Q: A variable plane which is at a constant distance $3p$ from the origin $O$ cuts the axes in the points $A, B, C$ respectively. Show that the locus of the centroid of the tetrahedron $OABC$ is $$9left(frac{1}{x^2} + frac{1}{y^2} + frac{1}{z^2}right) = frac{16}{p^2}.$$

Geometric Setup and Intercept Form Let the variable plane cut the three mutually perpendicular coordinate axes (the $x$-axis, $y$-axis, and $z$-axis) at the points $A, B$, and $C$ respectively. Let the intercepts made by the plane on the axes be $a, b, c$. Thus, the coordinates of the intercept points are: $$A = (a, 0, 0), qquad B = (0, b, 0), qquad C = (0, 0, c)$$ The origin is $O = (0, 0, 0)$. The equation of the variable plane in intercept form is: $$frac{x}{a} + frac{y}{b} + frac{z}{c} = 1$$ or in standard linear form: $$frac{1}{a}x + frac{1}{b}y + frac{1}{c}z - 1 = 0$$ --- Step 1: Perpendicular Distance from the Origin The perpendicular distance $d$ from a point $(x0, y0, z0)$ to the plane $A x + B y + C z + D = 0$ is given by: $$d = frac{|A x0 + B y0 + C z0 + D|}{sqrt{A^2 + B^2 + C^2}}$$ Applying this formula from the origin $O(0, 0, 0)$ to the variable plane: $$d = frac{|0 + 0 + 0 - 1|}{sqrt{left(frac{1}{a}right)^2 + left(frac{1}{b}right)^2 + left(frac{1}{c}right)^2}} = frac{1}{sqrt{frac{1}{a^2} + frac{1}{b^2} + frac{1}{c^2}}}$$ We are given that this distance is constant and equal to $3p$: $$frac{1}{sqrt{frac{1}{a^2} + frac{1}{b^2} + frac{1}{c^2}}} = 3p$$ Squaring both sides and taking the reciprocal: $$frac{1}{a^2} + frac{1}{b^2} + frac{1}{c^2} = frac{1}{(3p)^2} = frac{1}{9p^2} tag{1}$$ --- Step 2: Coordinates of the Centroid of Tetrahedron $OABC$ The tetrahedron $OABC$ has four vertices: $$O(0, 0, 0), quad A(a, 0, 0), quad B(0, b, 0), quad C(0, 0, c)$$ Let $G(bar{x}, bar{y}, bar{z})$ be the centroid of the tetrahedron $OABC$. The coordinates of the centroid of a tetrahedron with vertices $(xi, yi, zi)$ for $i = 1, 2, 3, 4$ are given by the arithmetic mean of the coordinates of its vertices: $$bar{x} = frac{0 + a + 0 + 0}{4} = frac{a}{4} implies a = 4bar{x} tag{2}$$ $$bar{y} = frac{0 + 0 + b + 0}{4} = frac{b}{4} implies b = 4bar{y} tag{3}$$ $$bar{z} = frac{0 + 0 + 0 + c}{4} = frac{c}{4} implies c = 4bar{z} tag{4}$$ --- Step 3: Derivation of the Locus Substi

Q: If the matrix of a linear transformation $T: mathbb{R}^3 to mathbb{R}^3$ relative to the basis ${(1, 0, 0), (0, 1, 0), (0, 0, 1)}$ is $$begin{bmatrix} 1 & 1 & 2 -1 & 2 & 1 0 & 1 & 3 end{bmatrix},$$ then find the matrix of $T$ relative to the basis ${(1, 1, 1), (0, 1, 1), (0, 0, 1)}$.

Change of Basis for a Linear Operator Let $V = mathbb{R}^3$ be a 3-dimensional vector space over $mathbb{R}$. Let $B1 = {e1, e2, e3}$ be the standard basis of $mathbb{R}^3$: $$e1 = begin{pmatrix} 1 0 0 end{pmatrix}, quad e2 = begin{pmatrix} 0 1 0 end{pmatrix}, quad e3 = begin{pmatrix} 0 0 1 end{pmatrix}$$ We are given the matrix of the linear transformation $T$ relative to the basis $B1$: $$A = [T]{B1} = begin{bmatrix} 1 & 1 & 2 -1 & 2 & 1 0 & 1 & 3 end{bmatrix}$$ Let $B2 = {v1, v2, v3}$ be the new basis of $mathbb{R}^3$, where: $$v1 = begin{pmatrix} 1 1 1 end{pmatrix}, quad v2 = begin{pmatrix} 0 1 1 end{pmatrix}, quad v3 = begin{pmatrix} 0 0 1 end{pmatrix}$$ We are required to compute $[T]{B2}$, the matrix of $T$ relative to $B2$. By the Change of Basis Theorem for linear operators: $$[T]{B2} = P^{-1} [T]{B1} P$$ where $P$ is the transition (change-of-basis) matrix from $B2$ to $B1$, whose columns are the coordinates of the vectors of $B2$ relative to $B1$. --- Step 1: Construction of Transition Matrix $P$ Since $B1$ is the standard basis: $$[v1]{B1} = begin{pmatrix} 1 1 1 end{pmatrix}, quad [v2]{B1} = begin{pmatrix} 0 1 1 end{pmatrix}, quad [v3]{B1} = begin{pmatrix} 0 0 1 end{pmatrix}$$ Thus: $$P = begin{bmatrix} 1 & 0 & 0 1 & 1 & 0 1 & 1 & 1 end{bmatrix}$$ --- Step 2: Inversion of Matrix $P$ Notice that $P$ is a lower triangular matrix with diagonal entries all equal to $1$. Hence $det(P) = 1 cdot 1 cdot 1 = 1 eq 0$, so $P$ is invertible. To compute $P^{-1}$, we solve the matrix equation $P X = I$, or equivalently solve for $(x, y, z)$ in terms of $(u, v, w)$ from: $$begin{bmatrix} 1 & 0 & 0 1 & 1 & 0 1 & 1 & 1 end{bmatrix} begin{pmatrix} x y z end{pmatrix} = begin{pmatrix} u v w end{pmatrix}$$ By forward substitution: $$begin{aligned} x &= u x + y &= v implies y = v - u = -u + v x + y + z &= w implies z = w - (x + y) = w - v = -v + w end{aligned}$$ Writing these relations in matrix form: $$begin{pmatrix} x y z end{pmatrix} = begin{bmatrix} 1 & 0 & 0 -1 & 1 & 0 0

Q: Evaluate the triple integral which gives the volume of the solid enclosed between the two paraboloids $Z = 5(x^2 + y^2)$ and $Z = 6 - 7x^2 - y^2$.

Mathematical Formulation of the Enclosed Solid We are required to compute the volume $V$ of the solid enclosed between two paraboloidal surfaces: 1. The lower paraboloid $S1$: $$Z1(x, y) = 5(x^2 + y^2) = 5x^2 + 5y^2$$ This is an elliptic (circular) paraboloid opening upwards with vertex at the origin $(0, 0, 0)$. 2. The upper paraboloid $S2$: $$Z2(x, y) = 6 - 7x^2 - y^2$$ This is an elliptic paraboloid opening downwards with vertex at $(0, 0, 6)$. The volume of the enclosed solid $Omega$ is given by the triple integral: $$V = iiintOmega dV = iintR left( int{Z1(x, y)}^{Z2(x, y)} dz right) dx,dy$$ where $R$ is the projection of the curve of intersection of the two surfaces onto the $xy$-plane. --- Step 1: Intersection of the Two Paraboloids and Projection Region $R$ The two surfaces intersect where $Z1(x, y) = Z2(x, y)$: $$5x^2 + 5y^2 = 6 - 7x^2 - y^2$$ Bringing all quadratic terms to the left-hand side: $$(5 + 7)x^2 + (5 + 1)y^2 = 6$$ $$12x^2 + 6y^2 = 6$$ Dividing through by $6$: $$2x^2 + y^2 = 1$$ or in standard elliptical form: $$frac{x^2}{(1/sqrt{2})^2} + frac{y^2}{1^2} = 1$$ Thus, the projection $R$ of the solid onto the $xy$-plane is an elliptical disk bounded by the ellipse: $$R = left{ (x, y) in mathbb{R}^2 : 2x^2 + y^2 le 1 right}$$ The semi-major axis is $b = 1$ (along the $y$-axis) and the semi-minor axis is $a = frac{1}{sqrt{2}}$ (along the $x$-axis). Inside this region $R$, $2x^2 + y^2 le 1 iff 12x^2 + 6y^2 le 6 iff 5x^2 + 5y^2 le 6 - 7x^2 - y^2$, confirming that $Z2(x, y) ge Z1(x, y)$. --- Step 2: Inner Integral with Respect to $z$ Evaluating the inner integral along the $z$-direction: $$I(x, y) = int{5x^2 + 5y^2}^{6 - 7x^2 - y^2} dz = (6 - 7x^2 - y^2) - (5x^2 + 5y^2) = 6 - 12x^2 - 6y^2$$ Factoring out $6$: $$I(x, y) = 6(1 - 2x^2 - y^2)$$ Hence, the volume integral reduces to the double integral: $$V = iintR 6(1 - 2x^2 - y^2) dx,dy$$ --- Step 3: Transformation to Elliptic Polar Coordinates To evaluate this integral over the elliptical domain $R$, we appl

Q: Show that the equation $2x^2 + 3y^2 - 8x + 6y - 12z + 11 = 0$ represents an elliptic paraboloid. Also find its principal axis and principal planes.

Standard Classification of Quadric Surfaces A general second-degree equation in three variables represents a paraboloid if the second-degree terms form a degenerate quadric cone consisting of two intersecting or parallel planes (i.e. having one zero eigenvalue in its quadratic form) and the linear term in the third variable cannot be eliminated by translation. Specifically, the standard canonical form of an elliptic paraboloid is: $$frac{X^2}{a^2} + frac{Y^2}{b^2} = 2k Z quad (k 0)$$ where the coefficients of $X^2$ and $Y^2$ have the same sign. --- Step 1: Reduction to Canonical Form via Translation of Axes The given equation is: $$2x^2 + 3y^2 - 8x + 6y - 12z + 11 = 0$$ Group the terms involving $x$, $y$, and $z$: $$[2x^2 - 8x] + [3y^2 + 6y] - 12z + 11 = 0$$ Factor out the coefficients of the squared terms: $$2(x^2 - 4x) + 3(y^2 + 2y) - 12z + 11 = 0$$ Complete the squares inside the parentheses: - $x^2 - 4x = (x - 2)^2 - 4$ - $y^2 + 2y = (y + 1)^2 - 1$ Substitute these back: $$2left[ (x - 2)^2 - 4 right] + 3left[ (y + 1)^2 - 1 right] - 12z + 11 = 0$$ $$2(x - 2)^2 - 8 + 3(y + 1)^2 - 3 - 12z + 11 = 0$$ $$2(x - 2)^2 + 3(y + 1)^2 - 11 - 12z + 11 = 0$$ The constant terms cancel out completely: $$2(x - 2)^2 + 3(y + 1)^2 - 12z = 0$$ $$2(x - 2)^2 + 3(y + 1)^2 = 12z$$ Dividing both sides by $12$: $$frac{(x - 2)^2}{6} + frac{(y + 1)^2}{4} = z$$ Now, define the translated coordinates $(X, Y, Z)$ by: $$X = x - 2, qquad Y = y + 1, qquad Z = z$$ Under this coordinate translation, the equation becomes: $$frac{X^2}{6} + frac{Y^2}{4} = Z quad text{or} quad frac{X^2}{(sqrt{6})^2} + frac{Y^2}{2^2} = Z$$ Since both denominators are positive and the signs of $X^2$ and $Y^2$ are identical, every horizontal plane $Z = c 0$ intersects the surface in an ellipse: $$frac{X^2}{6c} + frac{Y^2}{4c} = 1$$ and vertical planes $X = text{const}$ or $Y = text{const}$ intersect the surface in parabolas. Hence, the given equation represents an elliptic paraboloid. --- Step 2: Determination of Vertex, P

Q: The plane $frac{x}{a} + frac{y}{b} + frac{z}{c} = 1$ meets the coordinate axes in $A, B, C$ respectively. Prove that the equation of the cone generated by the lines drawn from the origin $O$ to meet the circle $ABC$ is $$yzleft(frac{b}{c} + frac{c}{b}right) + zxleft(frac{c}{a} + frac{a}{c}right) + x

Problem Analysis & Guiding Curve The plane meets the coordinate axes at the points: $$A = (a, 0, 0), qquad B = (0, b, 0), qquad C = (0, 0, c)$$ The origin is $O = (0, 0, 0)$. The given plane equation is: $$pi: frac{x}{a} + frac{y}{b} + frac{z}{c} = 1 tag{1}$$ The circle $ABC$ is a curve in three dimensions defined as the intersection of: 1. The plane $pi$ passing through $A, B, C$. 2. The sphere passing through the four points $O(0, 0, 0), A(a, 0, 0), B(0, b, 0), C(0, 0, c)$. The problem asks for the equation of the cone whose vertex is the origin $O(0, 0, 0)$ and whose guiding curve is the circle $ABC$. --- Step 1: Equation of the Sphere Passing Through $O, A, B, C$ The general equation of a sphere in $mathbb{R}^3$ is: $$x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0$$ 1. Since the sphere passes through the origin $O(0, 0, 0)$: $$0^2 + 0^2 + 0^2 + 0 + 0 + 0 + d = 0 implies d = 0$$ 2. Since it passes through $A(a, 0, 0)$: $$a^2 + 2ua = 0 implies 2u = -a$$ 3. Since it passes through $B(0, b, 0)$: $$b^2 + 2vb = 0 implies 2v = -b$$ 4. Since it passes through $C(0, 0, c)$: $$c^2 + 2wc = 0 implies 2w = -c$$ Thus, the equation of the sphere circumscribing tetrahedron $OABC$ is: $$S: x^2 + y^2 + z^2 - ax - by - cz = 0 tag{2}$$ The circle $ABC$ is the intersection of the sphere $S = 0$ and the plane $pi = 0$. --- Step 2: Generating the Cone with Vertex at the Origin A cone with vertex at the origin $O(0, 0, 0)$ is represented by a homogeneous equation in $x, y, z$. To find the homogeneous equation of the second degree representing the cone passing through the intersection of the second-degree surface $S = 0$ and the plane $pi = 0$, we homogenize the second-degree equation (2) using the linear equation of the plane (1): $$x^2 + y^2 + z^2 - (ax + by + cz) cdot 1 = 0$$ From equation (1), we have: $$1 = frac{x}{a} + frac{y}{b} + frac{z}{c}$$ Substituting this expression for $1$: $$x^2 + y^2 + z^2 - (ax + by + cz) left( frac{x}{a} + frac{y}{b} + frac{z}{c} right) = 0 tag{3}$$ --- Ste

Q: Let $A = begin{bmatrix} 1 & 0 & 0 1 & 0 & 1 0 & 1 & 0 end{bmatrix}$. (i) Verify the Cayley-Hamilton theorem for the matrix $A$. (ii) Show that $A^n = A^{n-2} + A^2 - I$ for $n ge 3$, where $I$ is the identity matrix of order 3. Hence, find $A^{40}$.

Statement of Cayley-Hamilton Theorem The Cayley-Hamilton Theorem asserts that every square matrix over a commutative ring (such as the field of real numbers $mathbb{R}$) satisfies its own characteristic equation. That is, if: $$chiA(lambda) = det(A - lambda I) = (-1)^n (lambda^n + c{n-1}lambda^{n-1} + dots + c1lambda + c0) = 0$$ is the characteristic polynomial of $A in mathcal{M}{ntimes n}(mathbb{R})$, then substituting the matrix $A$ in place of the scalar variable $lambda$ yields the zero matrix: $$chiA(A) = A^n + c{n-1} A^{n-1} + dots + c1 A + c0 I = mathbf{0}$$ We are given the $3 times 3$ real matrix: $$A = begin{bmatrix} 1 & 0 & 0 1 & 0 & 1 0 & 1 & 0 end{bmatrix}$$ --- Part (i): Characteristic Polynomial and Verification of Cayley-Hamilton Theorem 1. Characteristic Polynomial of $A$: $$chiA(lambda) = det(A - lambda I) = begin{vmatrix} 1 - lambda & 0 & 0 1 & -lambda & 1 0 & 1 & -lambda end{vmatrix}$$ Expanding along the first row (which contains two zeros): $$chiA(lambda) = (1 - lambda) begin{vmatrix} -lambda & 1 1 & -lambda end{vmatrix} - 0 + 0$$ $$chiA(lambda) = (1 - lambda) left[ (-lambda)(-lambda) - (1)(1) right] = (1 - lambda)(lambda^2 - 1)$$ Expanding the product: $$chiA(lambda) = lambda^2 - 1 - lambda^3 + lambda = -(lambda^3 - lambda^2 - lambda + 1)$$ Setting the characteristic equation $chiA(lambda) = 0$: $$lambda^3 - lambda^2 - lambda + 1 = 0$$ According to the Cayley-Hamilton theorem, $A$ must satisfy: $$A^3 - A^2 - A + I = mathbf{0} tag{1}$$ 2. Computation of Matrix Powers $A^2$ and $A^3$: - Computing $A^2$: $$A^2 = A cdot A = begin{bmatrix} 1 & 0 & 0 1 & 0 & 1 0 & 1 & 0 end{bmatrix} begin{bmatrix} 1 & 0 & 0 1 & 0 & 1 0 & 1 & 0 end{bmatrix}$$ $$text{Row 1:} quad [1(1)+0(1)+0(0), ; 1(0)+0(0)+0(1), ; 1(0)+0(1)+0(0)] = [1, 0, 0]$$ $$text{Row 2:} quad [1(1)+0(1)+1(0), ; 1(0)+0(0)+1(1), ; 1(0)+0(1)+1(0)] = [1, 1, 0]$$ $$text{Row 3:} quad [0(1)+1(1)+0(0), ; 0(0)+1(0)+0(1), ; 0(0)+1(1)+0(0)] = [1, 0, 1]$$ $$A^2 = begin{bmatrix} 1 & 0 & 0 1 & 1 & 0 1 & 0 &

Q: Justify whether $(0, 0)$ is an extreme point for the function $f(x, y) = 2x^4 - 3x^2 y + y^2$.

Theoretical Concept of Extreme Points Let $f: D subset mathbb{R}^2 to mathbb{R}$ be a function of two variables defined on an open neighborhood of a point $(x0, y0)$. - $(x0, y0)$ is said to be a local minimum of $f$ if there exists a disk $Bdelta(x0, y0)$ such that: $$f(x, y) ge f(x0, y0) quad forall (x, y) in Bdelta(x0, y0)$$ - $(x0, y0)$ is said to be a local maximum of $f$ if there exists a disk $Bdelta(x0, y0)$ such that: $$f(x, y) le f(x0, y0) quad forall (x, y) in Bdelta(x0, y0)$$ - $(x0, y0)$ is called an extreme point (or local extremum) if it is either a local minimum or a local maximum. - If in every open disk $Bdelta(x0, y0)$, the difference $f(x, y) - f(x0, y0)$ takes both strictly positive and strictly negative values, then $(x0, y0)$ is a saddle point and is not an extreme point. We are given the polynomial function: $$f(x, y) = 2x^4 - 3x^2 y + y^2$$ --- Step 1: Stationary (Critical) Point Analysis The value of the function at the origin is: $$f(0, 0) = 0$$ Computing the first partial derivatives: $$fx(x, y) = frac{partial f}{partial x} = 8x^3 - 6xy = 2x(4x^2 - 3y)$$ $$fy(x, y) = frac{partial f}{partial y} = -3x^2 + 2y$$ Evaluating at the origin $(0, 0)$: $$fx(0, 0) = 0, qquad fy(0, 0) = 0$$ Since both first-order partial derivatives vanish at $(0, 0)$, the origin is indeed a stationary (critical) point of $f$. --- Step 2: Second Derivative Test (Hessian Matrix) To investigate the nature of the critical point, we compute the second partial derivatives: $$f{xx}(x, y) = 24x^2 - 6y implies r = f{xx}(0, 0) = 0$$ $$f{xy}(x, y) = -6x implies s = f{xy}(0, 0) = 0$$ $$f{yy}(x, y) = 2 implies t = f{yy}(0, 0) = 2$$ The discriminant (Hessian determinant) at $(0, 0)$ is: $$Delta = r t - s^2 = (0)(2) - (0)^2 = 0$$ Since $Delta = 0$, the classical second derivative test fails (is inconclusive). We must therefore examine the behavior of $f(x, y) - f(0, 0)$ directly in neighborhoods of $(0, 0)$. --- Step 3: Behavior Along Straight Lines Through the Origin Let us test

Question 1

Let $V1 = (2, -1, 3, 2)$, $V2 = (-1, 1, 1, -3)$ and $V3 = (1, 1, 9, -5)$ be three vectors of the space $mathbb{R}^4$. Does $(3, -1, 0, -1) in text{span}{V1, V2, V3}$? Justify your answer.

Accepted Answer

Mathematical Context and Definition of Span Let $V$ be a vector space over the field $mathbb{R}$. The span of a set of vectors $S = {V1, V2, dots, Vk} subset V$, denoted by $text{span}{S}$ or $mathcal{L}(S)$, is the set of all linear combinations of the vectors in $S$: $$text{span}{V1, V2, dots, Vk} = left{ sum{i=1}^k ci Vi : ci in mathbb{R} right}$$ Here, we are given three vectors in $mathbb{R}^4$: $$V1 = (2, -1, 3, 2)$$ $$V2 = (-1, 1, 1, -3)$$ $$V3 = (1, 1, 9, -5)$$ and a target vector $W = (3, -1, 0, -1) in mathbb{R}^4$. The question asks whether $W in text{span}{V1, V2, V3}$. By definition, this holds if and only if there exist real scalars $c1, c2, c3 in mathbb{R}$ such that: $$c1 V1 + c2 V2 + c3 V3 = W$$ --- Step 1: Setting up the System of Linear Equations Substituting the vector coordinates into the vector equation: $$c1 begin{pmatrix} 2 -1 3 2 end{pmatrix} + c2 begin{pmatrix} -1 1 1 -3 end{pmatrix} + c3 begin{pmatrix} 1 1 9 -5 end{pmatrix} = begin{pmatrix} 3 -1 0 -1 end{pmatrix}$$ Equating corresponding components yields a non-homogeneous system of four linear equations in three unknowns ($c1, c2, c3$): $$begin{aligned} (1) quad & 2c1 - c2 + c3 = 3 (2) quad & -c1 + c2 + c3 = -1 (3) quad & 3c1 + c2 + 9c3 = 0 (4) quad & 2c1 - 3c2 - 5c3 = -1 end{aligned}$$ --- Step 2: Consistency Analysis via Direct Elimination Let us analyze equations (1) and (2): Adding equation (1) and equation (2): $$(2c1 - c2 + c3) + (-c1 + c2 + c3) = 3 + (-1)$$ $$c1 + 2c3 = 2 implies c1 = 2 - 2c3 tag{A}$$ Next, multiplying equation (2) by 2 and adding to equation (1): $$(2c1 - c2 + c3) + 2(-c1 + c2 + c3) = 3 + 2(-1)$$ $$c2 + 3c3 = 1 implies c2 = 1 - 3c3 tag{B}$$ Now, substitute expressions (A) and (B) into equation (3): $$3c1 + c2 + 9c3 = 3(2 - 2c3) + (1 - 3c3) + 9c3$$ $$= 6 - 6c3 + 1 - 3c3 + 9c3$$ $$= 7 + (-6 - 3 + 9)c3$$ $$= 7 + 0 cdot c3 = 7$$ However, according to equation (3), this expression must equal $0$: $$7 = 0$$ which is a blatant mathematical impossibility (contradiction). T

Question 2

Find the rank and nullity of the linear transformation: $T: mathbb{R}^3 to mathbb{R}^3$ given by $T(x, y, z) = (x + z, x + y + 2z, 2x + y + 3z)$.

Accepted Answer

Theoretical Framework Let $T: V to W$ be a linear transformation between finite-dimensional vector spaces over a field $mathbb{F}$. 1. The kernel (or null space) of $T$ is the subspace of $V$ defined by: $$ker(T) = {v in V : T(v) = 0W}$$ The dimension of $ker(T)$ is called the nullity of $T$, denoted by $text{nullity}(T)$. 2. The image (or range) of $T$ is the subspace of $W$ defined by: $$text{Im}(T) = {T(v) : v in V}$$ The dimension of $text{Im}(T)$ is called the rank of $T$, denoted by $text{rank}(T)$. 3. Rank-Nullity Theorem (Sylvester's Law of Nullity): $$dim(V) = text{rank}(T) + text{nullity}(T)$$ Here, $V = mathbb{R}^3$, so $dim(V) = 3$. --- Step 1: Matrix Representation of $T$ Let $B = {e1, e2, e3}$ be the standard ordered basis of $mathbb{R}^3$, where: $$e1 = (1, 0, 0), quad e2 = (0, 1, 0), quad e3 = (0, 0, 1)$$ Applying the given transformation $T(x, y, z) = (x + z, x + y + 2z, 2x + y + 3z)$ to the standard basis vectors: $$T(e1) = T(1, 0, 0) = (1 + 0, 1 + 0 + 0, 2(1) + 0 + 0) = (1, 1, 2)$$ $$T(e2) = T(0, 1, 0) = (0 + 0, 0 + 1 + 0, 0 + 1 + 0) = (0, 1, 1)$$ $$T(e3) = T(0, 0, 1) = (0 + 1, 0 + 0 + 2, 0 + 0 + 3) = (1, 2, 3)$$ Writing these image vectors as coordinate columns gives the matrix representation of $T$ relative to the standard basis: $$A = [T]B = begin{pmatrix} 1 & 0 & 1 1 & 1 & 2 2 & 1 & 3 end{pmatrix}$$ --- Step 2: Determining the Rank of $T$ The rank of the transformation $T$ is equal to the row rank (or column rank) of its matrix representation $A$. We compute the row-echelon form of $A$ via elementary row operations: $$A = begin{pmatrix} 1 & 0 & 1 1 & 1 & 2 2 & 1 & 3 end{pmatrix}$$ 1. Perform row operations to introduce zeros in the first column: - $R2 to R2 - R1$: $$R2 = (1 - 1, 1 - 0, 2 - 1) = (0, 1, 1)$$ - $R3 to R3 - 2R1$: $$R3 = (2 - 2, 1 - 0, 3 - 2) = (0, 1, 1)$$ The matrix becomes: $$begin{pmatrix} 1 & 0 & 1 0 & 1 & 1 0 & 1 & 1 end{pmatrix}$$ 2. Eliminate the second row entry in row 3: - $R3 to R3 - R2$: $$R3 = (0 - 0, 1 - 1, 1 - 1) = (0

Question 3

Find the values of $p$ and $q$ for which $lim{x to 0} frac{x(1 + pcos x) - qsin x}{x^3}$ exists and equals 1.

Accepted Answer

Problem Statement & Approach We are given the limit: $$L = lim{x to 0} frac{x(1 + pcos x) - qsin x}{x^3} = 1$$ We are required to determine the real parameters $p$ and $q$. We solve this problem using two rigorous methods: 1. Method I: Power Series (Taylor / Maclaurin expansion). 2. Method II: L'Hôpital's Rule. --- Method I: Evaluation via Maclaurin Series Expansion Recall the standard Maclaurin series expansions of $cos x$ and $sin x$ about $x = 0$: $$cos x = 1 - frac{x^2}{2!} + frac{x^4}{4!} - O(x^6) = 1 - frac{x^2}{2} + frac{x^4}{24} - O(x^6)$$ $$sin x = x - frac{x^3}{3!} + frac{x^5}{5!} - O(x^7) = x - frac{x^3}{6} + frac{x^5}{120} - O(x^7)$$ Let $N(x) = x(1 + pcos x) - qsin x$ denote the numerator. Substituting the expansions: $$N(x) = xleft[ 1 + pleft( 1 - frac{x^2}{2} + frac{x^4}{24} - O(x^6) right) right] - qleft[ x - frac{x^3}{6} + frac{x^5}{120} - O(x^7) right]$$ Distributing terms: $$N(x) = xleft[ (1 + p) - frac{p}{2}x^2 + frac{p}{24}x^4 - O(x^6) right] - left[ q x - frac{q}{6}x^3 + frac{q}{120}x^5 - O(x^7) right]$$ $$N(x) = (1 + p)x - frac{p}{2}x^3 + frac{p}{24}x^5 - q x + frac{q}{6}x^3 - frac{q}{120}x^5 + O(x^7)$$ Collecting like powers of $x$: $$N(x) = (1 + p - q) x + left( -frac{p}{2} + frac{q}{6} right) x^3 + left( frac{p}{24} - frac{q}{120} right) x^5 + O(x^7)$$ Now, consider the limit: $$L = lim{x to 0} frac{N(x)}{x^3} = lim{x to 0} left[ frac{(1 + p - q)}{x^2} + left( -frac{p}{2} + frac{q}{6} right) + left( frac{p}{24} - frac{q}{120} right) x^2 + O(x^4) right]$$ For the limit $L$ to exist as a finite real number: 1. The coefficient of the singular term $frac{1}{x^2}$ must vanish: $$1 + p - q = 0 implies q = 1 + p tag{1}$$ 2. Once $1 + p - q = 0$, the limit evaluates directly to the constant term: $$L = -frac{p}{2} + frac{q}{6}$$ We are given that $L = 1$: $$-frac{p}{2} + frac{q}{6} = 1 tag{2}$$ Multiplying equation (2) by $6$: $$-3p + q = 6 implies q - 3p = 6 tag{3}$$ Substituting equation (1), $q = 1 + p$, into equation (3): $$(1 + p) - 3p = 6$$ $

Question 4

Examine the convergence of the integral $int0^1 frac{log x}{1 + x} dx$.

Accepted Answer

Classification of the Integral We are required to examine the convergence of the definite integral: $$I = int0^1 frac{log x}{1 + x} dx$$ Let $f(x) = frac{log x}{1 + x}$ for $x in (0, 1]$. 1. At the upper limit $x = 1$: $$lim{x to 1^-} f(x) = frac{log 1}{1 + 1} = frac{0}{2} = 0$$ Hence, $x = 1$ is a regular (ordinary) point. 2. At the lower limit $x = 0$: $$lim{x to 0^+} f(x) = lim{x to 0^+} frac{log x}{1 + x} = frac{-infty}{1} = -infty$$ Thus, $f(x)$ has an infinite discontinuity at the lower limit $x = 0$. Therefore, the integral is an improper integral of the second kind with a single singularity at $x = 0$. --- Step 1: Sign of the Integrand and Convergence Definition For all $x in (0, 1)$: $$log x 0 implies f(x) 0 quad forall x in (0, 1)$$ By definition, the improper integral $int0^1 f(x) dx$ converges if and only if $int0^1 g(x) dx$ converges: $$int0^1 g(x) dx = lim{epsilon to 0^+} intepsilon^1 frac{-log x}{1 + x} dx$$ --- Step 2: Limit Comparison Test We compare $g(x) = frac{-log x}{1 + x}$ with the standard test function: $$phi(x) = frac{1}{x^p}$$ where $p$ is chosen such that $0 < p < 1$. Let us choose $p = frac{1}{2}$. The auxiliary integral: $$int0^1 phi(x) dx = int0^1 frac{1}{x^{1/2}} dx = left[ frac{x^{1/2}}{1/2} right]0^1 = 2 < infty$$ is known to converge because $p = frac{1}{2} < 1$. Now, we evaluate the limit of the ratio: $$L = lim{x to 0^+} frac{g(x)}{phi(x)} = lim{x to 0^+} frac{frac{-log x}{1 + x}}{frac{1}{x^{1/2}}} = lim{x to 0^+} frac{-x^{1/2}log x}{1 + x}$$ Using the algebra of limits: $$L = left( lim{x to 0^+} frac{1}{1 + x} right) cdot left( lim{x to 0^+} (-x^{1/2}log x) right) = 1 cdot lim{x to 0^+} frac{-log x}{x^{-1/2}}$$ Applying L'Hôpital's Rule to the $frac{infty}{infty}$ form: $$lim{x to 0^+} frac{-frac{1}{x}}{-frac{1}{2} x^{-3/2}} = lim{x to 0^+} frac{2 x^{3/2}}{x} = lim{x to 0^+} 2x^{1/2} = 0$$ Since $L = 0$, by the Comparison Test (Limit Form): If $lim{x to 0^+} frac{g(x)}{phi(x)} = 0$ and $int0^1 phi(x) dx$ converges, then $int0^1

Question 5

A variable plane which is at a constant distance $3p$ from the origin $O$ cuts the axes in the points $A, B, C$ respectively. Show that the locus of the centroid of the tetrahedron $OABC$ is $$9left(frac{1}{x^2} + frac{1}{y^2} + frac{1}{z^2}right) = frac{16}{p^2}.$$

Accepted Answer

Geometric Setup and Intercept Form Let the variable plane cut the three mutually perpendicular coordinate axes (the $x$-axis, $y$-axis, and $z$-axis) at the points $A, B$, and $C$ respectively. Let the intercepts made by the plane on the axes be $a, b, c$. Thus, the coordinates of the intercept points are: $$A = (a, 0, 0), qquad B = (0, b, 0), qquad C = (0, 0, c)$$ The origin is $O = (0, 0, 0)$. The equation of the variable plane in intercept form is: $$frac{x}{a} + frac{y}{b} + frac{z}{c} = 1$$ or in standard linear form: $$frac{1}{a}x + frac{1}{b}y + frac{1}{c}z - 1 = 0$$ --- Step 1: Perpendicular Distance from the Origin The perpendicular distance $d$ from a point $(x0, y0, z0)$ to the plane $A x + B y + C z + D = 0$ is given by: $$d = frac{|A x0 + B y0 + C z0 + D|}{sqrt{A^2 + B^2 + C^2}}$$ Applying this formula from the origin $O(0, 0, 0)$ to the variable plane: $$d = frac{|0 + 0 + 0 - 1|}{sqrt{left(frac{1}{a}right)^2 + left(frac{1}{b}right)^2 + left(frac{1}{c}right)^2}} = frac{1}{sqrt{frac{1}{a^2} + frac{1}{b^2} + frac{1}{c^2}}}$$ We are given that this distance is constant and equal to $3p$: $$frac{1}{sqrt{frac{1}{a^2} + frac{1}{b^2} + frac{1}{c^2}}} = 3p$$ Squaring both sides and taking the reciprocal: $$frac{1}{a^2} + frac{1}{b^2} + frac{1}{c^2} = frac{1}{(3p)^2} = frac{1}{9p^2} tag{1}$$ --- Step 2: Coordinates of the Centroid of Tetrahedron $OABC$ The tetrahedron $OABC$ has four vertices: $$O(0, 0, 0), quad A(a, 0, 0), quad B(0, b, 0), quad C(0, 0, c)$$ Let $G(bar{x}, bar{y}, bar{z})$ be the centroid of the tetrahedron $OABC$. The coordinates of the centroid of a tetrahedron with vertices $(xi, yi, zi)$ for $i = 1, 2, 3, 4$ are given by the arithmetic mean of the coordinates of its vertices: $$bar{x} = frac{0 + a + 0 + 0}{4} = frac{a}{4} implies a = 4bar{x} tag{2}$$ $$bar{y} = frac{0 + 0 + b + 0}{4} = frac{b}{4} implies b = 4bar{y} tag{3}$$ $$bar{z} = frac{0 + 0 + 0 + c}{4} = frac{c}{4} implies c = 4bar{z} tag{4}$$ --- Step 3: Derivation of the Locus Substi

Question 6

If the matrix of a linear transformation $T: mathbb{R}^3 to mathbb{R}^3$ relative to the basis ${(1, 0, 0), (0, 1, 0), (0, 0, 1)}$ is $$begin{bmatrix} 1 & 1 & 2 -1 & 2 & 1 0 & 1 & 3 end{bmatrix},$$ then find the matrix of $T$ relative to the basis ${(1, 1, 1), (0, 1, 1), (0, 0, 1)}$.

Accepted Answer

Change of Basis for a Linear Operator Let $V = mathbb{R}^3$ be a 3-dimensional vector space over $mathbb{R}$. Let $B1 = {e1, e2, e3}$ be the standard basis of $mathbb{R}^3$: $$e1 = begin{pmatrix} 1 0 0 end{pmatrix}, quad e2 = begin{pmatrix} 0 1 0 end{pmatrix}, quad e3 = begin{pmatrix} 0 0 1 end{pmatrix}$$ We are given the matrix of the linear transformation $T$ relative to the basis $B1$: $$A = [T]{B1} = begin{bmatrix} 1 & 1 & 2 -1 & 2 & 1 0 & 1 & 3 end{bmatrix}$$ Let $B2 = {v1, v2, v3}$ be the new basis of $mathbb{R}^3$, where: $$v1 = begin{pmatrix} 1 1 1 end{pmatrix}, quad v2 = begin{pmatrix} 0 1 1 end{pmatrix}, quad v3 = begin{pmatrix} 0 0 1 end{pmatrix}$$ We are required to compute $[T]{B2}$, the matrix of $T$ relative to $B2$. By the Change of Basis Theorem for linear operators: $$[T]{B2} = P^{-1} [T]{B1} P$$ where $P$ is the transition (change-of-basis) matrix from $B2$ to $B1$, whose columns are the coordinates of the vectors of $B2$ relative to $B1$. --- Step 1: Construction of Transition Matrix $P$ Since $B1$ is the standard basis: $$[v1]{B1} = begin{pmatrix} 1 1 1 end{pmatrix}, quad [v2]{B1} = begin{pmatrix} 0 1 1 end{pmatrix}, quad [v3]{B1} = begin{pmatrix} 0 0 1 end{pmatrix}$$ Thus: $$P = begin{bmatrix} 1 & 0 & 0 1 & 1 & 0 1 & 1 & 1 end{bmatrix}$$ --- Step 2: Inversion of Matrix $P$ Notice that $P$ is a lower triangular matrix with diagonal entries all equal to $1$. Hence $det(P) = 1 cdot 1 cdot 1 = 1 eq 0$, so $P$ is invertible. To compute $P^{-1}$, we solve the matrix equation $P X = I$, or equivalently solve for $(x, y, z)$ in terms of $(u, v, w)$ from: $$begin{bmatrix} 1 & 0 & 0 1 & 1 & 0 1 & 1 & 1 end{bmatrix} begin{pmatrix} x y z end{pmatrix} = begin{pmatrix} u v w end{pmatrix}$$ By forward substitution: $$begin{aligned} x &= u x + y &= v implies y = v - u = -u + v x + y + z &= w implies z = w - (x + y) = w - v = -v + w end{aligned}$$ Writing these relations in matrix form: $$begin{pmatrix} x y z end{pmatrix} = begin{bmatrix} 1 & 0 & 0 -1 & 1 & 0 0

Question 7

Evaluate the triple integral which gives the volume of the solid enclosed between the two paraboloids $Z = 5(x^2 + y^2)$ and $Z = 6 - 7x^2 - y^2$.

Accepted Answer

Mathematical Formulation of the Enclosed Solid We are required to compute the volume $V$ of the solid enclosed between two paraboloidal surfaces: 1. The lower paraboloid $S1$: $$Z1(x, y) = 5(x^2 + y^2) = 5x^2 + 5y^2$$ This is an elliptic (circular) paraboloid opening upwards with vertex at the origin $(0, 0, 0)$. 2. The upper paraboloid $S2$: $$Z2(x, y) = 6 - 7x^2 - y^2$$ This is an elliptic paraboloid opening downwards with vertex at $(0, 0, 6)$. The volume of the enclosed solid $Omega$ is given by the triple integral: $$V = iiintOmega dV = iintR left( int{Z1(x, y)}^{Z2(x, y)} dz right) dx,dy$$ where $R$ is the projection of the curve of intersection of the two surfaces onto the $xy$-plane. --- Step 1: Intersection of the Two Paraboloids and Projection Region $R$ The two surfaces intersect where $Z1(x, y) = Z2(x, y)$: $$5x^2 + 5y^2 = 6 - 7x^2 - y^2$$ Bringing all quadratic terms to the left-hand side: $$(5 + 7)x^2 + (5 + 1)y^2 = 6$$ $$12x^2 + 6y^2 = 6$$ Dividing through by $6$: $$2x^2 + y^2 = 1$$ or in standard elliptical form: $$frac{x^2}{(1/sqrt{2})^2} + frac{y^2}{1^2} = 1$$ Thus, the projection $R$ of the solid onto the $xy$-plane is an elliptical disk bounded by the ellipse: $$R = left{ (x, y) in mathbb{R}^2 : 2x^2 + y^2 le 1 right}$$ The semi-major axis is $b = 1$ (along the $y$-axis) and the semi-minor axis is $a = frac{1}{sqrt{2}}$ (along the $x$-axis). Inside this region $R$, $2x^2 + y^2 le 1 iff 12x^2 + 6y^2 le 6 iff 5x^2 + 5y^2 le 6 - 7x^2 - y^2$, confirming that $Z2(x, y) ge Z1(x, y)$. --- Step 2: Inner Integral with Respect to $z$ Evaluating the inner integral along the $z$-direction: $$I(x, y) = int{5x^2 + 5y^2}^{6 - 7x^2 - y^2} dz = (6 - 7x^2 - y^2) - (5x^2 + 5y^2) = 6 - 12x^2 - 6y^2$$ Factoring out $6$: $$I(x, y) = 6(1 - 2x^2 - y^2)$$ Hence, the volume integral reduces to the double integral: $$V = iintR 6(1 - 2x^2 - y^2) dx,dy$$ --- Step 3: Transformation to Elliptic Polar Coordinates To evaluate this integral over the elliptical domain $R$, we appl

Question 8

Show that the equation $2x^2 + 3y^2 - 8x + 6y - 12z + 11 = 0$ represents an elliptic paraboloid. Also find its principal axis and principal planes.

Accepted Answer

Standard Classification of Quadric Surfaces A general second-degree equation in three variables represents a paraboloid if the second-degree terms form a degenerate quadric cone consisting of two intersecting or parallel planes (i.e. having one zero eigenvalue in its quadratic form) and the linear term in the third variable cannot be eliminated by translation. Specifically, the standard canonical form of an elliptic paraboloid is: $$frac{X^2}{a^2} + frac{Y^2}{b^2} = 2k Z quad (k 0)$$ where the coefficients of $X^2$ and $Y^2$ have the same sign. --- Step 1: Reduction to Canonical Form via Translation of Axes The given equation is: $$2x^2 + 3y^2 - 8x + 6y - 12z + 11 = 0$$ Group the terms involving $x$, $y$, and $z$: $$[2x^2 - 8x] + [3y^2 + 6y] - 12z + 11 = 0$$ Factor out the coefficients of the squared terms: $$2(x^2 - 4x) + 3(y^2 + 2y) - 12z + 11 = 0$$ Complete the squares inside the parentheses: - $x^2 - 4x = (x - 2)^2 - 4$ - $y^2 + 2y = (y + 1)^2 - 1$ Substitute these back: $$2left[ (x - 2)^2 - 4 right] + 3left[ (y + 1)^2 - 1 right] - 12z + 11 = 0$$ $$2(x - 2)^2 - 8 + 3(y + 1)^2 - 3 - 12z + 11 = 0$$ $$2(x - 2)^2 + 3(y + 1)^2 - 11 - 12z + 11 = 0$$ The constant terms cancel out completely: $$2(x - 2)^2 + 3(y + 1)^2 - 12z = 0$$ $$2(x - 2)^2 + 3(y + 1)^2 = 12z$$ Dividing both sides by $12$: $$frac{(x - 2)^2}{6} + frac{(y + 1)^2}{4} = z$$ Now, define the translated coordinates $(X, Y, Z)$ by: $$X = x - 2, qquad Y = y + 1, qquad Z = z$$ Under this coordinate translation, the equation becomes: $$frac{X^2}{6} + frac{Y^2}{4} = Z quad text{or} quad frac{X^2}{(sqrt{6})^2} + frac{Y^2}{2^2} = Z$$ Since both denominators are positive and the signs of $X^2$ and $Y^2$ are identical, every horizontal plane $Z = c 0$ intersects the surface in an ellipse: $$frac{X^2}{6c} + frac{Y^2}{4c} = 1$$ and vertical planes $X = text{const}$ or $Y = text{const}$ intersect the surface in parabolas. Hence, the given equation represents an elliptic paraboloid. --- Step 2: Determination of Vertex, P

Question 9

The plane $frac{x}{a} + frac{y}{b} + frac{z}{c} = 1$ meets the coordinate axes in $A, B, C$ respectively. Prove that the equation of the cone generated by the lines drawn from the origin $O$ to meet the circle $ABC$ is $$yzleft(frac{b}{c} + frac{c}{b}right) + zxleft(frac{c}{a} + frac{a}{c}right) + x

Accepted Answer

Problem Analysis & Guiding Curve The plane meets the coordinate axes at the points: $$A = (a, 0, 0), qquad B = (0, b, 0), qquad C = (0, 0, c)$$ The origin is $O = (0, 0, 0)$. The given plane equation is: $$pi: frac{x}{a} + frac{y}{b} + frac{z}{c} = 1 tag{1}$$ The circle $ABC$ is a curve in three dimensions defined as the intersection of: 1. The plane $pi$ passing through $A, B, C$. 2. The sphere passing through the four points $O(0, 0, 0), A(a, 0, 0), B(0, b, 0), C(0, 0, c)$. The problem asks for the equation of the cone whose vertex is the origin $O(0, 0, 0)$ and whose guiding curve is the circle $ABC$. --- Step 1: Equation of the Sphere Passing Through $O, A, B, C$ The general equation of a sphere in $mathbb{R}^3$ is: $$x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0$$ 1. Since the sphere passes through the origin $O(0, 0, 0)$: $$0^2 + 0^2 + 0^2 + 0 + 0 + 0 + d = 0 implies d = 0$$ 2. Since it passes through $A(a, 0, 0)$: $$a^2 + 2ua = 0 implies 2u = -a$$ 3. Since it passes through $B(0, b, 0)$: $$b^2 + 2vb = 0 implies 2v = -b$$ 4. Since it passes through $C(0, 0, c)$: $$c^2 + 2wc = 0 implies 2w = -c$$ Thus, the equation of the sphere circumscribing tetrahedron $OABC$ is: $$S: x^2 + y^2 + z^2 - ax - by - cz = 0 tag{2}$$ The circle $ABC$ is the intersection of the sphere $S = 0$ and the plane $pi = 0$. --- Step 2: Generating the Cone with Vertex at the Origin A cone with vertex at the origin $O(0, 0, 0)$ is represented by a homogeneous equation in $x, y, z$. To find the homogeneous equation of the second degree representing the cone passing through the intersection of the second-degree surface $S = 0$ and the plane $pi = 0$, we homogenize the second-degree equation (2) using the linear equation of the plane (1): $$x^2 + y^2 + z^2 - (ax + by + cz) cdot 1 = 0$$ From equation (1), we have: $$1 = frac{x}{a} + frac{y}{b} + frac{z}{c}$$ Substituting this expression for $1$: $$x^2 + y^2 + z^2 - (ax + by + cz) left( frac{x}{a} + frac{y}{b} + frac{z}{c} right) = 0 tag{3}$$ --- Ste

Question 10

Let $A = begin{bmatrix} 1 & 0 & 0 1 & 0 & 1 0 & 1 & 0 end{bmatrix}$. (i) Verify the Cayley-Hamilton theorem for the matrix $A$. (ii) Show that $A^n = A^{n-2} + A^2 - I$ for $n ge 3$, where $I$ is the identity matrix of order 3. Hence, find $A^{40}$.

Accepted Answer

Statement of Cayley-Hamilton Theorem The Cayley-Hamilton Theorem asserts that every square matrix over a commutative ring (such as the field of real numbers $mathbb{R}$) satisfies its own characteristic equation. That is, if: $$chiA(lambda) = det(A - lambda I) = (-1)^n (lambda^n + c{n-1}lambda^{n-1} + dots + c1lambda + c0) = 0$$ is the characteristic polynomial of $A in mathcal{M}{ntimes n}(mathbb{R})$, then substituting the matrix $A$ in place of the scalar variable $lambda$ yields the zero matrix: $$chiA(A) = A^n + c{n-1} A^{n-1} + dots + c1 A + c0 I = mathbf{0}$$ We are given the $3 times 3$ real matrix: $$A = begin{bmatrix} 1 & 0 & 0 1 & 0 & 1 0 & 1 & 0 end{bmatrix}$$ --- Part (i): Characteristic Polynomial and Verification of Cayley-Hamilton Theorem 1. Characteristic Polynomial of $A$: $$chiA(lambda) = det(A - lambda I) = begin{vmatrix} 1 - lambda & 0 & 0 1 & -lambda & 1 0 & 1 & -lambda end{vmatrix}$$ Expanding along the first row (which contains two zeros): $$chiA(lambda) = (1 - lambda) begin{vmatrix} -lambda & 1 1 & -lambda end{vmatrix} - 0 + 0$$ $$chiA(lambda) = (1 - lambda) left[ (-lambda)(-lambda) - (1)(1) right] = (1 - lambda)(lambda^2 - 1)$$ Expanding the product: $$chiA(lambda) = lambda^2 - 1 - lambda^3 + lambda = -(lambda^3 - lambda^2 - lambda + 1)$$ Setting the characteristic equation $chiA(lambda) = 0$: $$lambda^3 - lambda^2 - lambda + 1 = 0$$ According to the Cayley-Hamilton theorem, $A$ must satisfy: $$A^3 - A^2 - A + I = mathbf{0} tag{1}$$ 2. Computation of Matrix Powers $A^2$ and $A^3$: - Computing $A^2$: $$A^2 = A cdot A = begin{bmatrix} 1 & 0 & 0 1 & 0 & 1 0 & 1 & 0 end{bmatrix} begin{bmatrix} 1 & 0 & 0 1 & 0 & 1 0 & 1 & 0 end{bmatrix}$$ $$text{Row 1:} quad [1(1)+0(1)+0(0), ; 1(0)+0(0)+0(1), ; 1(0)+0(1)+0(0)] = [1, 0, 0]$$ $$text{Row 2:} quad [1(1)+0(1)+1(0), ; 1(0)+0(0)+1(1), ; 1(0)+0(1)+1(0)] = [1, 1, 0]$$ $$text{Row 3:} quad [0(1)+1(1)+0(0), ; 0(0)+1(0)+0(1), ; 0(0)+1(1)+0(0)] = [1, 0, 1]$$ $$A^2 = begin{bmatrix} 1 & 0 & 0 1 & 1 & 0 1 & 0 &

Question 11

Justify whether $(0, 0)$ is an extreme point for the function $f(x, y) = 2x^4 - 3x^2 y + y^2$.

Accepted Answer

Theoretical Concept of Extreme Points Let $f: D subset mathbb{R}^2 to mathbb{R}$ be a function of two variables defined on an open neighborhood of a point $(x0, y0)$. - $(x0, y0)$ is said to be a local minimum of $f$ if there exists a disk $Bdelta(x0, y0)$ such that: $$f(x, y) ge f(x0, y0) quad forall (x, y) in Bdelta(x0, y0)$$ - $(x0, y0)$ is said to be a local maximum of $f$ if there exists a disk $Bdelta(x0, y0)$ such that: $$f(x, y) le f(x0, y0) quad forall (x, y) in Bdelta(x0, y0)$$ - $(x0, y0)$ is called an extreme point (or local extremum) if it is either a local minimum or a local maximum. - If in every open disk $Bdelta(x0, y0)$, the difference $f(x, y) - f(x0, y0)$ takes both strictly positive and strictly negative values, then $(x0, y0)$ is a saddle point and is not an extreme point. We are given the polynomial function: $$f(x, y) = 2x^4 - 3x^2 y + y^2$$ --- Step 1: Stationary (Critical) Point Analysis The value of the function at the origin is: $$f(0, 0) = 0$$ Computing the first partial derivatives: $$fx(x, y) = frac{partial f}{partial x} = 8x^3 - 6xy = 2x(4x^2 - 3y)$$ $$fy(x, y) = frac{partial f}{partial y} = -3x^2 + 2y$$ Evaluating at the origin $(0, 0)$: $$fx(0, 0) = 0, qquad fy(0, 0) = 0$$ Since both first-order partial derivatives vanish at $(0, 0)$, the origin is indeed a stationary (critical) point of $f$. --- Step 2: Second Derivative Test (Hessian Matrix) To investigate the nature of the critical point, we compute the second partial derivatives: $$f{xx}(x, y) = 24x^2 - 6y implies r = f{xx}(0, 0) = 0$$ $$f{xy}(x, y) = -6x implies s = f{xy}(0, 0) = 0$$ $$f{yy}(x, y) = 2 implies t = f{yy}(0, 0) = 2$$ The discriminant (Hessian determinant) at $(0, 0)$ is: $$Delta = r t - s^2 = (0)(2) - (0)^2 = 0$$ Since $Delta = 0$, the classical second derivative test fails (is inconclusive). We must therefore examine the behavior of $f(x, y) - f(0, 0)$ directly in neighborhoods of $(0, 0)$. --- Step 3: Behavior Along Straight Lines Through the Origin Let us test

Question 12

Find the equation of the sphere through the circle $x^2 + y^2 + z^2 - 4x - 6y + 2z - 16 = 0$; $3x + y + 3z - 4 = 0$ in the following two cases: (i) the point $(1, 0, -3)$ lies on the sphere. (ii) the given circle is a great circle of the sphere.

Accepted Answer

Family of Spheres Through a Given Circle A circle in three dimensions is represented as the intersection of a sphere $S = 0$ and a plane $U = 0$. Here, the given sphere and plane equations are: $$S: x^2 + y^2 + z^2 - 4x - 6y + 2z - 16 = 0 tag{1}$$ $$U: 3x + y + 3z - 4 = 0 tag{2}$$ The equation of any sphere passing through the common circle of intersection of $S = 0$ and $U = 0$ belongs to the linear pencil: $$S + lambda U = 0 tag{3}$$ where $lambda in mathbb{R}$ is an arbitrary parameter. Substituting equations (1) and (2) into (3): $$(x^2 + y^2 + z^2 - 4x - 6y + 2z - 16) + lambda(3x + y + 3z - 4) = 0$$ Collecting corresponding powers of coordinates: $$x^2 + y^2 + z^2 + (3lambda - 4)x + (lambda - 6)y + (3lambda + 2)z - (16 + 4lambda) = 0 tag{4}$$ Comparing equation (4) with the standard equation of a sphere: $$x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0$$ we identify: $$u = frac{3lambda - 4}{2}, qquad v = frac{lambda - 6}{2}, qquad w = frac{3lambda + 2}{2}$$ The center $C$ of the sphere is given by $(-u, -v, -w)$: $$C = left( -frac{3lambda - 4}{2}, ; -frac{lambda - 6}{2}, ; -frac{3lambda + 2}{2} right) = left( frac{4 - 3lambda}{2}, ; frac{6 - lambda}{2}, ; -frac{3lambda + 2}{2} right) tag{5}$$ --- Case (i): The Point $(1, 0, -3)$ Lies on the Sphere Since the point $P(1, 0, -3)$ lies on the sphere (3), its coordinates must satisfy $S(1, 0, -3) + lambda U(1, 0, -3) = 0$. 1. Evaluating $S$ at $(1, 0, -3)$: $$S(1, 0, -3) = (1)^2 + (0)^2 + (-3)^2 - 4(1) - 6(0) + 2(-3) - 16$$ $$= 1 + 0 + 9 - 4 - 0 - 6 - 16 = 10 - 26 = -16$$ 2. Evaluating $U$ at $(1, 0, -3)$: $$U(1, 0, -3) = 3(1) + 1(0) + 3(-3) - 4$$ $$= 3 + 0 - 9 - 4 = -10$$ Substituting these values into $S + lambda U = 0$: $$-16 + lambda(-10) = 0$$ $$-10lambda = 16 implies lambda = -frac{16}{10} = -frac{8}{5}$$ Substituting $lambda = -frac{8}{5}$ into the pencil equation $S - frac{8}{5} U = 0$: $$5 S - 8 U = 0$$ $$5left( x^2 + y^2 + z^2 - 4x - 6y + 2z - 16 right) - 8left( 3x + y + 3z - 4 right) = 0$$ Expanding and combin

Question 13

Find the rank of the matrix $$A = begin{bmatrix} 1 & 2 & -1 & 0 -1 & 3 & 0 & -4 2 & 1 & 3 & -2 1 & 1 & 1 & -1 end{bmatrix}$$ by reducing it to row-reduced echelon form.

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Definition of Row-Reduced Echelon Form (RREF) A matrix is in Row-Reduced Echelon Form (RREF) if: 1. All non-zero rows are above any rows of all zeros. 2. The leading entry (the first non-zero number from the left, called the pivot) of each non-zero row is $1$. 3. Each leading entry $1$ is strictly to the right of the leading entry of the row above it. 4. Each column containing a leading entry $1$ has zeros everywhere else in that column. The rank of the matrix is the number of non-zero rows (or equivalently, the number of pivot positions) in its row-reduced echelon form. We are given the $4 times 4$ real matrix: $$A = begin{bmatrix} 1 & 2 & -1 & 0 -1 & 3 & 0 & -4 2 & 1 & 3 & -2 1 & 1 & 1 & -1 end{bmatrix}$$ --- Step 1: Forward Elimination to Clear Column 1 The pivot in Row 1 is already $a{11} = 1$. We use Row 1 to eliminate the entries in column 1 below it: 1. $R2 to R2 + R1$: $$R2 = (-1 + 1, ; 3 + 2, ; 0 + (-1), ; -4 + 0) = (0, 5, -1, -4)$$ 2. $R3 to R3 - 2R1$: $$R3 = (2 - 2(1), ; 1 - 2(2), ; 3 - 2(-1), ; -2 - 2(0)) = (0, -3, 5, -2)$$ 3. $R4 to R4 - R1$: $$R4 = (1 - 1, ; 1 - 2, ; 1 - (-1), ; -1 - 0) = (0, -1, 2, -1)$$ The matrix becomes: $$begin{bmatrix} 1 & 2 & -1 & 0 0 & 5 & -1 & -4 0 & -3 & 5 & -2 0 & -1 & 2 & -1 end{bmatrix}$$ --- Step 2: Normalizing Column 2 Pivot To obtain a convenient pivot of $1$ in row 2 without introducing fractions, we swap $R2$ and $R4$: $$R2 longleftrightarrow R4$$ $$begin{bmatrix} 1 & 2 & -1 & 0 0 & -1 & 2 & -1 0 & -3 & 5 & -2 0 & 5 & -1 & -4 end{bmatrix}$$ Multiply the new row 2 by $-1$ ($R2 to -R2$): $$begin{bmatrix} 1 & 2 & -1 & 0 0 & 1 & -2 & 1 0 & -3 & 5 & -2 0 & 5 & -1 & -4 end{bmatrix}$$ --- Step 3: Clearing Column 2 Above and Below the Pivot Now use $R2 = (0, 1, -2, 1)$ to eliminate all other entries in Column 2: 1. $R1 to R1 - 2R2$: $$R1 = (1 - 0, ; 2 - 2(1), ; -1 - 2(-2), ; 0 - 2(1)) = (1, 0, 3, -2)$$ 2. $R3 to R3 + 3R2$: $$R3 = (0 + 0, ; -3 + 3(1), ; 5 + 3(-2), ; -2 + 3(1)) = (0, 0, -1, 1)$$ 3. $R4 to R4 - 5R2$: $$R4 = (0 -

Question 14

Trace the curve $y^2(x^2 - 1) = 2x - 1$.

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Introduction & General Equation The given algebraic curve is: $$y^2(x^2 - 1) = 2x - 1$$ Solving explicitly for $y^2$: $$y^2 = frac{2x - 1}{x^2 - 1} = frac{2x - 1}{(x - 1)(x + 1)} tag{1}$$ We follow the standard, rigorous UPSC curve-tracing framework: 1. Symmetry 2. Origin and Intercepts 3. Asymptotes (Parallel and Oblique) 4. Region of Existence (Domain) 5. Critical Points and Monotonicity 6. Curvature and Shape --- Step 1: Symmetry - Symmetry about the $x$-axis: Replacing $y$ by $-y$ gives: $$(-y)^2(x^2 - 1) = y^2(x^2 - 1) = 2x - 1$$ The equation remains unchanged. Hence, the curve is symmetric about the $x$-axis. - Symmetry about the $y$-axis: Replacing $x$ by $-x$ gives $y^2(x^2 - 1) = -2x - 1 eq 2x - 1$. The curve is not symmetric about the $y$-axis. - Symmetry about the Origin: Replacing $(x, y)$ by $(-x, -y)$ changes the equation. Hence, the curve is not symmetric about the origin. --- Step 2: Origin, Intercepts, and Tangents 1. Origin: Putting $x = 0$ and $y = 0$: $$0^2(0^2 - 1) = 0 eq 2(0) - 1 = -1$$ The curve does not pass through the origin. 2. $y$-intercepts (putting $x = 0$): $$y^2(-1) = -1 implies y^2 = 1 implies y = pm 1$$ Thus, the curve cuts the $y$-axis at two points: $$(0, 1) quad text{and} quad (0, -1)$$ 3. $x$-intercepts (putting $y = 0$): $$0 = 2x - 1 implies x = frac{1}{2}$$ Thus, the curve cuts the $x$-axis at the single point: $$left( frac{1}{2}, 0 right)$$ 4. Tangent at $left( frac{1}{2}, 0 right)$: Implicit differentiation of $y^2(x^2 - 1) = 2x - 1$: $$2y frac{dy}{dx}(x^2 - 1) + 2x y^2 = 2 implies frac{dy}{dx} = frac{2(1 - x y^2)}{2y(x^2 - 1)} = frac{1 - x y^2}{y(x^2 - 1)}$$ At $x = frac{1}{2}, y = 0$: $$left. frac{dy}{dx} right|{left(frac{1}{2}, 0right)} = frac{1 - 0}{0 cdot (-3/4)} = infty$$ Therefore, the tangent to the curve at the vertex $left( frac{1}{2}, 0 right)$ is vertical (parallel to the $y$-axis). --- Step 3: Asymptotes 1. Asymptotes parallel to the $y$-axis (Vertical Asymptotes): Equating the coefficient of the highest power o

Question 15

Prove that the locus of a line which meets the lines $y = mx, z = c$; $y = -mx, z = -c$ and the circle $x^2 + y^2 = a^2, z = 0$ is $$c^2 m^2 (cy - mzx)^2 + c^2 (yz - cmx)^2 = a^2 m^2 (z^2 - c^2)^2.$$

Accepted Answer

Geometric Setup and Three Guiding Curves We are required to determine the locus of a variable straight line that simultaneously intersects three given spatial curves/lines: 1. The straight line $L1$: $$y = mx, qquad z = c tag{1}$$ 2. The straight line $L2$: $$y = -mx, qquad z = -c tag{2}$$ 3. The circle $C$ in the $xy$-plane ($z = 0$): $$x^2 + y^2 = a^2, qquad z = 0 tag{3}$$ --- Step 1: Parametric Representation of Points on $L1$ and $L2$ - Any point $P1$ lying on the line $L1$ can be parameterized by its $x$-coordinate, say $x1$: $$P1 = (x1, ; mx1, ; c)$$ - Any point $P2$ lying on the line $L2$ can be parameterized by its $x$-coordinate, say $x2$: $$P2 = (x2, ; -mx2, ; -c)$$ The variable line is the line passing through $P1$ and $P2$. --- Step 2: Coordinates of a General Point $(x, y, z)$ on the Variable Line Using the two-point form of a line in $mathbb{R}^3$, any point $(x, y, z)$ on the line joining $P1$ and $P2$ can be written parametrically as: $$frac{x - x1}{x2 - x1} = frac{y - mx1}{-mx2 - mx1} = frac{z - c}{-c - c} = t$$ From the $z$-coordinate equation: $$frac{z - c}{-2c} = t implies z - c = -2ct implies t = frac{c - z}{2c}$$ Consequently: $$1 - t = 1 - frac{c - z}{2c} = frac{c + z}{2c}$$ Expressing $x$ and $y$ using the parameter $t$: $$x = (1 - t)x1 + t x2 = left( frac{c + z}{2c} right) x1 + left( frac{c - z}{2c} right) x2 tag{4}$$ $$y = (1 - t)(mx1) + t(-mx2) = m left[ left( frac{c + z}{2c} right) x1 - left( frac{c - z}{2c} right) x2 right] tag{5}$$ Multiplying equations (4) and (5) by $2c$: $$2cx = (c + z)x1 + (c - z)x2 tag{6}$$ $$frac{2cy}{m} = (c + z)x1 - (c - z)x2 tag{7}$$ --- Step 3: Solving for the Parameters $x1$ and $x2$ 1. Adding equations (6) and (7): $$2cx + frac{2cy}{m} = 2(c + z)x1$$ $$cleft( x + frac{y}{m} right) = (c + z)x1$$ $$frac{c(mx + y)}{m} = (c + z)x1 implies x1 = frac{c(mx + y)}{m(c + z)} tag{8}$$ 2. Subtracting equation (7) from equation (6): $$2cx - frac{2cy}{m} = 2(c - z)x2$$ $$cleft( x - frac{y}{m} right) = (c - z)x2$$ $$frac{c

Question 16

When a particle is projected from a point $O1$ on the sea level with a velocity $v$ and angle of projection $theta$ with the horizon in a vertical plane, its horizontal range is $R1$. If it is further projected from a point $O2$, which is vertically above $O1$ at a height $h$ in the same vertical pl

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Projectile Kinematics and Coordinate Formulation Consider a Cartesian coordinate system in the vertical plane of projection: - The horizontal sea level is chosen as the line $y = 0$. - The point of projection on the sea level is $O1 = (0, 0)$. - The elevated point of projection is $O2 = (0, h)$, which is vertically above $O1$ at height $h$. - In both cases, the initial velocity is $v$ and the inclination to the horizontal is $theta$. - The horizontal and initial vertical components of velocity are: $$ux = v costheta, qquad uy = v sintheta$$ - Neglecting air resistance, the equations of motion under gravity $g$ are: $$x(t) = x0 + (v costheta) t$$ $$y(t) = y0 + (v sintheta) t - frac{1}{2} g t^2$$ --- Step 1: Motion from Sea Level Point $O1(0, 0)$ Here $x0 = 0$ and $y0 = 0$. The equation of vertical displacement is: $$y(t) = (v sintheta) t - frac{1}{2} g t^2$$ The particle lands on the sea level when $y(t) = 0$: $$t left( v sintheta - frac{1}{2} g t right) = 0$$ Since $t 0$ at landing, the time of flight $T1$ is: $$T1 = frac{2v sintheta}{g} tag{1}$$ The horizontal range $R1$ is the horizontal distance traveled during time $T1$: $$R1 = (v costheta) T1 = (v costheta) left( frac{2v sintheta}{g} right) = frac{2v^2 sintheta costheta}{g} = frac{v^2 sin 2theta}{g} tag{2}$$ Notice that from equation (2): $$frac{v^2 sintheta costheta}{g} = frac{1}{2} R1 tag{3}$$ --- Step 2: Motion from Elevated Point $O2(0, h)$ Here $x0 = 0$ and $y0 = h$. The equation of vertical displacement is: $$y(t) = h + (v sintheta) t - frac{1}{2} g t^2$$ The particle strikes the sea level when $y(t) = 0$: $$h + (v sintheta) t - frac{1}{2} g t^2 = 0$$ Multiplying by $-2$ and rearranging: $$g t^2 - 2(v sintheta) t - 2h = 0 tag{4}$$ This is a quadratic equation in $t$. By the quadratic formula: $$t = frac{2v sintheta pm sqrt{4v^2 sin^2theta - 4(g)(-2h)}}{2g} = frac{2v sintheta pm sqrt{4v^2 sin^2theta + 8gh}}{2g}$$ $$t = frac{v sintheta pm sqrt{v^2 sin^2theta + 2gh}}{g}$$ Since the time of flight must be str

Question 17

Obtain the solution of the initial-value problem $$frac{dy}{dx} - 2xy = 2, qquad y(0) = 1$$ in the form $y = e^{x^2}[1 + sqrt{pi}operatorname{erf}(x)]$.

Accepted Answer

Standard Form & Classification of the Differential Equation We are given the first-order ordinary differential equation with initial condition: $$frac{dy}{dx} - 2xy = 2, qquad y(0) = 1 tag{1}$$ This equation is a non-homogeneous, first-order linear differential equation of the standard Leibniz form: $$frac{dy}{dx} + P(x)y = Q(x)$$ where: $$P(x) = -2x, qquad Q(x) = 2$$ Both $P(x)$ and $Q(x)$ are continuous everywhere on $mathbb{R}$. --- Step 1: Integrating Factor Determination The integrating factor $I(x)$ is given by: $$I(x) = expleft( int P(x) dx right) = expleft( int (-2x) dx right) = e^{-x^2}$$ Multiplying both sides of the differential equation (1) by $I(x) = e^{-x^2}$: $$e^{-x^2} frac{dy}{dx} - 2x e^{-x^2} y = 2 e^{-x^2}$$ By the product rule of differentiation, the left-hand side collapses into the exact derivative: $$frac{d}{dx} left[ y e^{-x^2} right] = 2 e^{-x^2} tag{2}$$ --- Step 2: Integration and Definition of the Gauss Error Function Integrating equation (2) with respect to $t$ from $t = 0$ to $t = x$: $$int0^x frac{d}{dt} left[ y(t) e^{-t^2} right] dt = int0^x 2 e^{-t^2} dt$$ By the Fundamental Theorem of Calculus: $$left[ y(t) e^{-t^2} right]0^x = 2 int0^x e^{-t^2} dt$$ $$y(x) e^{-x^2} - y(0) e^0 = 2 int0^x e^{-t^2} dt tag{3}$$ Recall the formal definition of the standard Gauss Error Function, $operatorname{erf}(x)$: $$operatorname{erf}(x) = frac{2}{sqrt{pi}} int0^x e^{-t^2} dt$$ Multiplying by $sqrt{pi}$: $$2 int0^x e^{-t^2} dt = sqrt{pi} operatorname{erf}(x) tag{4}$$ --- Step 3: Imposition of the Initial Condition We are given the initial condition: $$y(0) = 1$$ Substituting $y(0) = 1$ and equation (4) into equation (3): $$y(x) e^{-x^2} - 1 = sqrt{pi} operatorname{erf}(x)$$ $$y(x) e^{-x^2} = 1 + sqrt{pi} operatorname{erf}(x)$$ Multiplying both sides by $e^{x^2}$: $$y(x) = e^{x^2} left[ 1 + sqrt{pi} operatorname{erf}(x) right]$$ --- Step 4: Verification of the Solution 1. Verification of the Initial Condition: At $x = 0$: $$y(0) = e^0 left[ 1 + sqrt{

Question 18

Given that $mathcal{L}{f(t); p} = F(p)$. Show that $int0^infty frac{f(t)}{t} dt = int0^infty F(x) dx$. Hence evaluate the integral $int0^infty frac{e^{-t} - e^{-3t}}{t} dt$.

Accepted Answer

Theoretical Foundation: Integration Property of Laplace Transforms The Laplace transform of an original function $f(t)$ defined for $t ge 0$ is given by: $$F(p) = mathcal{L}{f(t); p} = int0^infty e^{-pt} f(t) dt tag{1}$$ provided the improper integral converges for $operatorname{Re}(p) p0$. We are required to establish the operational theorem: $$int0^infty frac{f(t)}{t} dt = int0^infty F(x) dx$$ and apply it to evaluate the improper integral $int0^infty frac{e^{-t} - e^{-3t}}{t} dt$. --- Step 1: Proof of the Integral Property From definition (1), for any real parameter $x 0$: $$F(x) = int0^infty e^{-xt} f(t) dt$$ We integrate both sides with respect to $x$ over the semi-infinite interval $[0, infty)$: $$int0^infty F(x) dx = int0^infty left[ int0^infty e^{-xt} f(t) dt right] dx tag{2}$$ Assuming $f(t)$ is piecewise continuous and of exponential order, and $lim{t to 0^+} frac{f(t)}{t}$ exists finitely (ensuring absolute convergence), we apply Fubini's Theorem / Dirichlet's integral rule to reverse the order of integration: $$int0^infty F(x) dx = int0^infty f(t) left[ int0^infty e^{-xt} dx right] dt tag{3}$$ We evaluate the inner definite integral with respect to $x$ for any fixed $t 0$: $$int0^infty e^{-xt} dx = lim{M to infty} int0^M e^{-xt} dx = lim{M to infty} left[ frac{e^{-xt}}{-t} right]0^M$$ Since $t 0$, as $M to infty$, $e^{-Mt} to 0$: $$int0^infty e^{-xt} dx = 0 - left( -frac{1}{t} right) = frac{1}{t} tag{4}$$ Substituting result (4) back into equation (3): $$int0^infty F(x) dx = int0^infty f(t) cdot frac{1}{t} dt = int0^infty frac{f(t)}{t} dt$$ This proves the identity: $$int0^infty frac{f(t)}{t} dt = int0^infty F(x) dx tag{5}$$ --- Step 2: Evaluation of $int0^infty frac{e^{-t} - e^{-3t}}{t} dt$ Let the numerator function be: $$f(t) = e^{-t} - e^{-3t}$$ 1. Check existence of the integral: - As $t to 0^+$: $$lim{t to 0^+} frac{e^{-t} - e^{-3t}}{t} = lim{t to 0^+} frac{(1 - t + O(t^2)) - (1 - 3t + O(t^2))}{t} = lim{t to 0^+} frac{2t}{t} = 2 < infty$$ Thus, the

Question 19

A cylinder of radius '$a$' touches a vertical wall along a generating line. Axis of the cylinder is fixed horizontally. A uniform flat beam of length '$l$' and weight '$W$' rests with its extremities in contact with the wall and the cylinder, making an angle of $45^circ$ with the vertical. If fricti

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Statics Formulation and Coordinate Geometry Let the cross-section of the system in the vertical plane perpendicular to the horizontal cylinder axis be represented in Cartesian coordinates: - The smooth vertical wall is the line $x = 0$. - The cylinder touches the vertical wall along a horizontal generating line, which corresponds to the point $(0, 0)$ in cross-section. - Since the cylinder of radius $a$ lies in the region $x ge 0$ and touches the line $x = 0$ at $(0, 0)$, its center $C$ is located at: $$C = (a, 0)$$ The equation of the circular cross-section of the cylinder is: $$(x - a)^2 + y^2 = a^2 tag{1}$$ The beam $AB$ has length $l$, weight $W$, and makes an angle of $45^circ$ with the vertical wall: - Extremity $A$ is in contact with the smooth vertical wall: $xA = 0$, so $A = (0, yA)$. - Extremity $B$ is in contact with the smooth surface of the cylinder: $B = (xB, yB)$. - Since the beam is inclined at an angle of $45^circ$ to the vertical: $$xB - xA = l sin(45^circ) = frac{l}{sqrt{2}}$$ Since $xA = 0$: $$xB = frac{l}{sqrt{2}} tag{2}$$ Let $s = frac{l}{sqrt{2}}$. Then $xB = s$. The vertical coordinate difference is: $$yA - yB = l cos(45^circ) = frac{l}{sqrt{2}} = s implies yA = yB + s tag{3}$$ --- Step 1: Equilibrium of Three Coplanar Forces Since friction is completely neglected, the beam is in static equilibrium under three forces: 1. Wall Reaction $vec{R}A$: Acting at $A(0, yA)$ perpendicular to the smooth vertical wall, hence horizontally: $$vec{R}A = RA hat{i}$$ 2. Weight $vec{W}$: Acting vertically downwards at the midpoint $G$ of the beam: $$xG = frac{xA + xB}{2} = frac{s}{2}, qquad vec{W} = -W hat{j}$$ 3. Cylinder Reaction $vec{R}B$: Acting at the contact point $B$ perpendicular to the surface of the cylinder. Since every normal to a circular cylinder passes through its axis/center $C(a, 0)$, the line of action of $vec{R}B$ must pass through the center $C(a, 0)$. By the Three-Force Concurrency Theorem in statics, three non-parallel coplanar forces in

Question 20

A particle is moving under Simple Harmonic Motion of period $T$ about a centre $O$. It passes through the point $P$ with velocity $v$ along the direction $OP$ and $OP = p$. Find the time that elapses before the particle returns to the point $P$. What will be the value of $p$ when the elapsed time is

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Theoretical Model of Simple Harmonic Motion Simple Harmonic Motion (SHM) is rectilinear motion under an attractive force directed toward a fixed center $O$, with magnitude proportional to the displacement $x$ from $O$. The governing differential equation is: $$frac{d^2 x}{dt^2} = -mu x tag{1}$$ where $mu 0$ is the intensity of the force per unit mass. 1. Relation between Period $T$ and $mu$: $$T = frac{2pi}{sqrt{mu}} implies sqrt{mu} = frac{2pi}{T} tag{2}$$ 2. First Integral (Velocity-Displacement Relation): $$frac{d^2 x}{dt^2} = v frac{dv}{dx} = -mu x$$ Integrating with respect to $x$: $$v^2 = mu (A^2 - x^2) tag{3}$$ where $A$ is the amplitude of oscillation (the maximum displacement where $v = 0$). --- Step 1: Determining Amplitude $A$ in terms of Given Data At the point $P$, the displacement from the center $O$ is $x = p$, and the particle's speed is $v$. Substituting $x = p$ into equation (3): $$v^2 = mu (A^2 - p^2)$$ $$A^2 - p^2 = frac{v^2}{mu} implies A = sqrt{p^2 + frac{v^2}{mu}} tag{4}$$ Using $sqrt{mu} = frac{2pi}{T}$ from equation (2): $$A = sqrt{p^2 + frac{v^2 T^2}{4pi^2}} tag{5}$$ --- Step 2: Time Taken to Return to Point $P$ The particle passes through $P(x = p)$ with velocity $v$ directed along $OP$ (i.e., moving away from the center $O$ towards the positive extreme position $x = A$). The particle travels from $x = p$ to the extreme position $x = A$, momentarily comes to rest ($v = 0$), and then travels back from $x = A$ to $x = p$. By symmetry of SHM: $$text{Total elapsed time } t{text{return}} = 2 times (text{time taken to travel from } x = p text{ to } x = A)$$ Let $t1$ be the time from $x = p$ to $x = A$. During outward motion ($v 0$): $$frac{dx}{dt} = sqrt{mu} sqrt{A^2 - x^2} implies dt = frac{dx}{sqrt{mu}sqrt{A^2 - x^2}}$$ Integrating from $x = p$ to $x = A$: $$t1 = intp^A frac{dx}{sqrt{mu}sqrt{A^2 - x^2}} = frac{1}{sqrt{mu}} left[ sin^{-1}left(frac{x}{A}right) right]p^A$$ $$t1 = frac{1}{sqrt{mu}} left[ sin^{-1}(1) - sin^{-1}left(frac{p}{A}right)

Question 21

If $vec{a} = sintheta,hat{i} + costheta,hat{j} + theta,hat{k}$, $vec{b} = costheta,hat{i} - sintheta,hat{j} - 3,hat{k}$, $vec{c} = 2,hat{i} + 3,hat{j} - 3,hat{k}$, then find the values of the derivative of the vector function $vec{a} times (vec{b} times vec{c})$ w.r.t. $theta$ at $theta = frac{pi}{2

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Vector Triple Product Identity Let $vec{a}, vec{b}, vec{c}$ be differentiable vector functions of a scalar parameter $theta$. The vector triple product is given by the standard identity: $$vec{V}(theta) = vec{a} times (vec{b} times vec{c}) = (vec{a} cdot vec{c})vec{b} - (vec{a} cdot vec{b})vec{c} tag{1}$$ We are given: $$vec{a}(theta) = sintheta,hat{i} + costheta,hat{j} + theta,hat{k}$$ $$vec{b}(theta) = costheta,hat{i} - sintheta,hat{j} - 3,hat{k}$$ $$vec{c} = 2,hat{i} + 3,hat{j} - 3,hat{k} quad (text{constant vector})$$ --- Step 1: Evaluating the Scalar Dot Products 1. Compute $vec{a} cdot vec{b}$: $$vec{a} cdot vec{b} = (sintheta)(costheta) + (costheta)(-sintheta) + (theta)(-3)$$ $$vec{a} cdot vec{b} = sinthetacostheta - sinthetacostheta - 3theta = -3theta tag{2}$$ 2. Compute $vec{a} cdot vec{c}$: $$vec{a} cdot vec{c} = (sintheta)(2) + (costheta)(3) + (theta)(-3)$$ $$vec{a} cdot vec{c} = 2sintheta + 3costheta - 3theta tag{3}$$ --- Step 2: Explicit Expression for $vec{V}(theta)$ Substituting equations (2) and (3) into identity (1): $$vec{V}(theta) = (2sintheta + 3costheta - 3theta)vec{b}(theta) - (-3theta)vec{c}$$ $$vec{V}(theta) = (2sintheta + 3costheta - 3theta)vec{b}(theta) + 3thetavec{c} tag{4}$$ --- Step 3: Differentiation with Respect to $theta$ Differentiating $vec{V}(theta)$ using the product rule: $$frac{dvec{V}}{dtheta} = frac{d}{dtheta}(2sintheta + 3costheta - 3theta) cdot vec{b}(theta) + (2sintheta + 3costheta - 3theta) cdot frac{dvec{b}}{dtheta} + 3vec{c} tag{5}$$ We evaluate the individual derivatives: - Derivative of the scalar coefficient: $$frac{d}{dtheta}(2sintheta + 3costheta - 3theta) = 2costheta - 3sintheta - 3$$ - Derivative of $vec{b}(theta)$: $$frac{dvec{b}}{dtheta} = frac{d}{dtheta}(costheta,hat{i} - sintheta,hat{j} - 3,hat{k}) = -sintheta,hat{i} - costheta,hat{j}$$ Substituting into equation (5): $$frac{dvec{V}}{dtheta} = (2costheta - 3sintheta - 3)vec{b}(theta) + (2sintheta + 3costheta - 3theta)(-sintheta,hat{i} - costheta,hat{j}) + 3(2h

Question 22

Solve the differential equation: $$frac{d^3y}{dx^3} - 3frac{d^2y}{dx^2} + 4frac{dy}{dx} - 2y = e^x + cos x.$$

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Standard Operator Form of the Differential Equation Let $D = frac{d}{dx}$ denote the differential operator with respect to $x$. The given third-order linear ordinary differential equation with constant coefficients is: $$(D^3 - 3D^2 + 4D - 2)y = e^x + cos x tag{1}$$ The complete general solution consists of the sum of the complementary function $yc(x)$ and the particular integral $yp(x)$: $$y(x) = yc(x) + yp(x)$$ --- Step 1: Determination of the Complementary Function ($yc$) The homogeneous equation corresponding to (1) is: $$(D^3 - 3D^2 + 4D - 2)y = 0$$ The auxiliary (characteristic) equation is: $$m^3 - 3m^2 + 4m - 2 = 0 tag{2}$$ We find the roots of this cubic equation: Testing $m = 1$: $$1^3 - 3(1)^2 + 4(1) - 2 = 1 - 3 + 4 - 2 = 0$$ Hence, $(m - 1)$ is a factor of the polynomial. Dividing $m^3 - 3m^2 + 4m - 2$ by $(m - 1)$ using synthetic division: $$m^3 - 3m^2 + 4m - 2 = (m - 1)(m^2 - 2m + 2) = 0$$ For the quadratic factor $m^2 - 2m + 2 = 0$: $$m = frac{-(-2) pm sqrt{(-2)^2 - 4(1)(2)}}{2} = frac{2 pm sqrt{4 - 8}}{2} = frac{2 pm 2i}{2} = 1 pm i$$ Thus, the roots of the auxiliary equation are: $$m1 = 1, qquad m{2, 3} = 1 pm i$$ Corresponding to the real root $m1 = 1$, the solution is $c1 e^x$. Corresponding to the complex conjugate pair $m = 1 pm i$, the solution is $e^x(c2 cos x + c3 sin x)$. Therefore, the Complementary Function is: $$yc(x) = c1 e^x + e^x (c2 cos x + c3 sin x) tag{3}$$ where $c1, c2, c3$ are arbitrary constants of integration. --- Step 2: Determination of the Particular Integral ($yp$) By the linearity of the inverse differential operator: $$yp(x) = frac{1}{D^3 - 3D^2 + 4D - 2} left( e^x + cos x right) = y{p1}(x) + y{p2}(x)$$ 1. Evaluation of $y{p1}(x) = frac{1}{f(D)} e^x$: Here $f(D) = D^3 - 3D^2 + 4D - 2$. Evaluating $f(1)$: $$f(1) = 1 - 3 + 4 - 2 = 0$$ This is a case of failure (resonance) because $m = 1$ is a root of multiplicity $1$. Using the standard operational rule for a simple root: $$y{p1}(x) = frac{1}{f(D)} e^{ax} = frac{x}{f'(a)} e

Question 23

For a scalar point function $phi$ and vector point function $vec{f}$, prove the identity $$ abla cdot (phi,vec{f}) = ablaphi cdot vec{f} + phi( abla cdot vec{f}).$$ Also find the value of $ abla cdot left(frac{f(r)}{r} vec{r}right)$ and then verify stated identity.

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Part 1: Proof of the Vector Identity Let $phi = phi(x, y, z)$ be a differentiable scalar field, and let $vec{f} = f1,hat{i} + f2,hat{j} + f3,hat{k}$ be a continuously differentiable vector field in $mathbb{R}^3$. Then: $$phi,vec{f} = (phi f1),hat{i} + (phi f2),hat{j} + (phi f3),hat{k}$$ The divergence of $phi,vec{f}$ is: $$ abla cdot (phi,vec{f}) = frac{partial}{partial x}(phi f1) + frac{partial}{partial y}(phi f2) + frac{partial}{partial z}(phi f3) tag{1}$$ Applying the product rule of ordinary differentiation to each component: $$frac{partial}{partial x}(phi f1) = frac{partialphi}{partial x} f1 + phi frac{partial f1}{partial x}$$ $$frac{partial}{partial y}(phi f2) = frac{partialphi}{partial y} f2 + phi frac{partial f2}{partial y}$$ $$frac{partial}{partial z}(phi f3) = frac{partialphi}{partial z} f3 + phi frac{partial f3}{partial z}$$ Substituting these expansions into equation (1) and regrouping: $$ abla cdot (phi,vec{f}) = left( frac{partialphi}{partial x} f1 + frac{partialphi}{partial y} f2 + frac{partialphi}{partial z} f3 right) + phi left( frac{partial f1}{partial x} + frac{partial f2}{partial y} + frac{partial f3}{partial z} right) tag{2}$$ Recognizing the definitions of the gradient of a scalar field and divergence of a vector field: $$ ablaphi = frac{partialphi}{partial x},hat{i} + frac{partialphi}{partial y},hat{j} + frac{partialphi}{partial z},hat{k}$$ $$ abla cdot vec{f} = frac{partial f1}{partial x} + frac{partial f2}{partial y} + frac{partial f3}{partial z}$$ The first group in (2) is the dot product: $$frac{partialphi}{partial x} f1 + frac{partialphi}{partial y} f2 + frac{partialphi}{partial z} f3 = ablaphi cdot vec{f}$$ The second group in (2) is: $$phi left( frac{partial f1}{partial x} + frac{partial f2}{partial y} + frac{partial f3}{partial z} right) = phi ( abla cdot vec{f})$$ Therefore: $$ abla cdot (phi,vec{f}) = ablaphi cdot vec{f} + phi( abla cdot vec{f}) tag{3}$$ This proves the identity. --- Part 2: Direct Evaluation of $ abla cdot left( fra

Question 24

Evaluate the integral $iintS (3y^2 z^2,hat{i} + 4z^2 x^2,hat{j} + z^2 y^2,hat{k}) cdot hat{n},dS$, where $S$ is the upper part of the surface $4x^2 + 4y^2 + 4z^2 = 1$ above the plane $z = 0$ and bounded by the $xy$-plane. Hence, verify Gauss-Divergence theorem.

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Statement of Gauss-Divergence Theorem Let $V$ be a closed, bounded region in $mathbb{R}^3$ with piecewise smooth boundary surface $partial V$ oriented with outward-pointing unit normal vector $hat{n}$. If $vec{F}$ is a continuously differentiable vector field on $V$, then: $$iint{partial V} vec{F} cdot hat{n} , dS = iiintV ( abla cdot vec{F}) , dV$$ --- Step 1: Geometry and Boundary Surfaces We are given the vector field: $$vec{F} = 3y^2 z^2,hat{i} + 4z^2 x^2,hat{j} + z^2 y^2,hat{k}$$ The closed solid region $V$ is bounded by two surfaces $partial V = S cup S1$: 1. $S$: The curved upper hemisphere $4x^2 + 4y^2 + 4z^2 = 1$ ($z ge 0$). In standard form: $$x^2 + y^2 + z^2 = R^2, qquad text{where } R = frac{1}{2}$$ 2. $S1$: The flat circular base lying in the $xy$-plane ($z = 0$), bounded by $x^2 + y^2 le R^2 = frac{1}{4}$. --- Step 2: Evaluation of the Volume Divergence Integral We compute the divergence of $vec{F}$: $$ abla cdot vec{F} = frac{partial}{partial x}(3y^2 z^2) + frac{partial}{partial y}(4z^2 x^2) + frac{partial}{partial z}(z^2 y^2)$$ $$ abla cdot vec{F} = 0 + 0 + 2z y^2 = 2z y^2 tag{1}$$ We evaluate the triple integral $iiintV 2z y^2 dV$ using spherical coordinates $(r, phi, theta)$: $$x = rsinphicostheta, quad y = rsinphisintheta, quad z = rcosphi$$ where: $$0 le r le R = frac{1}{2}, qquad 0 le phi le frac{pi}{2}, qquad 0 le theta le 2pi$$ The volume element is $dV = r^2sinphi , dr,dphi,dtheta$. Expressing the integrand in spherical coordinates: $$2z y^2 = 2(rcosphi)(rsinphisintheta)^2 = 2r^3 cosphisin^2phisin^2theta$$ Thus: $$iiintV ( abla cdot vec{F}) dV = int0^{2pi} int0^{pi/2} int0^{1/2} (2r^3 cosphisin^2phisin^2theta) (r^2sinphi) , dr,dphi,dtheta$$ $$= 2 left( int0^{2pi} sin^2theta , dtheta right) left( int0^{pi/2} sin^3phicosphi , dphi right) left( int0^{1/2} r^5 , dr right) tag{2}$$ We evaluate each integral factor: 1. $int0^{2pi} sin^2theta , dtheta = pi$ 2. $int0^{pi/2} sin^3phicosphi , dphi = left[ frac{sin^4phi}{4} right]0^{pi/2} = frac{1}{4}$

Question 25

Find the solution of the differential equation: $$frac{dy}{dx} = -frac{2xy^3 + 2}{3x^2 y^2 + 8e^{4y}}.$$

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Classification of the Differential Equation We are given the first-order differential equation: $$frac{dy}{dx} = -frac{2xy^3 + 2}{3x^2 y^2 + 8e^{4y}} tag{1}$$ Cross-multiplying to put the equation into the standard Pfaffian differential form: $$(2xy^3 + 2) dx + (3x^2 y^2 + 8e^{4y}) dy = 0 tag{2}$$ This is of the form: $$M(x, y) dx + N(x, y) dy = 0$$ where: $$M(x, y) = 2xy^3 + 2$$ $$N(x, y) = 3x^2 y^2 + 8e^{4y}$$ --- Step 1: Test for Exactness A necessary and sufficient condition for the differential form $M dx + N dy$ to be an exact differential $dPhi$ on a simply connected domain is: $$frac{partial M}{partial y} = frac{partial N}{partial x}$$ Computing the partial derivatives: 1. Differentiating $M(x, y)$ with respect to $y$ (treating $x$ as constant): $$frac{partial M}{partial y} = frac{partial}{partial y}(2xy^3 + 2) = 2x(3y^2) + 0 = 6xy^2$$ 2. Differentiating $N(x, y)$ with respect to $x$ (treating $y$ as constant): $$frac{partial N}{partial x} = frac{partial}{partial x}(3x^2 y^2 + 8e^{4y}) = 3(2x)y^2 + 0 = 6xy^2$$ Since: $$frac{partial M}{partial y} = 6xy^2 = frac{partial N}{partial x}$$ the given differential equation is strictly EXACT. --- Step 2: Determination of the General Solution Since the equation is exact, there exists a potential function $Phi(x, y)$ such that: $$dPhi = frac{partial Phi}{partial x} dx + frac{partial Phi}{partial y} dy = M dx + N dy = 0$$ and the general solution is given by: $$Phi(x, y) = C$$ We determine $Phi(x, y)$ using the standard formula: $$Phi(x, y) = int M(x, y) dx ;; (text{treating } y text{ as a constant}) + int left[ text{terms in } N(x, y) text{ not containing } x right] dy = C$$ 1. Integrating $M$ with respect to $x$: $$int (2xy^3 + 2) dx = 2y^3 left( frac{x^2}{2} right) + 2x = x^2 y^3 + 2x$$ 2. Integrating terms in $N$ independent of $x$: $$N(x, y) = 3x^2 y^2 + 8e^{4y}$$ - The term $3x^2 y^2$ contains $x$ and is omitted. - The term independent of $x$ is $8e^{4y}$. $$int 8e^{4y} dy = 8 left( frac{e^{4y}}{4} right) = 2e^{4y

Question 26

Reduce the equation $x^2 p^2 + y(2x + y)p + y^2 = 0$ to Clairaut's form by the substitution $y = u$ and $xy = v$. Hence solve the equation and show that $y + 4x = 0$ is a singular solution of the differential equation.

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Given Differential Equation & Transformation of Variables We are given the non-linear first-order differential equation: $$x^2 p^2 + y(2x + y)p + y^2 = 0 tag{1}$$ where $p = frac{dy}{dx}$. We are directed to apply the coordinate transformation: $$u = y tag{2}$$ $$v = xy tag{3}$$ --- Step 1: Relating the Derivative $p = frac{dy}{dx}$ to $P = frac{dv}{du}$ Differentiating the transformation equations: $$du = dy$$ $$dv = d(xy) = x,dy + y,dx$$ Now consider the derivative $P = frac{dv}{du}$: $$P = frac{dv}{du} = frac{x,dy + y,dx}{dy} = x + y frac{dx}{dy} = x + frac{y}{p} tag{4}$$ From equation (4): $$P - x = frac{y}{p} implies p = frac{y}{P - x} tag{5}$$ Expressing $x$ and $y$ in terms of the new variables $u$ and $v$: - From (2): $y = u$ - From (3): $x = frac{v}{y} = frac{v}{u}$ Substituting these into the denominator of (5): $$P - x = P - frac{v}{u} = frac{uP - v}{u}$$ Thus: $$p = frac{u}{frac{uP - v}{u}} = frac{u^2}{uP - v} tag{6}$$ --- Step 2: Reduction to Clairaut's Form Dividing the given differential equation (1) by $y^2 = u^2$: $$left(frac{x}{y}right)^2 p^2 + left(frac{2x}{y} + 1right) p + 1 = 0 tag{7}$$ Observe that: $$frac{x}{y} p = left( frac{v/u}{u} right) left( frac{u^2}{uP - v} right) = frac{v}{u^2} cdot frac{u^2}{uP - v} = frac{v}{uP - v}$$ Substituting this into equation (7): $$left( frac{v}{uP - v} right)^2 + left( frac{2v}{u^2} + 1 right) left( frac{u^2}{uP - v} right) + 1 = 0$$ $$frac{v^2}{(uP - v)^2} + frac{2v + u^2}{uP - v} + 1 = 0$$ Multiplying the entire equation by $(uP - v)^2$: $$v^2 + (2v + u^2)(uP - v) + (uP - v)^2 = 0$$ $$v^2 + 2v(uP - v) + (uP - v)^2 + u^2(uP - v) = 0$$ The first three terms form a perfect square: $$[ v + (uP - v) ]^2 + u^2(uP - v) = 0$$ $$(uP)^2 + u^2(uP - v) = 0$$ $$u^2 P^2 + u^2(uP - v) = 0$$ Since $u = y eq 0$, dividing by $u^2$: $$P^2 + uP - v = 0$$ $$v = uP + P^2 tag{8}$$ Equation (8) is strictly of Clairaut's Form: $$v = uP + f(P) qquad text{where } f(P) = P^2$$ --- Step 3: General Solution Differentiating Clairaut's e

Question 27

A solid hemisphere is supported by a string fixed to a point on its rim and to a point on a smooth vertical wall with which the curved surface is in contact. If $theta$ is the angle of inclination of the string with vertical and $phi$ is the angle of inclination of the plane base of the hemisphere t

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Statics & Coordinate Geometry of the Hemisphere Let $a$ be the radius of the solid hemisphere: - The curved spherical surface is in contact with the smooth vertical wall at point $P$. - Since the wall is vertical and smooth, the normal reaction $R$ at $P$ is purely horizontal. - By the geometric property of a sphere, every normal to the spherical surface passes through its center of curvature $O$. - Therefore, the line segment $OP$ is perpendicular to the vertical wall, which means $OP$ is horizontal, and $OP = a$. - Hence, the center $O$ of the plane circular base lies at a constant horizontal distance $a$ from the vertical wall. --- Step 1: Center of Gravity and Point of Concurrency Let the plane circular base of the hemisphere be inclined at an angle $phi$ to the vertical: 1. Center of Gravity $G$: The center of gravity $G$ of a uniform solid hemisphere of radius $a$ lies on its axis of symmetry (the normal to the flat base through $O$) at a distance: $$OG = frac{3}{8} a$$ Since the base makes an angle $phi$ with the vertical, the axis of symmetry $OG$ makes an angle $phi$ with the horizontal. The horizontal distance of $G$ from the center $O$ is: $$xG - xO = OG sinphi = frac{3}{8} a sinphi tag{1}$$ 2. Equilibrium of Three Forces: The hemisphere is in static equilibrium under three forces: - Horizontal normal reaction $vec{R}$ from the wall at $P$, acting along the horizontal line through $O$. - Weight $vec{W}$ acting vertically downwards through the center of gravity $G$. - Tension $vec{T}$ in the string $AS$, attached to a point $A$ on the rim of the base and a fixed point $S$ on the wall, making an angle $theta$ with the vertical. By the Three-Force Concurrency Principle, the lines of action of $vec{R}, vec{W}, vec{T}$ must intersect at a single point $K$: - The line of action of $vec{R}$ is the horizontal line through $O$ ($y = yO$). - The line of action of $vec{W}$ is the vertical line through $G$ ($x = xG$). - Therefore, the concurrence point $K$ has coordi

Question 28

If the tangent to a curve makes a constant angle $theta$ with a fixed line, then prove that the ratio of radius of torsion to radius of curvature is proportional to $tantheta$. Further prove that if this ratio is constant, then the tangent makes a constant angle with a fixed direction.

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Statement & Significance: Lancret's Theorem for Helices A space curve whose tangent makes a constant angle with a fixed straight line is called a cylindrical helix (or a curve of constant slope). The classical theorem of Lancret (1802) establishes that a necessary and sufficient condition for a space curve of non-vanishing curvature to be a cylindrical helix is that the ratio of its curvature to torsion (or equivalently, radius of torsion to radius of curvature) is constant. --- Mathematical Notation & Frenet-Serret Apparatus Let $vec{r} = vec{r}(s)$ be a smooth space curve parameterized by arc length $s$. At any point on the curve, let: - $vec{t} = frac{dvec{r}}{ds}$ be the unit tangent vector. - $vec{n}$ be the principal unit normal vector. - $vec{b} = vec{t} times vec{n}$ be the unit binormal vector. - $kappa(s) 0$ be the curvature, and $rho = frac{1}{kappa}$ be the radius of curvature. - $tau(s)$ be the torsion, and $sigma = frac{1}{tau}$ be the radius of torsion. The Frenet-Serret Formulas are: $$frac{dvec{t}}{ds} = kappa vec{n} tag{1}$$ $$frac{dvec{n}}{ds} = -kappa vec{t} + tau vec{b} tag{2}$$ $$frac{dvec{b}}{ds} = -tau vec{n} tag{3}$$ --- Part 1: Direct Statement (Tangent Makes Constant Angle $theta$) Let $vec{a}$ be a fixed unit vector along the given line. Since $vec{a}$ is constant in both magnitude and direction: $$|vec{a}| = 1, qquad frac{dvec{a}}{ds} = vec{0} tag{4}$$ We are given that the tangent $vec{t}$ makes a constant angle $theta$ with $vec{a}$: $$vec{t} cdot vec{a} = costheta = text{constant} tag{5}$$ Differentiating equation (5) with respect to arc length $s$: $$frac{d}{ds}(vec{t} cdot vec{a}) = frac{dvec{t}}{ds} cdot vec{a} + vec{t} cdot frac{dvec{a}}{ds} = 0$$ Using equation (1) and $frac{dvec{a}}{ds} = vec{0}$: $$(kappa vec{n}) cdot vec{a} + 0 = 0$$ $$kappa (vec{n} cdot vec{a}) = 0$$ Since the curvature $kappa eq 0$: $$vec{n} cdot vec{a} = 0 tag{6}$$ Equation (6) shows that the fixed vector $vec{a}$ is everywhere orthogonal to the principal n

Question 29

Solve the following initial value problem by using Laplace transform technique: $$frac{d^2y}{dt^2} - 4frac{dy}{dt} + 3y(t) = f(t), qquad y(0) = 1, quad y'(0) = 0$$ and $f(t)$ is a given function of $t$.

Accepted Answer

Initial Value Problem Formulation We are given the second-order linear non-homogeneous differential equation: $$frac{d^2y}{dt^2} - 4frac{dy}{dt} + 3y(t) = f(t) tag{1}$$ subject to the initial conditions: $$y(0) = 1, qquad y'(0) = 0 tag{2}$$ where $f(t)$ is an arbitrary given function of $t$ for $t ge 0$. --- Step 1: Application of the Laplace Transform Let $mathcal{L}{y(t)} = Y(s)$ and $mathcal{L}{f(t)} = F(s)$. Applying the Laplace transform to both sides of equation (1): $$mathcal{L}{y''(t)} - 4mathcal{L}{y'(t)} + 3mathcal{L}{y(t)} = mathcal{L}{f(t)}$$ Using the standard differentiation properties: $$mathcal{L}{y'(t)} = s Y(s) - y(0)$$ $$mathcal{L}{y''(t)} = s^2 Y(s) - s y(0) - y'(0)$$ Substituting these expressions: $$left[ s^2 Y(s) - s y(0) - y'(0) right] - 4left[ s Y(s) - y(0) right] + 3 Y(s) = F(s)$$ Substituting the initial conditions $y(0) = 1$ and $y'(0) = 0$: $$left[ s^2 Y(s) - s(1) - 0 right] - 4left[ s Y(s) - 1 right] + 3 Y(s) = F(s)$$ $$s^2 Y(s) - s - 4s Y(s) + 4 + 3 Y(s) = F(s)$$ $$(s^2 - 4s + 3) Y(s) - s + 4 = F(s)$$ Isolating the transform $Y(s)$: $$(s^2 - 4s + 3) Y(s) = (s - 4) + F(s)$$ $$Y(s) = frac{s - 4}{s^2 - 4s + 3} + frac{F(s)}{s^2 - 4s + 3} tag{3}$$ Factoring the characteristic polynomial in the denominator: $$s^2 - 4s + 3 = (s - 1)(s - 3)$$ Thus: $$Y(s) = frac{s - 4}{(s - 1)(s - 3)} + frac{F(s)}{(s - 1)(s - 3)} tag{4}$$ --- Step 2: Inverse Laplace Transform of the Homogeneous Term Let $Y1(s) = frac{s - 4}{(s - 1)(s - 3)}$. We resolve into partial fractions: $$frac{s - 4}{(s - 1)(s - 3)} = frac{A}{s - 1} + frac{B}{s - 3}$$ Using the Heaviside cover-up method: $$A = lim{s to 1} frac{s - 4}{s - 3} = frac{1 - 4}{1 - 3} = frac{-3}{-2} = frac{3}{2}$$ $$B = lim{s to 3} frac{s - 4}{s - 1} = frac{3 - 4}{3 - 1} = frac{-1}{2} = -frac{1}{2}$$ Therefore: $$Y1(s) = frac{3}{2} left( frac{1}{s - 1} right) - frac{1}{2} left( frac{1}{s - 3} right)$$ Taking the inverse Laplace transform: $$y1(t) = mathcal{L}^{-1}{Y1(s)} = frac{3}{2} e^t - frac{1}{2} e^{3t} tag

Question 30

A particle is projected from an apse at a distance $sqrt{c}$ from the centre of force with a velocity $sqrt{frac{2lambda}{3} c^3}$ and is moving with central acceleration $lambda(r^5 - c^2 r)$. Find the path of motion of this particle. Will that be the curve $x^4 + y^4 = c^2$?

Accepted Answer

Formulation of the Central Orbit Problem A particle moves in a plane under a central attractive force per unit mass: $$P = lambda(r^5 - c^2 r) tag{1}$$ directed toward the fixed center of force $O$. Let the polar coordinates of the particle be $(r, theta)$ with origin at $O$. The particle is projected from an apse at distance: $$r0 = sqrt{c} tag{2}$$ with velocity: $$v0 = sqrt{frac{2lambda}{3} c^3} tag{3}$$ By definition of an apse, the velocity vector is perpendicular to the radius vector, so $frac{dr}{dt} = 0$, and the areal velocity constant $h$ is: $$h = r0 v0 = sqrt{c} sqrt{frac{2lambda}{3} c^3} = c^2 sqrt{frac{2lambda}{3}} tag{4}$$ $$h^2 = frac{2lambda}{3} c^4 tag{5}$$ --- Step 1: Energy Integral & Evaluation of Constant The velocity $v$ at distance $r$ under a central force $P$ directed toward the center satisfies: $$v^2 = C - 2 int P dr = C - 2lambda int (r^5 - c^2 r) dr$$ $$v^2 = C - 2lambda left( frac{r^6}{6} - frac{c^2 r^2}{2} right) = C - lambda left( frac{r^6}{3} - c^2 r^2 right) tag{6}$$ At the apse $r = r0 = sqrt{c}$, $v = v0$: $$v0^2 = frac{2lambda}{3} c^3$$ Substituting $r = sqrt{c}$ into equation (6): $$v0^2 = C - lambda left( frac{(sqrt{c})^6}{3} - c^2(sqrt{c})^2 right) = C - lambda left( frac{c^3}{3} - c^3 right) = C - lambda left( -frac{2}{3} c^3 right) = C + frac{2lambda}{3} c^3$$ Equating to $v0^2 = frac{2lambda}{3} c^3$: $$frac{2lambda}{3} c^3 = C + frac{2lambda}{3} c^3 implies C = 0 tag{7}$$ Therefore: $$v^2 = -lambda left( frac{r^6}{3} - c^2 r^2 right) = frac{lambda}{3}(3c^2 r^2 - r^6) = frac{lambda r^2}{3}(3c^2 - r^4) tag{8}$$ --- Step 2: Differential Equation of the Path in $u = frac{1}{r}$ The standard differential equation of a central orbit is: $$frac{d^2 u}{dtheta^2} + u = frac{P}{h^2 u^2} tag{9}$$ where $u = frac{1}{r}$, so $r = frac{1}{u}$. Expressing $P$ in terms of $u$: $$P = lambda(r^5 - c^2 r) = lambda left( frac{1}{u^5} - frac{c^2}{u} right) = frac{lambda(1 - c^2 u^4)}{u^5}$$ Substitute $P$ and $h^2 = frac{2lambda}{3} c^4$ into

Maths Optional 2023 Paper I

30 Questions Available

Maths Optional 2023 Paper I

30 Questions Available

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Question 15: Prove that the locus of a line which meets the lines $y = mx, z = c$; $y = -mx, z = -c$ and the circle $x^2 + y^2 = a^2, z = 0$ is $$c^2 m^2 (cy - mzx

Question 16: When a particle is projected from a point $O1$ on the sea level with a velocity $v$ and angle of projection $theta$ with the horizon in a vertical pla

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Question 22: Solve the differential equation: $$frac{d^3y}{dx^3} - 3frac{d^2y}{dx^2} + 4frac{dy}{dx} - 2y = e^x + cos x.$$

Question 23: For a scalar point function $phi$ and vector point function $vec{f}$, prove the identity $$ abla cdot (phi,vec{f}) = ablaphi cdot vec{f} + phi( abla c

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Question 29: Solve the following initial value problem by using Laplace transform technique: $$frac{d^2y}{dt^2} - 4frac{dy}{dt} + 3y(t) = f(t), qquad y(0) = 1, qua

Question 30: A particle is projected from an apse at a distance $sqrt{c}$ from the centre of force with a velocity $sqrt{frac{2lambda}{3} c^3}$ and is moving with