R(x)}{W(x)} , dx$$ 4. Complete Solution: $$y(x) = yc(x) + yp(x) = [c1 + u(x)] y1(x) + [c2 + v(x)] y2(x)$$ --- 2. Standard Form of the Differential Equation The given equation is: $$(x + 2) y'' - (2x + 5) y' + 2y = (1 + x) e^x$$ Dividing through by $(x + 2)$ to reduce to standard form $y'' + P(x) y' + Q(x) y = R(x)$: $$y'' - frac{2x + 5}{x + 2} y' + frac{2}{x + 2} y = frac{1 + x}{x + 2} e^x$$ Here: $$P(x) = -frac{2x + 5}{x + 2}, quad Q(x) = frac{2}{x + 2}, quad R(x) = frac{1 + x}{x + 2} e^x$$ --- 3. Finding Linearly Independent Solutions of the Homogeneous Equation Consider the homogeneous equation: $$(x + 2) y'' - (2x + 5) y' + 2y = 0$$ Let us test standard inspection rules for finding one integral belonging to the complementary function: 1. Rule for $y = e^{mx}$: $m^2 + m P + Q = 0$. Let us test $m = 2$: $$(x + 2)(2^2) - (2x + 5)(2) + 2 = 4x + 8 - 4x - 10 + 2 = (4 - 4)x + (8 - 10 + 2) = 0$$ This identity vanishes identically! Therefore, $y1(x)

Question 33

8. (b) Verify Gauss's divergence theorem for F = (x^2 - yz) i + (y^2 - zx) j + (z^2 - xy) k, taken over the rectangular parallelopiped 0 <= x <= a, 0 <= y <= b, 0 <= z <= c. (15 Marks)

Accepted Answer

1. Conceptual Framework & Gauss's Divergence Theorem Gauss's Divergence Theorem is one of the crowning results of vector integral calculus. It establishes that the outward flux of a continuously differentiable vector field $vec{F}$ across a closed, orientable surface $S$ bounding a spatial region $V$ is equal to the volume integral of the divergence of $vec{F}$ throughout $V$: $$iintS vec{F} cdot hat{n} , dS = iiintV ( abla cdot vec{F}) , dV$$ where $hat{n}$ is the outward-pointing unit normal vector to the surface $S$. To verify the theorem: 1. Compute the volume integral $iiintV ( abla cdot vec{F}) , dV$ directly. 2. Compute the surface integral $iintS vec{F} cdot hat{n} , dS$ across all 6 faces of the rectangular parallelopiped. 3. Show that both independent calculations give identical results. Here, the vector field is: $$vec{F} = (x^2 - yz)hat{i} + (y^2 - zx)hat{j} + (z^2 - xy)hat{k}$$ and the volume $V$ is the rectangular parallelopiped: $$V = { (x, y, z) mid 0 le x le a, quad 0 le y le b, quad 0 le z le c }$$ --- 2. Evaluation of the Volume Integral (RHS) First, compute the divergence of $vec{F}$: $$begin{aligned} abla cdot vec{F} &= frac{partial}{partial x}(x^2 - yz) + frac{partial}{partial y}(y^2 - zx) + frac{partial}{partial z}(z^2 - xy) &= 2x + 2y + 2z = 2(x + y + z) end{aligned}$$ Now, evaluate the volume integral over $V$: $$IV = iiintV 2(x + y + z) , dx , dy , dz = 2 int0^a int0^b int0^c (x + y + z) , dz , dy , dx$$ We split this into three symmetric integrals: $$iiintV x , dV = left(int0^a x , dxright) left(int0^b dyright) left(int0^c dzright) = left(frac{a^2}{2}right) (b) (c) = frac{a^2 b c}{2}$$ $$iiintV y , dV = left(int0^a dxright) left(int0^b y , dyright) left(int0^c dzright) = (a) left(frac{b^2}{2}right) (c) = frac{a b^2 c}{2}$$ $$iiintV z , dV = left(int0^a dxright) left(int0^b dyright) left(int0^c z , dzright) = (a) (b) left(frac{c^2}{2}right) = frac{a b c^2}{2}$$ Adding these three components: $$iiintV (x + y + z) , dV = frac{a b c}{2} (a + b

Maths Optional 2025 Paper I

33 Questions Available

Maths Optional 2025 Paper I

33 Questions Available

Question 1: 1. (d) Given that f(x + y) = f(x) f(y) for all real x, y, f(x) != 0 for any real x and f'(0) = 2. Show that for all real x, f'(x) = 2f(x). Hence find

Question 2: 1. (c) A rectangular sheet of metal of length 6 meters and width 2 meters is given. Four equal squares are removed from the four corners. The sides of

Question 3: 1. (b) Find the range, rank, kernel and nullity of the linear transformation T : R^4 - R^3 given by T(x, y, z, w) = (x - w, y + z, z - w). (10 Marks)

Question 4: 1. (a) Can the set {(0, 0, 0, 3), (1, 1, 0, 0), (0, 1, -1, 0)} be extended to form a basis of the vector space R^4 ? Justify your answer. (10 Marks)

Question 5: 1. (e) Find the equation of the cone whose vertex is the point (1, 1, 0) and whose guiding curve is y = 0, x^2 + z^2 = 4. (10 Marks)

Question 6: 2. (a) Let T : R^3 - R^2 be a linear transformation such that T(1, 1, -1) = (1, 0), T(4, 1, 1) = (0, 1) and T(1, -1, 2) = (1, 1). Find T. (15 Marks)

Question 7: 2. (b) Using Mean Value Theorem, prove that pi/6 + sqrt(3)/15 < sin^(-1)(3/5) < pi/6 + 1/8. (15 Marks)

Question 8: 2. (c) (i) Find the equation of the cylinder whose generators are parallel to the line x/1 = y/2 = z/3 and that passes through the curve x^2 + y^2 = 1

Question 9: 2. (c) (ii) Find the shortest distance between the straight lines (x - 3)/3 = (y - 8)/(-1) = (z - 3)/1 and (x + 3)/(-3) = (y + 7)/2 = (z - 6)/4. (10 M

Question 10: 3. (c) (ii) If u(x, y) = x f(y/x) + g(y/x), where f and g are arbitrary functions, then show that: I. x (du/dx) + y (du/dy) = x f(y/x), II. x^2 (d^2u/

Question 11: 3. (c) (i) Evaluate iintR y dx dy, where R is the region bounded by y = x and y = 4x - x^2. (10 Marks)

Question 12: 3. (b) Find the equations of the spheres which pass through the circle x^2 + y^2 + z^2 - 2x + 2y + 4z - 3 = 0, 2x + y + z = 4 and touch the plane 3x +

Question 13: 3. (a) Reduce the following matrix to echelon form: A = [[2, -2, 2, 1], [-3, 6, 0, -1], [1, -7, 10, 2]]. (15 Marks)

Question 14: 4. (a) Show that there is no tangent plane to the sphere x^2 + y^2 + z^2 - 4x + 2y - 4z + 4 = 0 that can be passed through the straight line (x + 6)/2

Question 15: 4. (c) (i) Find the eigenvalues and the corresponding eigenvectors of the matrix A = [[1, 2, 0], [2, 1, -6], [2, -2, 3]]. (12 Marks)

Question 16: 4. (b) If f(x, y) = xy (x^2 - y^2)/(x^2 + y^2) when (x, y) != (0, 0) and 0 when (x, y) = (0, 0), then find fxy(0, 0) and fyx(0, 0). (15 Marks)

Question 17: 4. (c) (ii) Let Pn denote the vector space of all polynomials of degree <= n over R. Verify that dim(P4 / P2) = dim P4 - dim P2. (8 Marks)

Question 18: 5. (b) Form the differential equation of all ellipses whose axes coincide with coordinate axes. (10 Marks)

Question 19: 5. (c) Prove that the time taken by the Earth to travel over half of its orbit, which is separated by the minor axis and is remote from the Sun, when

Question 20: 5. (d) Given that A and B are two points in the same horizontal line distant 2a apart. AO and BO are two equal heavy strings tied together at O and ca

Question 21: 5. (e) If u = x + y + z, v = x^2 + y^2 + z^2 and w = xy + yz + zx, then show that grad u, grad v and grad w are coplanar. (10 Marks)

Question 22: 5. (a) Solve: (1 - y^2 + y^4/x^2)(dy/dx)^2 - 2(y/x)(dy/dx) + y^2/x^2 = 0. (10 Marks)

Question 23: 6. (c) (i) Find the absolute value of the directional derivative of phi(x, y, z) = x^2 y^2 z^2 at the point (1, 1, -1) in the direction of the tangent

Question 24: 6. (c) (ii) If del . E = 0, del . H = 0, del x E = -dH/dt and del x H = dE/dt, then show that del^2 H = d^2H/dt^2 and del^2 E = d^2E/dt^2. (10 Marks)

Question 25: 6. (a) If F(s) and G(s) are Laplace transforms of f(t) and g(t) respectively, then prove that L[int0^t f(x) g(t - x) dx] = F(s) G(s). Using this resul

Question 26: 6. (b) One end of an elastic string, having natural length a, is fixed at some point O and a heavy particle is attached to the other end of the string

Question 27: 7. (b) Verify Green's theorem in the plane for ointC [(xy + y^2) dx + x^2 dy], where C is the boundary of the region bounded by the curves y = x and y

Question 28: 7. (c) (ii) Find the complete solution of x^3 (d^3y/dx^3) + 3x^2 (d^2y/dx^2) + x (dy/dx) + y = x log x. (10 Marks)

Question 29: 7. (a) A solid sphere rests inside a fixed rough and hemispherical bowl of twice its radius. If a large amount of weight, whatsoever, is attached to t

Question 30: 7. (c) (i) Find the general solution and singular solution of the differential equation (1 + dy/dx)^3 = (27/8a) (x + y) (1 - dy/dx)^3. (10 Marks)

Question 31: 8. (c) A particle is projected inside a fixed smooth cylinder with circular cross-section in a vertical plane from the lowest point with initial horiz

Question 32: 8. (a) Solve the differential equation (x + 2) (d^2y/dx^2) - (2x + 5) (dy/dx) + 2y = (1 + x) e^x by the method of variation of parameters. (15 Marks)

Question 33: 8. (b) Verify Gauss's divergence theorem for F = (x^2 - yz) i + (y^2 - zx) j + (z^2 - xy) k, taken over the rectangular parallelopiped 0 <= x <= a, 0