Q: 4. (a) Let $A = begin{pmatrix} 3 & 2 & 4 2 & 0 & 2 4 & 2 & 3 end{pmatrix}$ be a $3 times 3$ matrix. Find the eigenvalues and the corresponding eigenvectors of $A$. Hence find the eigenvalues and the corresponding eigenvectors of $A^{-15}$, where $A^{-15} = (A^{-1})^{15}$.

Spectral Theory and Mathematical Preliminaries Let $A$ be a real symmetric matrix of size $3 times 3$: $$A = begin{pmatrix} 3 & 2 & 4 2 & 0 & 2 4 & 2 & 3 end{pmatrix}$$ Because $A = A^T$, the Spectral Theorem guarantees that all eigenvalues of $A$ are real, and eigenvectors corresponding to distinct eigenvalues are mutually orthogonal. Furthermore, by the Spectral Mapping Theorem, if $lambda$ is an eigenvalue of $A$ with eigenvector $v$, then for any non-singular matrix $A$ and integer $k$: $$A^k v = lambda^k v$$ Specifically, for $k = -15$: $$A^{-15} v = lambda^{-15} v$$ Thus, the eigenvalues of $A^{-15}$ are $lambda^{-15}$, while the eigenvectors remain completely unchanged. --- Step 1: Finding the Characteristic Equation and Eigenvalues of $A$ The characteristic equation of $A$ is: $$det(A - lambda I) = 0$$ $$begin{vmatrix} 3 - lambda & 2 & 4 2 & -lambda & 2 4 & 2 & 3 - lambda end{vmatrix} = 0$$ Using standard invariants for a $3 times 3$ matrix: $$det(A - lambda I) = -lambda^3 + S1 lambda^2 - S2 lambda + S3 = 0$$ where: 1. $S1 = operatorname{Tr}(A) = 3 + 0 + 3 = 6$ 2. $S2 = M{11} + M{22} + M{33}$: - $M{11} = (0)(3) - (2)(2) = -4$ - $M{22} = (3)(3) - (4)(4) = 9 - 16 = -7$ - $M{33} = (3)(0) - (2)(2) = -4$ $$S2 = -4 - 7 - 4 = -15$$ 3. $S3 = det(A)$: $$det(A) = 3(0 - 4) - 2(6 - 8) + 4(4 - 0) = -12 - 2(-2) + 16 = -12 + 4 + 16 = 8$$ Thus, the characteristic polynomial is: $$lambda^3 - 6lambda^2 - 15lambda - 8 = 0$$ To find the roots, test $lambda = -1$: $$(-1)^3 - 6(-1)^2 - 15(-1) - 8 = -1 - 6 + 15 - 8 = 0$$ Thus, $(lambda + 1)$ is a factor. Dividing by $(lambda + 1)$: $$lambda^3 - 6lambda^2 - 15lambda - 8 = (lambda + 1)(lambda^2 - 7lambda - 8) = (lambda + 1)(lambda + 1)(lambda - 8) = (lambda + 1)^2(lambda - 8) = 0$$ The eigenvalues of $A$ are: $$lambda1 = 8, quad lambda2 = -1 ; (text{with algebraic multiplicity } 2)$$ --- Step 2: Determining Eigenvectors of $A$ Case 1: Eigenvector for $lambda1 = 8$ We solve $(A - 8I)X = mathbf{0}$: $$begin{pmatrix} -5 & 2 & 4 2 & -8

Q: 3. (a) Let $V = M{2times 2}(mathbb{R})$ denote a vector space over the field of real numbers. Find the matrix of the linear mapping $phi : V to V$ given by $phi(v) = begin{pmatrix} 1 & 2 3 & -1 end{pmatrix} v$ with respect to standard basis of $M{2times 2}(mathbb{R})$, and hence find the rank of $ph

Mathematical Setting and Framework We consider the real vector space $V = M{2times 2}(mathbb{R})$ of all $2 times 2$ matrices with real entries, which has dimension $dim(V) = 2 times 2 = 4$. The standard ordered basis $mathcal{B} = {E{11}, E{12}, E{21}, E{22}}$ for $M{2times 2}(mathbb{R})$ consists of the elementary matrices: $$E{11} = begin{pmatrix} 1 & 0 0 & 0 end{pmatrix}, quad E{12} = begin{pmatrix} 0 & 1 0 & 0 end{pmatrix}, quad E{21} = begin{pmatrix} 0 & 0 1 & 0 end{pmatrix}, quad E{22} = begin{pmatrix} 0 & 0 0 & 1 end{pmatrix}$$ The linear operator $phi : V to V$ is defined by left-multiplication by a fixed matrix $M$: $$phi(v) = M v, quad text{where } M = begin{pmatrix} 1 & 2 3 & -1 end{pmatrix}$$ --- Step 1: Determining the Images of the Basis Elements We evaluate $phi$ on each of the four basis matrices: 1. For $E{11} = begin{pmatrix} 1 & 0 0 & 0 end{pmatrix}$: $$phi(E{11}) = begin{pmatrix} 1 & 2 3 & -1 end{pmatrix} begin{pmatrix} 1 & 0 0 & 0 end{pmatrix} = begin{pmatrix} 1 & 0 3 & 0 end{pmatrix} = 1 E{11} + 0 E{12} + 3 E{21} + 0 E{22}$$ The coordinate column vector is: $$[phi(E{11})]mathcal{B} = begin{pmatrix} 1 0 3 0 end{pmatrix}$$ 2. For $E{12} = begin{pmatrix} 0 & 1 0 & 0 end{pmatrix}$: $$phi(E{12}) = begin{pmatrix} 1 & 2 3 & -1 end{pmatrix} begin{pmatrix} 0 & 1 0 & 0 end{pmatrix} = begin{pmatrix} 0 & 1 0 & 3 end{pmatrix} = 0 E{11} + 1 E{12} + 0 E{21} + 3 E{22}$$ The coordinate column vector is: $$[phi(E{12})]mathcal{B} = begin{pmatrix} 0 1 0 3 end{pmatrix}$$ 3. For $E{21} = begin{pmatrix} 0 & 0 1 & 0 end{pmatrix}$: $$phi(E{21}) = begin{pmatrix} 1 & 2 3 & -1 end{pmatrix} begin{pmatrix} 0 & 0 1 & 0 end{pmatrix} = begin{pmatrix} 2 & 0 -1 & 0 end{pmatrix} = 2 E{11} + 0 E{12} - 1 E{21} + 0 E{22}$$ The coordinate column vector is: $$[phi(E{21})]mathcal{B} = begin{pmatrix} 2 0 -1 0 end{pmatrix}$$ 4. For $E{22} = begin{pmatrix} 0 & 0 0 & 1 end{pmatrix}$: $$phi(E{22}) = begin{pmatrix} 1 & 2 3 & -1 end{pmatrix} begin{pmatrix} 0 & 0 0 & 1 end{pmatrix} = begin{pm

Q: 3. (c) Find the vertex of the cone $4x^2 - y^2 + 2z^2 + 2xy - 3yz + 12x - 11y + 6z + 4 = 0$.

Mathematical Theory of Cones and Their Vertices The general second-degree equation in three variables: $$F(x, y, z) = ax^2 + by^2 + cz^2 + 2fyz + 2gzx + 2hxy + 2ux + 2vy + 2wz + d = 0$$ represents a cone if and only if there exists a point $V(alpha, beta, gamma)$ (called the vertex) such that shifting the origin to $V$ eliminates all linear terms and the constant term, rendering the equation homogeneous of degree two. To find the vertex systematically, we homogenize the equation by introducing an auxiliary variable $t$: $$Phi(x, y, z, t) = ax^2 + by^2 + cz^2 + 2fyz + 2gzx + 2hxy + 2uxt + 2vyt + 2wzt + dt^2 = 0$$ By Euler's theorem on homogeneous functions, $Phi$ satisfies: $$xfrac{partialPhi}{partial x} + yfrac{partialPhi}{partial y} + zfrac{partialPhi}{partial z} + tfrac{partialPhi}{partial t} = 2Phi$$ The vertex $(alpha, beta, gamma)$ of the cone corresponds to the singular point where the partial derivatives with respect to all four variables vanish simultaneously at $t = 1$: $$left.frac{partialPhi}{partial x}right|{t=1} = 0, quad left.frac{partialPhi}{partial y}right|{t=1} = 0, quad left.frac{partialPhi}{partial z}right|{t=1} = 0, quad left.frac{partialPhi}{partial t}right|{t=1} = 0$$ --- Step 1: Identification of Coefficients and Partial Derivatives The given equation is: $$F(x, y, z) = 4x^2 - y^2 + 2z^2 + 2xy - 3yz + 12x - 11y + 6z + 4 = 0$$ Homogenizing with $t$: $$Phi(x, y, z, t) = 4x^2 - y^2 + 2z^2 + 2xy - 3yz + 12xt - 11yt + 6zt + 4t^2$$ Setting the partial derivatives to zero and dividing by 2 (with $t = 1$): 1. From $frac{1}{2}frac{partialPhi}{partial x} = 0$: $$4x + y + 6 = 0 implies 4x + y = -6 quad text{--- (1)}$$ 2. From $frac{1}{2}frac{partialPhi}{partial y} = 0$: $$-y + x - frac{3}{2}z - frac{11}{2} = 0 implies 2x - 2y - 3z = 11 quad text{--- (2)}$$ 3. From $frac{1}{2}frac{partialPhi}{partial z} = 0$: $$2z - frac{3}{2}y + 3 = 0 implies -3y + 4z = -6 quad text{--- (3)}$$ 4. From $frac{1}{2}frac{partialPhi}{partial t} = 0$: $$6x - frac{11}{2}y + 3z +

Question 1

2. (a) Consider a linear operator $T$ on $mathbb{R}^3$ over $mathbb{R}$ defined by $T(x, y, z) = (2x, 4x - y, 2x + 3y - z)$. Is $T$ invertible? If yes, justify your answer and find $T^{-1}$.

Accepted Answer

Formulation and Definitions We consider the vector space $mathbb{R}^3$ over the field of real numbers $mathbb{R}$. A linear operator $T: mathbb{R}^3 to mathbb{R}^3$ is given by: $$T(x, y, z) = (2x, 4x - y, 2x + 3y - z)$$ A linear operator on a finite-dimensional vector space is invertible (bijective) if and only if any of the following equivalent conditions hold: 1. $det([T]) eq 0$, where $[T]$ is the matrix representation of $T$ with respect to any chosen basis. 2. The null space is trivial: $ker(T) = {mathbf{0}}$. 3. The rank of $T$ equals the dimension of the domain: $operatorname{rank}(T) = 3$. --- Step 1: Matrix Representation of $T$ Let $mathcal{E} = {e1, e2, e3}$ denote the standard ordered basis of $mathbb{R}^3$: $$e1 = (1, 0, 0), quad e2 = (0, 1, 0), quad e3 = (0, 0, 1)$$ Applying the linear transformation $T$ to the standard basis vectors: $$T(e1) = T(1, 0, 0) = (2(1), 4(1) - 0, 2(1) + 3(0) - 0) = (2, 4, 2)$$ $$T(e2) = T(0, 1, 0) = (2(0), 4(0) - 1, 2(0) + 3(1) - 0) = (0, -1, 3)$$ $$T(e3) = T(0, 0, 1) = (2(0), 4(0) - 0, 2(0) + 3(0) - 1) = (0, 0, -1)$$ Forming the matrix $[T]$ with these column vectors: $$[T] = begin{pmatrix} 2 & 0 & 0 4 & -1 & 0 2 & 3 & -1 end{pmatrix}$$ Notice that $[T]$ is a lower triangular matrix. --- Step 2: Testing for Invertibility The determinant of a triangular matrix is the product of its main diagonal entries: $$det([T]) = 2 times (-1) times (-1) = 2$$ Since $det([T]) = 2 eq 0$, the matrix $[T]$ is non-singular. By the fundamental theorem of invertible matrices: - $T$ is one-to-one (injective), - $T$ is onto (surjective), and - Therefore, $T$ is invertible. --- Step 3: Finding $T^{-1}$ Let $(u, v, w) in mathbb{R}^3$ be an arbitrary vector in the codomain such that: $$T(x, y, z) = (u, v, w)$$ This yields the linear system: $$begin{cases} 2x = u & quad text{--- (1)} 4x - y = v & quad text{--- (2)} 2x + 3y - z = w & quad text{--- (3)} end{cases}$$ We solve this system for $(x, y, z)$ in terms of $(u, v, w)$ using forward substitutio

Question 2

2. (b) If $u = (x + y)/(1 - xy)$ and $v = tan^{-1} x + tan^{-1} y$, then find $frac{partial(u, v)}{partial(x, y)}$. Are $u$ and $v$ functionally related? If yes, find the relationship.

Accepted Answer

Theoretical Framework: Functional Dependence and Jacobians Let $u = u(x, y)$ and $v = v(x, y)$ be two continuously differentiable functions of two independent variables $x$ and $y$. The Jacobian determinant of $u$ and $v$ with respect to $x$ and $y$ is defined as: $$J = frac{partial(u, v)}{partial(x, y)} = begin{vmatrix} dfrac{partial u}{partial x} & dfrac{partial u}{partial y} [2ex] dfrac{partial v}{partial x} & dfrac{partial v}{partial y} end{vmatrix} = frac{partial u}{partial x}frac{partial v}{partial y} - frac{partial u}{partial y}frac{partial v}{partial x}$$ Theorem (Cauchy-Jacobian Criterion for Functional Dependence): Two continuously differentiable functions $u(x, y)$ and $v(x, y)$ in a connected domain are functionally dependent (meaning there exists a non-trivial relation $F(u, v) = 0$) if and only if their Jacobian vanishes identically throughout the domain: $$frac{partial(u, v)}{partial(x, y)} equiv 0$$ --- Step 1: Calculating Partial Derivatives of $u$ We are given: $$u = frac{x + y}{1 - xy}$$ 1. Differentiating $u$ partially with respect to $x$ (treating $y$ as constant): Using the quotient rule: $$frac{partial u}{partial x} = frac{(1 - xy)frac{partial}{partial x}(x + y) - (x + y)frac{partial}{partial x}(1 - xy)}{(1 - xy)^2}$$ $$frac{partial u}{partial x} = frac{(1 - xy)(1) - (x + y)(-y)}{(1 - xy)^2} = frac{1 - xy + xy + y^2}{(1 - xy)^2} = frac{1 + y^2}{(1 - xy)^2}$$ 2. Differentiating $u$ partially with respect to $y$ (treating $x$ as constant): By symmetry of $u$ in $x$ and $y$: $$frac{partial u}{partial y} = frac{1 + x^2}{(1 - xy)^2}$$ --- Step 2: Calculating Partial Derivatives of $v$ We are given: $$v = tan^{-1} x + tan^{-1} y$$ 1. Differentiating $v$ partially with respect to $x$: $$frac{partial v}{partial x} = frac{1}{1 + x^2} + 0 = frac{1}{1 + x^2}$$ 2. Differentiating $v$ partially with respect to $y$: $$frac{partial v}{partial y} = 0 + frac{1}{1 + y^2} = frac{1}{1 + y^2}$$ --- Step 3: Evaluating the Jacobian $frac{partial(u, v)}{partial(x, y)

Question 3

1. (e) Find the equation of the right circular cylinder which passes through the circle $x^2 + y^2 + z^2 = 9$, $x - y + z = 3$.

Accepted Answer

Axis: x = -y = z C(1, -1, 1) R = √6 Origin O(0,0,0) d = √3 Plane: x - y + z = 3 Figure: Right circular cylinder of radius $R = sqrt{6}$ coaxial with line $(1, -1, 1)$ through circle center $C(1, -1, 1)$. Geometric Formulation and Core Principles A right circular cylinder is uniquely characterized by: 1. Its axis of symmetry (a line $L$), 2. Its radius $R$. We are given that the cylinder passes through the circle formed by the intersection of: - The sphere: $S: x^2 + y^2 + z^2 = 9$ (centered at $O(0, 0, 0)$ with radius $rho = 3$), - The plane: $Pi: x - y + z = 3$. Because the cylinder is a right circular cylinder and has this circle as a cross-section, the plane of the circle $Pi$ must be perpendicular to the axis of the cylinder. Consequently, the direction of the cylinder's axis is given by the normal vector to the plane $Pi$. --- Step 1: Axis and Radius of the Guiding Circle The normal vector to the plane $x - y + z = 3$ is: $$mathbf{n} = (1, -1, 1)$$ The distance $p$ from the center of the sphere $O(0, 0, 0)$ to the plane $Pi$ is: $$p = frac{|0 - 0 + 0 - 3|}{sqrt{1^2 + (-1)^2 + 1^2}} = frac{3}{sqrt{3}} = sqrt{3}$$ By the Pythagorean theorem for the sphere-plane intersection, the radius $r$ of the guiding circle is: $$r = sqrt{rho^2 - p^2} = sqrt{3^2 - (sqrt{3})^2} = sqrt{9 - 3} = sqrt{6}$$ Since the guiding circle lies in a plane orthogonal to the cylinder's axis, the radius of the right circular cylinder is equal to the radius of this circle: $$R = r = sqrt{6}$$ Now, find the center $C(x0, y0, z0)$ of the circle. The center $C$ is the projection of the sphere's center $O(0, 0, 0)$ onto the plane $Pi$. The line passing through $O$ along normal $mathbf{n}$ is: $$x = t, quad y = -t, quad z = t$$ Substituting into the plane equation $x - y + z = 3$: $$t - (-t) + t = 3 implies 3t = 3 implies t = 1$$ Therefore, the center of the circle is: $$C = (1, -1, 1)$$ --- Step 2: Equation of the Axis of the Cylinder The axis of the cylinder passes through the center $C(1, -1, 1

Question 4

2. (c) Find the image of the line $x = 3 - 6t, y = 2t, z = 3 + 2t$ in the plane $3x + 4y - 5z + 26 = 0$.

Accepted Answer

Problem Formulation and Strategy We are given: - A line $L$ in parametric form: $$x = 3 - 6t, quad y = 2t, quad z = 3 + 2t$$ - A plane $Pi$: $$3x + 4y - 5z + 26 = 0$$ We seek to determine the equation of the reflected line $L'$ (the mirror image of line $L$ in the plane $Pi$). The reflection of a line across a plane is also a straight line. To uniquely determine the image line $L'$, it suffices to find: 1. The point of intersection $P0$ of the line $L$ with the plane $Pi$ (since any point lying on the mirror plane is its own reflection). 2. The reflection $A'$ of any other chosen point $A$ on $L$ across the plane $Pi$. Then $L'$ is the line passing through $P0$ and $A'$. --- Step 1: Intersection Point $P0$ of Line $L$ and Plane $Pi$ Substitute the parametric coordinates of line $L$ into the plane equation: $$3(3 - 6t) + 4(2t) - 5(3 + 2t) + 26 = 0$$ Expanding and collecting like terms: $$9 - 18t + 8t - 15 - 10t + 26 = 0$$ $$(9 - 15 + 26) + (-18 + 8 - 10)t = 0$$ $$20 - 20t = 0 implies 20t = 20 implies t = 1$$ Substitute $t = 1$ back into the parametric equations of the line: $$x = 3 - 6(1) = -3$$ $$y = 2(1) = 2$$ $$z = 3 + 2(1) = 5$$ Thus, the line intersects the plane at the point: $$P0 = (-3, 2, 5)$$ Since $P0$ lies on the plane $Pi$, its reflection is $P0$ itself, so $P0 in L'$. --- Step 2: Reflection of a Second Point $A$ on Line $L$ Choose a convenient point on line $L$ by setting $t = 0$: $$A = (x1, y1, z1) = (3, 0, 3)$$ Let $A'(x', y', z')$ be the reflection of $A$ across the plane $ax + by + cz + d = 0$, where: $$a = 3, quad b = 4, quad c = -5, quad d = 26$$ The standard formula for the reflection $(x', y', z')$ of a point $(x1, y1, z1)$ across a plane $ax + by + cz + d = 0$ is: $$frac{x' - x1}{a} = frac{y' - y1}{b} = frac{z' - z1}{c} = -2 frac{ax1 + by1 + cz1 + d}{a^2 + b^2 + c^2}$$ Let us compute the numerator and denominator: 1. Numerator evaluation: $$ax1 + by1 + cz1 + d = 3(3) + 4(0) - 5(3) + 26 = 9 + 0 - 15 + 26 = 20$$ 2. Denominator evaluation: $$a^2 +

Question 5

1. (a) Let $H$ be a subspace of $mathbb{R}^4$ spanned by the vectors $v1 = (1, -2, 5, -3)$, $v2 = (2, 3, 1, -4)$, $v3 = (3, 8, -3, -5)$. Then find a basis and dimension of $H$, and extend the basis of $H$ to a basis of $mathbb{R}^4$.

Accepted Answer

Problem Formulation and Theoretical Framework We are given three vectors in the 4-dimensional Euclidean space $mathbb{R}^4$: $$v1 = (1, -2, 5, -3), quad v2 = (2, 3, 1, -4), quad v3 = (3, 8, -3, -5)$$ and $H = operatorname{span}{v1, v2, v3}$. We are required to: 1. Determine a basis for the subspace $H$ and compute the dimension $dim(H)$. 2. Extend the basis of $H$ to construct a complete basis for $mathbb{R}^4$. A subset of vectors forms a basis of a vector subspace if they are linearly independent and span the subspace. To extract a basis from a spanning set, we form a matrix whose rows are the spanning vectors and reduce it to Row Echelon Form (REF). The non-zero rows of the echelon matrix (or the corresponding original vectors) will form a linearly independent spanning set, hence a basis. --- Step 1: Finding a Basis and Dimension of $H$ Form the matrix $A$ having $v1, v2, v3$ as its rows: $$A = begin{pmatrix} 1 & -2 & 5 & -3 2 & 3 & 1 & -4 3 & 8 & -3 & -5 end{pmatrix}$$ We now perform elementary row operations to reduce $A$ to row echelon form: 1. Subtract $2R1$ from $R2$ ($R2 to R2 - 2R1$): $$R2 - 2R1 = (2 - 2(1), 3 - 2(-2), 1 - 2(5), -4 - 2(-3)) = (0, 7, -9, 2)$$ 2. Subtract $3R1$ from $R3$ ($R3 to R3 - 3R1$): $$R3 - 3R1 = (3 - 3(1), 8 - 3(-2), -3 - 3(5), -5 - 3(-3)) = (0, 14, -18, 4)$$ After these operations, the matrix becomes: $$begin{pmatrix} 1 & -2 & 5 & -3 0 & 7 & -9 & 2 0 & 14 & -18 & 4 end{pmatrix}$$ 3. Now eliminate the second non-zero entry of row 3 by $R3 to R3 - 2R2$: $$R3 - 2R2 = (0 - 0, 14 - 2(7), -18 - 2(-9), 4 - 2(2)) = (0, 0, 0, 0)$$ The row echelon form of $A$ is: $$operatorname{REF}(A) = begin{pmatrix} 1 & -2 & 5 & -3 0 & 7 & -9 & 2 0 & 0 & 0 & 0 end{pmatrix}$$ Since there are exactly 2 non-zero rows in the row echelon form, the rank of the matrix is 2. This implies: $$v3 = v1 + 2v2$$ Indeed: $$v1 + 2v2 = (1 + 4, -2 + 6, 5 + 2, -3 - 8) = (5, 4, 7, -11) eq v3$$ Wait, let us check the linear relation directly from row operations: $$R3 - 3R1 - 2

Question 6

1. (b) Let $T : mathbb{R}^3 to mathbb{R}^3$ be a linear operator and $B = {v1, v2, v3}$ be a basis of $mathbb{R}^3$ over $mathbb{R}$. Suppose that $T v1 = (1, 1, 0)$, $T v2 = (1, 0, -1)$, $T v3 = (2, 1, -1)$. Find a basis for the range space and null space of $T$.

Accepted Answer

Problem Formulation and Fundamentals We are given a linear operator $T: mathbb{R}^3 to mathbb{R}^3$ defined on a given basis $B = {v1, v2, v3}$ of $mathbb{R}^3$ with the images: $$T(v1) = (1, 1, 0)$$ $$T(v2) = (1, 0, -1)$$ $$T(v3) = (2, 1, -1)$$ Our objective is to determine: 1. A basis for the range space (image) of $T$, denoted $operatorname{Range}(T) = operatorname{Im}(T)$. 2. A basis for the null space (kernel) of $T$, denoted $operatorname{Null}(T) = ker(T)$. --- Step 1: Range Space and its Basis By definition, the range of a linear transformation $T$ is the subspace spanned by the images of the basis vectors: $$operatorname{Range}(T) = operatorname{span}{T(v1), T(v2), T(v3)} = operatorname{span}{(1, 1, 0), (1, 0, -1), (2, 1, -1)}$$ Let us test the linear dependence among these image vectors: Notice that: $$T(v1) + T(v2) = (1, 1, 0) + (1, 0, -1) = (1 + 1, 1 + 0, 0 - 1) = (2, 1, -1) = T(v3)$$ Thus, $T(v3)$ is a linear combination of $T(v1)$ and $T(v2)$: $$T(v3) = T(v1) + T(v2)$$ Consequently: $$operatorname{Range}(T) = operatorname{span}{(1, 1, 0), (1, 0, -1)}$$ Now we check if ${(1, 1, 0), (1, 0, -1)}$ is linearly independent: Suppose $c1(1, 1, 0) + c2(1, 0, -1) = (0, 0, 0)$. Comparing components: $$begin{cases} c1 + c2 = 0 c1 = 0 -c2 = 0 end{cases} implies c1 = 0, ; c2 = 0$$ Hence, the vectors $T(v1)$ and $T(v2)$ are linearly independent. Therefore: - Basis for Range Space: $$mathcal{B}{operatorname{Range}(T)} = {(1, 1, 0), (1, 0, -1)}$$ - Rank of $T$: $$operatorname{rank}(T) = dim(operatorname{Range}(T)) = 2$$ --- Step 2: Null Space and its Basis The null space of $T$ is defined as: $$operatorname{Null}(T) = {v in mathbb{R}^3 : T(v) = mathbf{0}}$$ By the Rank-Nullity Theorem (Dimension Theorem for Linear Transformations): $$dim(mathbb{R}^3) = operatorname{rank}(T) + operatorname{nullity}(T)$$ $$3 = 2 + operatorname{nullity}(T) implies operatorname{nullity}(T) = 1$$ Thus, the null space has dimension 1, which means any non-zero vector in $ker(T)$ will form a b

Question 7

1. (c) Discuss the continuity of the function $$f(x) = begin{cases} frac{1}{1 - e^{-1/x}}, & x eq 0 0, & x = 0 end{cases}$$ for all values of $x$.

Accepted Answer

Problem Statement We are tasked with investigating the continuity of the real-valued function: $$f(x) = begin{cases} dfrac{1}{1 - e^{-1/x}}, & x eq 0 [2ex] 0, & x = 0 end{cases}$$ across its entire domain $mathbb{R} = (-infty, infty)$. --- Part 1: Continuity for $x eq 0$ Let $x in mathbb{R}$ such that $x eq 0$. The component functions are: 1. $g(x) = -dfrac{1}{x}$, which is a rational function continuous on $(-infty, 0) cup (0, infty)$. 2. The natural exponential function $h(u) = e^u$, which is continuous and strictly positive everywhere on $mathbb{R}$. 3. Therefore, their composition $h(g(x)) = e^{-1/x}$ is continuous for all $x eq 0$. Next, consider the denominator $D(x) = 1 - e^{-1/x}$. For the quotient $frac{1}{D(x)}$ to be continuous, we must verify that $D(x) eq 0$ for all real $x eq 0$: $$1 - e^{-1/x} = 0 iff e^{-1/x} = 1 iff -frac{1}{x} = 0$$ Since $-frac{1}{x} = 0$ has no finite real solution for $x in mathbb{R} setminus {0}$, the denominator $1 - e^{-1/x}$ is never zero on $(-infty, 0) cup (0, infty)$. Specifically: - For $x 0$: $frac{1}{x} 0 implies -frac{1}{x} 0$. - For $x 0 implies e^{-1/x} 1$, so $1 - e^{-1/x} 0$): $$frac{1}{x} to +infty implies -frac{1}{x} to -infty$$ Since $lim{u to -infty} e^u = 0$, we have: $$lim{x to 0^+} e^{-1/x} = 0$$ Substituting this into the expression for $f(x)$: $$lim{x to 0^+} f(x) = lim{x to 0^+} frac{1}{1 - e^{-1/x}} = frac{1}{1 - 0} = 1$$ Left-Hand Limit (LHL) as $x to 0^-$: As $x to 0$ from the left ($x 0$ and $h to 0$. Then: $$-frac{1}{x} = -frac{1}{-h} = frac{1}{h} to +infty$$ Since $lim{u to +infty} e^u = +infty$, we have: $$lim{x to 0^-} e^{-1/x} = +infty$$ Substituting this into the expression for $f(x)$: $$lim{x to 0^-} f(x) = lim{h to 0^+} frac{1}{1 - e^{1/h}} = 0$$ Comparison and Nature of Discontinuity: We observe: $$lim{x to 0^+} f(x) = 1$$ $$lim{x to 0^-} f(x) = 0$$ $$f(0) = 0$$ Although the left-hand limit equals the function value (so $f$ is left-continuous at $0$): $$lim{x to 0^-} f(x) = f(0) = 0$$ the tw

Question 8

1. (d) Expand $ln(x)$ in powers of $(x - 1)$ by Taylor's theorem and hence find the value of $ln(1.1)$ correct up to four decimal places.

Accepted Answer

Theoretical Foundation: Taylor's Theorem Taylor's theorem states that if a function $f(x)$ possesses continuous derivatives up to order $n$ on an interval containing the point $a$, then for any $x$ in that interval: $$f(x) = f(a) + f'(a)(x - a) + frac{f''(a)}{2!}(x - a)^2 + frac{f'''(a)}{3!}(x - a)^3 + cdots + frac{f^{(n)}(a)}{n!}(x - a)^n + Rn(x)$$ where $Rn(x)$ is the remainder term. When $f(x)$ is infinitely differentiable and the remainder tends to zero as $n to infty$, $f(x)$ is represented by its Taylor series expansion about $x = a$: $$f(x) = sum{k=0}^{infty} frac{f^{(k)}(a)}{k!}(x - a)^k$$ Here, we are given $f(x) = ln(x)$, and we wish to expand in powers of $(x - 1)$, meaning $a = 1$. --- Step 1: Successive Differentiation and Evaluation at $a = 1$ Let us compute the derivatives of $f(x) = ln(x)$: 1. $f(x) = ln(x) implies f(1) = ln(1) = 0$ 2. $f'(x) = x^{-1} implies f'(1) = 1$ 3. $f''(x) = -x^{-2} implies f''(1) = -1$ 4. $f'''(x) = 2x^{-3} implies f'''(1) = 2 = 2!$ 5. $f^{(4)}(x) = -6x^{-4} implies f^{(4)}(1) = -6 = -3!$ 6. $f^{(5)}(x) = 24x^{-5} implies f^{(5)}(1) = 24 = 4!$ By mathematical induction, the $n$-th derivative for $n ge 1$ is: $$f^{(n)}(x) = (-1)^{n-1} (n - 1)! x^{-n}$$ Evaluating at $x = 1$: $$f^{(n)}(1) = (-1)^{n-1} (n - 1)!$$ --- Step 2: Constructing the Taylor Series Expansion Substituting the values of the derivatives into Taylor's formula about $a = 1$: $$f(x) = f(1) + sum{n=1}^{infty} frac{f^{(n)}(1)}{n!} (x - 1)^n$$ $$ln(x) = 0 + sum{n=1}^{infty} frac{(-1)^{n-1} (n - 1)!}{n!} (x - 1)^n$$ Since $frac{(n - 1)!}{n!} = frac{1}{n}$, we obtain: $$ln(x) = sum{n=1}^{infty} frac{(-1)^{n-1}}{n} (x - 1)^n$$ Writing out the first few terms explicitly: $$ln(x) = (x - 1) - frac{(x - 1)^2}{2} + frac{(x - 1)^3}{3} - frac{(x - 1)^4}{4} + frac{(x - 1)^5}{5} - cdots$$ This series is valid for $|x - 1| < 1$, which corresponds to the interval of convergence $0 < x le 2$. --- Step 3: Computing $ln(1.1)$ Correct to Four Decimal Places Substitute $x = 1.1$ in

Question 9

4. (a) Let $A = begin{pmatrix} 3 & 2 & 4 2 & 0 & 2 4 & 2 & 3 end{pmatrix}$ be a $3 times 3$ matrix. Find the eigenvalues and the corresponding eigenvectors of $A$. Hence find the eigenvalues and the corresponding eigenvectors of $A^{-15}$, where $A^{-15} = (A^{-1})^{15}$.

Accepted Answer

Spectral Theory and Mathematical Preliminaries Let $A$ be a real symmetric matrix of size $3 times 3$: $$A = begin{pmatrix} 3 & 2 & 4 2 & 0 & 2 4 & 2 & 3 end{pmatrix}$$ Because $A = A^T$, the Spectral Theorem guarantees that all eigenvalues of $A$ are real, and eigenvectors corresponding to distinct eigenvalues are mutually orthogonal. Furthermore, by the Spectral Mapping Theorem, if $lambda$ is an eigenvalue of $A$ with eigenvector $v$, then for any non-singular matrix $A$ and integer $k$: $$A^k v = lambda^k v$$ Specifically, for $k = -15$: $$A^{-15} v = lambda^{-15} v$$ Thus, the eigenvalues of $A^{-15}$ are $lambda^{-15}$, while the eigenvectors remain completely unchanged. --- Step 1: Finding the Characteristic Equation and Eigenvalues of $A$ The characteristic equation of $A$ is: $$det(A - lambda I) = 0$$ $$begin{vmatrix} 3 - lambda & 2 & 4 2 & -lambda & 2 4 & 2 & 3 - lambda end{vmatrix} = 0$$ Using standard invariants for a $3 times 3$ matrix: $$det(A - lambda I) = -lambda^3 + S1 lambda^2 - S2 lambda + S3 = 0$$ where: 1. $S1 = operatorname{Tr}(A) = 3 + 0 + 3 = 6$ 2. $S2 = M{11} + M{22} + M{33}$: - $M{11} = (0)(3) - (2)(2) = -4$ - $M{22} = (3)(3) - (4)(4) = 9 - 16 = -7$ - $M{33} = (3)(0) - (2)(2) = -4$ $$S2 = -4 - 7 - 4 = -15$$ 3. $S3 = det(A)$: $$det(A) = 3(0 - 4) - 2(6 - 8) + 4(4 - 0) = -12 - 2(-2) + 16 = -12 + 4 + 16 = 8$$ Thus, the characteristic polynomial is: $$lambda^3 - 6lambda^2 - 15lambda - 8 = 0$$ To find the roots, test $lambda = -1$: $$(-1)^3 - 6(-1)^2 - 15(-1) - 8 = -1 - 6 + 15 - 8 = 0$$ Thus, $(lambda + 1)$ is a factor. Dividing by $(lambda + 1)$: $$lambda^3 - 6lambda^2 - 15lambda - 8 = (lambda + 1)(lambda^2 - 7lambda - 8) = (lambda + 1)(lambda + 1)(lambda - 8) = (lambda + 1)^2(lambda - 8) = 0$$ The eigenvalues of $A$ are: $$lambda1 = 8, quad lambda2 = -1 ; (text{with algebraic multiplicity } 2)$$ --- Step 2: Determining Eigenvectors of $A$ Case 1: Eigenvector for $lambda1 = 8$ We solve $(A - 8I)X = mathbf{0}$: $$begin{pmatrix} -5 & 2 & 4 2 & -8

Question 10

3. (a) Let $V = M{2times 2}(mathbb{R})$ denote a vector space over the field of real numbers. Find the matrix of the linear mapping $phi : V to V$ given by $phi(v) = begin{pmatrix} 1 & 2 3 & -1 end{pmatrix} v$ with respect to standard basis of $M{2times 2}(mathbb{R})$, and hence find the rank of $ph

Accepted Answer

Mathematical Setting and Framework We consider the real vector space $V = M{2times 2}(mathbb{R})$ of all $2 times 2$ matrices with real entries, which has dimension $dim(V) = 2 times 2 = 4$. The standard ordered basis $mathcal{B} = {E{11}, E{12}, E{21}, E{22}}$ for $M{2times 2}(mathbb{R})$ consists of the elementary matrices: $$E{11} = begin{pmatrix} 1 & 0 0 & 0 end{pmatrix}, quad E{12} = begin{pmatrix} 0 & 1 0 & 0 end{pmatrix}, quad E{21} = begin{pmatrix} 0 & 0 1 & 0 end{pmatrix}, quad E{22} = begin{pmatrix} 0 & 0 0 & 1 end{pmatrix}$$ The linear operator $phi : V to V$ is defined by left-multiplication by a fixed matrix $M$: $$phi(v) = M v, quad text{where } M = begin{pmatrix} 1 & 2 3 & -1 end{pmatrix}$$ --- Step 1: Determining the Images of the Basis Elements We evaluate $phi$ on each of the four basis matrices: 1. For $E{11} = begin{pmatrix} 1 & 0 0 & 0 end{pmatrix}$: $$phi(E{11}) = begin{pmatrix} 1 & 2 3 & -1 end{pmatrix} begin{pmatrix} 1 & 0 0 & 0 end{pmatrix} = begin{pmatrix} 1 & 0 3 & 0 end{pmatrix} = 1 E{11} + 0 E{12} + 3 E{21} + 0 E{22}$$ The coordinate column vector is: $$[phi(E{11})]mathcal{B} = begin{pmatrix} 1 0 3 0 end{pmatrix}$$ 2. For $E{12} = begin{pmatrix} 0 & 1 0 & 0 end{pmatrix}$: $$phi(E{12}) = begin{pmatrix} 1 & 2 3 & -1 end{pmatrix} begin{pmatrix} 0 & 1 0 & 0 end{pmatrix} = begin{pmatrix} 0 & 1 0 & 3 end{pmatrix} = 0 E{11} + 1 E{12} + 0 E{21} + 3 E{22}$$ The coordinate column vector is: $$[phi(E{12})]mathcal{B} = begin{pmatrix} 0 1 0 3 end{pmatrix}$$ 3. For $E{21} = begin{pmatrix} 0 & 0 1 & 0 end{pmatrix}$: $$phi(E{21}) = begin{pmatrix} 1 & 2 3 & -1 end{pmatrix} begin{pmatrix} 0 & 0 1 & 0 end{pmatrix} = begin{pmatrix} 2 & 0 -1 & 0 end{pmatrix} = 2 E{11} + 0 E{12} - 1 E{21} + 0 E{22}$$ The coordinate column vector is: $$[phi(E{21})]mathcal{B} = begin{pmatrix} 2 0 -1 0 end{pmatrix}$$ 4. For $E{22} = begin{pmatrix} 0 & 0 0 & 1 end{pmatrix}$: $$phi(E{22}) = begin{pmatrix} 1 & 2 3 & -1 end{pmatrix} begin{pmatrix} 0 & 0 0 & 1 end{pmatrix} = begin{pm

Question 11

4. (b) Using double integration, find the area lying inside the cardioid $r = a(1 + costheta)$ and outside the circle $r = a$.

Accepted Answer

Problem Setting and Geometric Description We are required to determine the area of the region $R$ in polar coordinates that satisfies: 1. Inside the cardioid: $r le a(1 + costheta)$ 2. Outside the circle: $r ge a$ Here $a 0$ is a positive constant. The cardioid $r = a(1 + costheta)$ is symmetric about the initial line $theta = 0$. --- θ = 0 θ = π/2 θ = -π/2 O Circle: r = a Cardioid: r = a(1 + cos θ) (a, π/2) (a, -π/2) (2a, 0) Area = a²(2 + π/4) Figure: Shaded region lying inside cardioid $r = a(1 + cos heta)$ and outside circle $r = a$. Step 1: Finding Points of Intersection To establish the angular limits of integration, equate the two boundary curves: $$a(1 + costheta) = a$$ $$1 + costheta = 1 implies costheta = 0$$ In the principal interval $[-pi, pi]$, $costheta = 0$ yields: $$theta = -frac{pi}{2} quad text{and} quad theta = frac{pi}{2}$$ For $theta in left(-frac{pi}{2}, frac{pi}{2}right)$, $costheta 0$, so: $$a(1 + costheta) a$$ Thus, the cardioid lies strictly outside the circle in the interval $theta in left(-frac{pi}{2}, frac{pi}{2}right)$. --- Step 2: Setting up the Double Integral In polar coordinates, the area element is: $$dA = r , dr , dtheta$$ By symmetry of both curves about the horizontal axis (initial line $theta = 0$), the total area is twice the area in the upper half-plane: $$text{Area} = 2 iint{R^+} r , dr , dtheta$$ where the upper-half region $R^+$ is defined by: $$0 le theta le frac{pi}{2}, quad a le r le a(1 + costheta)$$ Hence: $$text{Area} = 2 int{0}^{pi/2} left( int{a}^{a(1 + costheta)} r , dr right) dtheta$$ --- Step 3: Evaluating the Inner Integral with Respect to $r$ Evaluating the inner integral: $$int{a}^{a(1 + costheta)} r , dr = left[ frac{r^2}{2} right]{a}^{a(1 + costheta)} = frac{1}{2} left[ a^2(1 + costheta)^2 - a^2 right]$$ $$= frac{a^2}{2} left[ (1 + 2costheta + cos^2theta) - 1 right] = frac{a^2}{2} left( 2costheta + cos^2theta right)$$ --- Step 4: Evaluating the Outer Integral with Respect to $theta$ Now substitute this back

Question 12

3. (b) Find the volume of the greatest cylinder which can be inscribed in a cone of height $h$ and semi-vertical angle $alpha$.

Accepted Answer

Geometric Setup and Formulation Consider a right circular cone of: - Height $h$, - Semi-vertical angle $alpha$, - Apex $V$ at the origin or top, and a circular base of radius $R = h tanalpha$. Let a right circular cylinder be inscribed inside this cone such that its base rests on the base of the cone and its upper circular edge touches the curved interior surface of the cone. Let: - $r$ be the radius of the inscribed cylinder ($0 V (Apex) α y = h/3 r = ⅔R R = h tan α Height h Figure: Maximum volume cylinder inscribed in a cone of height $h$ and semi-vertical angle $lpha$. Step 1: Geometric Constraint via Similar Triangles Consider a vertical cross-section through the common axis of the cone and the cylinder. The cone cross-section is an isosceles triangle with vertical altitude $h$ and semi-base $R = htanalpha$. The cylinder intersects this cross-section as a rectangle of width $2r$ and height $y$. The upper portion of the cone, above the flat top of the cylinder, is a smaller cone of height $(h - y)$ and base radius $r$. By the similarity of the upper small cone triangle and the entire cone triangle: $$frac{h - y}{h} = frac{r}{R}$$ Substituting $R = htanalpha$: $$frac{h - y}{h} = frac{r}{htanalpha}$$ $$1 - frac{y}{h} = frac{r}{htanalpha} implies y = h - rcotalpha = hleft(1 - frac{r}{htanalpha}right)$$ --- Step 2: The Volume Function The volume $V$ of a cylinder of radius $r$ and height $y$ is given by: $$V = pi r^2 y$$ Substituting the expression for $y$ in terms of $r$: $$V(r) = pi r^2 left(h - rcotalpharight) = pi h r^2 - pi cotalpha , r^3$$ This is a cubic function in $r$ defined on the physical domain $r in (0, htanalpha)$. --- Step 3: Finding the Critical Points (First Derivative Test) To find the radius that maximizes the volume, differentiate $V(r)$ with respect to $r$: $$frac{dV}{dr} = frac{d}{dr}left(pi h r^2 - pi cotalpha , r^3right) = 2pi h r - 3pi cotalpha , r^2$$ Setting the first derivative to zero for stationary points: $$2pi h r - 3pi cotalpha , r^

Question 13

3. (c) Find the vertex of the cone $4x^2 - y^2 + 2z^2 + 2xy - 3yz + 12x - 11y + 6z + 4 = 0$.

Accepted Answer

Mathematical Theory of Cones and Their Vertices The general second-degree equation in three variables: $$F(x, y, z) = ax^2 + by^2 + cz^2 + 2fyz + 2gzx + 2hxy + 2ux + 2vy + 2wz + d = 0$$ represents a cone if and only if there exists a point $V(alpha, beta, gamma)$ (called the vertex) such that shifting the origin to $V$ eliminates all linear terms and the constant term, rendering the equation homogeneous of degree two. To find the vertex systematically, we homogenize the equation by introducing an auxiliary variable $t$: $$Phi(x, y, z, t) = ax^2 + by^2 + cz^2 + 2fyz + 2gzx + 2hxy + 2uxt + 2vyt + 2wzt + dt^2 = 0$$ By Euler's theorem on homogeneous functions, $Phi$ satisfies: $$xfrac{partialPhi}{partial x} + yfrac{partialPhi}{partial y} + zfrac{partialPhi}{partial z} + tfrac{partialPhi}{partial t} = 2Phi$$ The vertex $(alpha, beta, gamma)$ of the cone corresponds to the singular point where the partial derivatives with respect to all four variables vanish simultaneously at $t = 1$: $$left.frac{partialPhi}{partial x}right|{t=1} = 0, quad left.frac{partialPhi}{partial y}right|{t=1} = 0, quad left.frac{partialPhi}{partial z}right|{t=1} = 0, quad left.frac{partialPhi}{partial t}right|{t=1} = 0$$ --- Step 1: Identification of Coefficients and Partial Derivatives The given equation is: $$F(x, y, z) = 4x^2 - y^2 + 2z^2 + 2xy - 3yz + 12x - 11y + 6z + 4 = 0$$ Homogenizing with $t$: $$Phi(x, y, z, t) = 4x^2 - y^2 + 2z^2 + 2xy - 3yz + 12xt - 11yt + 6zt + 4t^2$$ Setting the partial derivatives to zero and dividing by 2 (with $t = 1$): 1. From $frac{1}{2}frac{partialPhi}{partial x} = 0$: $$4x + y + 6 = 0 implies 4x + y = -6 quad text{--- (1)}$$ 2. From $frac{1}{2}frac{partialPhi}{partial y} = 0$: $$-y + x - frac{3}{2}z - frac{11}{2} = 0 implies 2x - 2y - 3z = 11 quad text{--- (2)}$$ 3. From $frac{1}{2}frac{partialPhi}{partial z} = 0$: $$2z - frac{3}{2}y + 3 = 0 implies -3y + 4z = -6 quad text{--- (3)}$$ 4. From $frac{1}{2}frac{partialPhi}{partial t} = 0$: $$6x - frac{11}{2}y + 3z +

Question 14

4. (c) Find the equation of the sphere which touches the plane $3x + 2y - z + 2 = 0$ at the point $(1, -2, 1)$ and cuts orthogonally the sphere $x^2 + y^2 + z^2 - 4x + 6y + 4 = 0$.

Accepted Answer

Problem Statement and Formulation We seek the equation of a sphere $S$ satisfying two conditions: 1. $S$ touches the given plane $Pi: 3x + 2y - z + 2 = 0$ at the specific point of contact $A(1, -2, 1)$. 2. $S$ cuts the sphere $S2: x^2 + y^2 + z^2 - 4x + 6y + 4 = 0$ orthogonally. First, verify that $A(1, -2, 1)$ lies on plane $Pi$: $$3(1) + 2(-2) - 1 + 2 = 3 - 4 - 1 + 2 = 0$$ Indeed, $A in Pi$. --- Step 1: Family of Spheres Touching the Plane at $A(1, -2, 1)$ A sphere touching the plane $Pi$ at point $A$ must have its center along the normal line to $Pi$ passing through $A$. The normal vector to the plane $Pi$ is: $$mathbf{n} = (3, 2, -1)$$ Alternatively, using the family of spheres touching a plane at a point, any such sphere can be represented in the form: $$(x - x1)^2 + (y - y1)^2 + (z - z1)^2 + lambda (ax + by + cz + d) = 0$$ where $(x - x1)^2 + (y - y1)^2 + (z - z1)^2 = 0$ represents the point sphere at $(1, -2, 1)$. Substituting the point $(1, -2, 1)$ and the plane equation: $$(x - 1)^2 + (y + 2)^2 + (z - 1)^2 + lambda (3x + 2y - z + 2) = 0$$ Expanding this equation: $$(x^2 - 2x + 1) + (y^2 + 4y + 4) + (z^2 - 2z + 1) + lambda (3x + 2y - z + 2) = 0$$ $$x^2 + y^2 + z^2 + (3lambda - 2)x + (2lambda + 4)y - (lambda + 2)z + (6 + 2lambda) = 0 quad text{--- (1)}$$ --- Step 2: Condition for Orthogonality of Two Spheres Recall that two spheres: $$S1: x^2 + y^2 + z^2 + 2u1 x + 2v1 y + 2w1 z + d1 = 0$$ $$S2: x^2 + y^2 + z^2 + 2u2 x + 2v2 y + 2w2 z + d2 = 0$$ cut each other orthogonally if and only if: $$2u1 u2 + 2v1 v2 + 2w1 w2 = d1 + d2$$ From our family of spheres (Equation 1): $$2u1 = 3lambda - 2 implies u1 = frac{3lambda - 2}{2}$$ $$2v1 = 2lambda + 4 implies v1 = lambda + 2$$ $$2w1 = -(lambda + 2) implies w1 = -frac{lambda + 2}{2}$$ $$d1 = 6 + 2lambda$$ From the given sphere $S2$: $$x^2 + y^2 + z^2 - 4x + 6y + 4 = 0$$ we have: $$2u2 = -4 implies u2 = -2$$ $$2v2 = 6 implies v2 = 3$$ $$2w2 = 0 implies w2 = 0$$ $$d2 = 4$$ --- Step 3: Determining the Parameter $lambda$ Sub

Question 15

5. (a) Find the orthogonal trajectories of the family of curves $r = c(sectheta + tantheta)$, where $c$ is a parameter.

Accepted Answer

Theory of Orthogonal Trajectories in Polar Coordinates Let a family of plane curves be given in polar coordinates by: $$F(r, theta, c) = 0$$ where $c$ is an arbitrary parameter. Two curves in polar coordinates are orthogonal at an intersection point if their tangents at that point are perpendicular. If $phi1$ and $phi2$ denote the angles that the tangents make with the radius vector, then: $$phi2 = phi1 pm frac{pi}{2} implies tanphi2 = -cotphi1 = -frac{1}{tanphi1}$$ In polar coordinates, the angle $phi$ between the radius vector and the tangent is given by: $$tanphi = rfrac{dtheta}{dr} = frac{r}{frac{dr}{dtheta}}$$ Thus, the differential equation of the orthogonal trajectories is obtained by replacing: $$rfrac{dtheta}{dr} quad text{with} quad -frac{1}{r}frac{dr}{dtheta}$$ or equivalently replacing $frac{dr}{dtheta}$ with $-r^2frac{dtheta}{dr}$. --- Step 1: Formulating the Differential Equation of the Given Family The given family of curves is: $$r = c(sectheta + tantheta) quad text{--- (1)}$$ To eliminate the parameter $c$, we first take the natural logarithm of both sides: $$ln r = ln c + ln(sectheta + tantheta)$$ Differentiating both sides with respect to $theta$: $$frac{1}{r}frac{dr}{dtheta} = 0 + frac{frac{d}{dtheta}(sectheta + tantheta)}{sectheta + tantheta}$$ Recall the standard derivatives of trigonometric functions: $$frac{d}{dtheta}(sectheta) = secthetatantheta, quad frac{d}{dtheta}(tantheta) = sec^2theta$$ Therefore: $$frac{1}{r}frac{dr}{dtheta} = frac{secthetatantheta + sec^2theta}{sectheta + tantheta} = frac{sectheta(tantheta + sectheta)}{sectheta + tantheta} = sectheta$$ Thus, the differential equation governing the given family of curves is: $$frac{1}{r}frac{dr}{dtheta} = sectheta quad text{--- (2)}$$ --- Step 2: Setting up the Differential Equation for the Orthogonal Family To obtain the differential equation of the orthogonal trajectories, replace: $$frac{dr}{dtheta} longrightarrow -r^2frac{dtheta}{dr}$$ Substituting into Equation (2): $$frac{1}{r}le

Question 16

5. (b) Solve the integral equation $y(t) = cos t + int0^t y(x)cos(t - x),dx$ using Laplace transform.

Accepted Answer

Theory of Volterra Integral Equations and Laplace Convolution The given equation: $$y(t) = cos t + int0^t y(x)cos(t - x),dx$$ is a Volterra integral equation of the second kind of convolution type. The convolution of two functions $f(t)$ and $g(t)$ defined for $t ge 0$ is: $$(f g)(t) = int0^t f(x)g(t - x),dx$$ By the Laplace Convolution Theorem: $$mathcal{L}{(f g)(t)} = F(s) cdot G(s)$$ where $F(s) = mathcal{L}{f(t)}$ and $G(s) = mathcal{L}{g(t)}$. The integral in the given equation is precisely the convolution of $y(t)$ and $cos t$: $$int0^t y(x)cos(t - x),dx = (y cos)(t)$$ Thus, the equation can be written compactly as: $$y(t) = cos t + (y cos)(t)$$ --- Step 1: Taking the Laplace Transform Apply the Laplace transform $mathcal{L}$ to both sides of the equation: $$mathcal{L}{y(t)} = mathcal{L}{cos t} + mathcal{L}{(y cos)(t)}$$ Let $Y(s) = mathcal{L}{y(t)}$. Recall the standard transform: $$mathcal{L}{cos(at)} = frac{s}{s^2 + a^2} implies mathcal{L}{cos t} = frac{s}{s^2 + 1}$$ Using the convolution theorem: $$mathcal{L}{(y cos)(t)} = Y(s) cdot frac{s}{s^2 + 1}$$ Substituting these expressions: $$Y(s) = frac{s}{s^2 + 1} + Y(s) cdot frac{s}{s^2 + 1}$$ --- Step 2: Solving for $Y(s)$ in the Frequency Domain Isolate $Y(s)$ on the left-hand side: $$Y(s) - Y(s) left(frac{s}{s^2 + 1}right) = frac{s}{s^2 + 1}$$ $$Y(s) left( 1 - frac{s}{s^2 + 1} right) = frac{s}{s^2 + 1}$$ $$Y(s) left( frac{s^2 - s + 1}{s^2 + 1} right) = frac{s}{s^2 + 1}$$ Assuming $s^2 + 1 eq 0$, multiply both sides by $(s^2 + 1)$: $$Y(s) (s^2 - s + 1) = s$$ $$Y(s) = frac{s}{s^2 - s + 1}$$ --- Step 3: Inverse Laplace Transform The denominator is an irreducible quadratic over the reals: $$s^2 - s + 1 = left(s - frac{1}{2}right)^2 + frac{3}{4} = left(s - frac{1}{2}right)^2 + left(frac{sqrt{3}}{2}right)^2$$ Now rewrite the numerator in terms of $left(s - frac{1}{2}right)$: $$s = left(s - frac{1}{2}right) + frac{1}{2} = left(s - frac{1}{2}right) + frac{1}{sqrt{3}} left(frac{sqrt{3}}{2}right)$$ Thus: $$Y(s) = frac

Question 17

5. (c) At any time $t$ (in seconds), the coterminous edges of a variable parallelepiped are represented by the vectors $vec{alpha} = that{i} + (t + 1)hat{j} + (2t + 1)hat{k}$, $vec{beta} = 2that{i} + (3t - 1)hat{j} + that{k}$, $vec{gamma} = hat{i} + 3that{j} + hat{k}$. What is the rate of change of

Accepted Answer

Mathematical Preliminaries and Vector Formulations We are given three time-dependent vector functions representing the coterminous edges of a variable parallelepiped: $$vec{alpha}(t) = that{i} + (t + 1)hat{j} + (2t + 1)hat{k}$$ $$vec{beta}(t) = 2that{i} + (3t - 1)hat{j} + that{k}$$ $$vec{gamma}(t) = hat{i} + 3that{j} + hat{k}$$ 1. The vectorial area $vec{A}(t)$ of the parallelogram formed by the coterminous edges $vec{alpha}$ and $vec{gamma}$ is defined as the cross product: $$vec{A}(t) = vec{alpha}(t) times vec{gamma}(t)$$ The rate of change of the vectorial area is the time derivative $frac{dvec{A}}{dt}$. 2. The volume $V(t)$ of the parallelepiped formed by coterminous edges $vec{alpha}, vec{beta}, vec{gamma}$ is given by the scalar triple product: $$V(t) = |[vec{alpha}, vec{beta}, vec{gamma}]| = |(vec{alpha} times vec{gamma}) cdot vec{beta}|$$ --- Step 1: Determining the Vectorial Area $vec{A}(t)$ Evaluate the cross product $vec{alpha} times vec{gamma}$: $$vec{A}(t) = begin{vmatrix} hat{i} & hat{j} & hat{k} t & t+1 & 2t+1 1 & 3t & 1 end{vmatrix}$$ Expanding by components: 1. $hat{i}$-component: $$Ax = (t + 1)(1) - (2t + 1)(3t) = t + 1 - (6t^2 + 3t) = -6t^2 - 2t + 1$$ 2. $hat{j}$-component: $$Ay = -[t(1) - (2t + 1)(1)] = -(t - 2t - 1) = -(-t - 1) = t + 1$$ 3. $hat{k}$-component: $$Az = t(3t) - (t + 1)(1) = 3t^2 - t - 1$$ Thus, the vectorial area is: $$vec{A}(t) = (-6t^2 - 2t + 1)hat{i} + (t + 1)hat{j} + (3t^2 - t - 1)hat{k}$$ --- Step 2: Rate of Change of the Vectorial Area $frac{dvec{A}}{dt}$ Differentiating each component with respect to time $t$: $$frac{dvec{A}}{dt} = frac{d}{dt}(-6t^2 - 2t + 1)hat{i} + frac{d}{dt}(t + 1)hat{j} + frac{d}{dt}(3t^2 - t - 1)hat{k}$$ $$frac{dvec{A}}{dt} = (-12t - 2)hat{i} + hat{j} + (6t - 1)hat{k}$$ At any time $t$, this gives the exact rate of change of the vectorial area. In particular, at $t = 1$: $$left.frac{dvec{A}}{dt}right|{t=1} = (-12(1) - 2)hat{i} + hat{j} + (6(1) - 1)hat{k} = -14hat{i} + hat{j} + 5hat{k}$$ --- Step 3: Det

Question 18

5. (e) (i) Let $C$ be a plane curve $vec{r}(t) = f(t)hat{i} + g(t)hat{j}$, where $f$ and $g$ have second-order derivatives. Show that the curvature at a point is given by $$kappa = frac{|f'(t)g''(t) - g'(t)f''(t)|}{([f'(t)]^2 + [g'(t)]^2)^{3/2}}$$ What is the value of torsion $tau$ at any point of t

Accepted Answer

Theory of Curves in Differential Geometry Let a regular curve $C$ in Euclidean 3-space be parameterized by an arbitrary parameter $t$: $$vec{r}(t) = x(t)hat{i} + y(t)hat{j} + z(t)hat{k}$$ Let primes denote derivatives with respect to $t$. The velocity vector is $vec{r}'(t)$, and the acceleration vector is $vec{r}''(t)$. The speed along the curve is: $$v = frac{ds}{dt} = |vec{r}'(t)|$$ By the Frenet-Serret apparatus, the unit tangent vector is $vec{t} = frac{vec{r}'}{|vec{r}'|}$. Differentiating with respect to arc length $s$: $$frac{dvec{t}}{ds} = kappa vec{n}$$ In terms of the general parameter $t$: $$kappa = frac{|vec{r}'(t) times vec{r}''(t)|}{|vec{r}'(t)|^3}$$ --- Step 1: Evaluating Velocity and Acceleration Vectors For the given plane curve in the $xy$-plane: $$vec{r}(t) = f(t)hat{i} + g(t)hat{j} + 0hat{k}$$ 1. The first derivative (velocity vector) is: $$vec{r}'(t) = f'(t)hat{i} + g'(t)hat{j}$$ Its magnitude (speed) is: $$|vec{r}'(t)| = sqrt{[f'(t)]^2 + [g'(t)]^2}$$ Thus: $$|vec{r}'(t)|^3 = left( [f'(t)]^2 + [g'(t)]^2 right)^{3/2}$$ 2. The second derivative (acceleration vector) is: $$vec{r}''(t) = f''(t)hat{i} + g''(t)hat{j}$$ --- Step 2: Evaluating the Cross Product $vec{r}' times vec{r}''$ Compute the cross product: $$vec{r}'(t) times vec{r}''(t) = begin{vmatrix} hat{i} & hat{j} & hat{k} f'(t) & g'(t) & 0 f''(t) & g''(t) & 0 end{vmatrix}$$ Expanding along the third column: $$vec{r}'(t) times vec{r}''(t) = left( f'(t)g''(t) - g'(t)f''(t) right) hat{k}$$ The magnitude of this vector is: $$|vec{r}'(t) times vec{r}''(t)| = |f'(t)g''(t) - g'(t)f''(t)|$$ --- Step 3: Expression for Curvature $kappa$ Substituting the magnitude of the cross product and the cube of the speed into the general curvature formula: $$kappa = frac{|vec{r}'(t) times vec{r}''(t)|}{|vec{r}'(t)|^3} = frac{|f'(t)g''(t) - g'(t)f''(t)|}{left( [f'(t)]^2 + [g'(t)]^2 right)^{3/2}}$$ This establishes the required expression for curvature. --- Step 4: Value of Torsion $tau$ Torsion measures the rate a

Question 19

5. (e) (ii) Show that the principal normals at two consecutive points of a curve do not intersect unless torsion $tau$ is zero.

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Differential Geometric Setup Let a space curve $C$ be parameterized by its arc length parameter $s$: $$vec{r} = vec{r}(s)$$ At any point $P(s)$, let ${vec{t}, vec{n}, vec{b}}$ denote the standard orthonormal Frenet-Serret triad: - $vec{t}(s) = frac{dvec{r}}{ds}$ is the unit tangent vector, - $vec{n}(s)$ is the unit principal normal vector, - $vec{b}(s) = vec{t}(s) times vec{n}(s)$ is the unit binormal vector. The Frenet-Serret equations describe the rate of change of the triad with respect to $s$: $$frac{dvec{t}}{ds} = kappa vec{n}$$ $$frac{dvec{n}}{ds} = -kappa vec{t} + tau vec{b}$$ $$frac{dvec{b}}{ds} = -tau vec{n}$$ where $kappa$ is the curvature and $tau$ is the torsion of the curve. --- Step 1: Principal Normal Lines at Consecutive Points Consider two neighboring points on the curve: 1. Point $P$ corresponding to arc length $s$, with position vector $vec{r}(s)$ and principal normal $vec{n}(s)$. The straight line along the principal normal at $P$ is: $$vec{R}1(lambda) = vec{r}(s) + lambda vec{n}(s)$$ 2. A consecutive point $Q$ corresponding to arc length $s + delta s$, with position vector $vec{r}(s + delta s)$ and principal normal $vec{n}(s + delta s)$. The straight line along the principal normal at $Q$ is: $$vec{R}2(mu) = vec{r}(s + delta s) + mu vec{n}(s + delta s)$$ --- Step 2: Condition for Two Straight Lines to Intersect Two straight lines in 3-dimensional space: - Line 1 passing through $vec{r}1$ with direction $vec{d}1$, - Line 2 passing through $vec{r}2$ with direction $vec{d}2$, are coplanar (and hence either intersect or are parallel) if and only if the shortest distance between them is zero. The shortest distance is given by: $$D = frac{|(vec{r}2 - vec{r}1) cdot (vec{d}1 times vec{d}2)|}{|vec{d}1 times vec{d}2|}$$ Thus, the condition for intersection is that the scalar triple product vanishes: $$(vec{r}2 - vec{r}1) cdot (vec{d}1 times vec{d}2) = [vec{r}2 - vec{r}1, ; vec{d}1, ; vec{d}2] = 0$$ For our two consecutive principal normals: $$vec{r}2 - ve

Question 20

6. (a) A regular tetrahedron, formed of six light rods, each of length $l$, rests on a smooth horizontal plane. A ring of weight $W$ and radius $r$ is supported by the slant sides. Using the principle of virtual work, find the stress in any of the horizontal sides.

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Physical and Geometric Model Let $ABCD$ be a regular tetrahedron formed by six light, smoothly jointed rods, each of unstrained length $l$. The base triangle $ABC$ consists of three rods of length $l$ resting horizontally on a smooth plane. The vertex $D$ lies vertically above the centroid of $triangle ABC$. A heavy circular ring of weight $W$ and radius $r$ is placed horizontally over the tetrahedron, resting in contact with the three slant edges $DA, DB, DC$. By symmetry, the tensions (or thrusts) in the three horizontal rods forming the base are equal. Let $T$ denote the tension (stress) in each of the three horizontal base rods. --- Step 1: Geometry of the Regular Tetrahedron For a regular tetrahedron of edge length $l$: 1. The base $triangle ABC$ is equilateral with side $l$. The distance of each vertex from the centroid $O$ of the base is: $$R{text{base}} = OA = OB = OC = frac{l}{sqrt{3}}$$ 2. The height $H$ of the apex $D$ above the base plane is: $$H = sqrt{l^2 - R{text{base}}^2} = sqrt{l^2 - frac{l^2}{3}} = sqrt{frac{2}{3}} l$$ 3. The angle $theta$ which each slant edge makes with the vertical axis of symmetry satisfies: $$sintheta = frac{R{text{base}}}{l} = frac{l/sqrt{3}}{l} = frac{1}{sqrt{3}}$$ $$costheta = sqrt{1 - sin^2theta} = sqrt{1 - frac{1}{3}} = sqrt{frac{2}{3}}$$ $$tantheta = frac{sintheta}{costheta} = frac{1/sqrt{3}}{sqrt{2/sqrt{3}}} = frac{1}{sqrt{2}}$$ --- Step 2: Position of the Ring The circular ring of radius $r$ rests horizontally against the three slant edges. By symmetry, the three points of contact lie at the vertices of an equilateral triangle inscribed in the ring. The distance of each contact point from the vertical axis through $D$ is equal to the radius $r$ of the ring. Therefore, the vertical distance $z$ of the ring below the apex $D$ is: $$z = r cottheta = r sqrt{2}$$ Thus, the height $y$ of the ring above the horizontal base plane is: $$y = H - z = sqrt{frac{2}{3}} l - rsqrt{2}$$ Notice that the radius $r$ of the ring is fixed

Question 21

5. (d) A solid hemisphere rests in equilibrium on a solid sphere of equal radius. Determine the stability of the equilibrium in the two situations—(i) when the curved surface and (ii) when the flat surface of the hemisphere rests on the sphere.

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Theory of Stability of Equilibrium for Rolling Bodies Consider a body whose lower surface has principal radius of curvature $rho1$ resting in equilibrium on the summit of a fixed body whose upper surface has principal radius of curvature $rho2$. Let: - $h$ be the height of the Center of Gravity ($G$) of the resting body above the point of contact $A$. - $rho1$ be the radius of curvature of the resting body at $A$ (taken positive when convex downwards). - $rho2$ be the radius of curvature of the fixed lower surface at $A$ (taken positive when convex upwards). By the Classical Stability Criterion for Rolling Equilibrium (derived from potential energy variation): The equilibrium is: 1. STABLE if: $$frac{1}{h} frac{1}{rho1} + frac{1}{rho2}$$ 2. UNSTABLE if: $$frac{1}{h} (i) Curved Surface (UNSTABLE) Fixed Sphere (ρ₂ = R) A (Contact) G (CG) h = ⅝R 1/h (1.6/R) < 1/ρ₁ + 1/ρ₂ (2/R) (ii) Flat Surface (STABLE) Fixed Sphere (ρ₂ = R) A (Contact) G (CG) h = ⅜R 1/h (2.67/R) > 1/ρ₁ + 1/ρ₂ (1/R) Figure: Equilibrium stability analysis for solid hemisphere resting on sphere: (i) Unstable, (ii) Stable. Situation (i): Curved Surface of the Hemisphere Rests on the Sphere In this configuration: - The point of contact $A$ is the vertex $V$ of the curved surface of the hemisphere. - The resting body is convex at the contact point, so its radius of curvature is $rho1 = R$. - The lower fixed body is a solid sphere of radius $R$, convex upwards, so its radius of curvature is $rho2 = R$. - The height of the Center of Gravity $G$ above the point of contact $A$ is: $$h = VG = frac{5}{8}R$$ Now, evaluate the stability condition: 1. LHS: $$frac{1}{h} = frac{1}{frac{5}{8}R} = frac{8}{5R} = frac{1.6}{R}$$ 2. RHS: $$frac{1}{rho1} + frac{1}{rho2} = frac{1}{R} + frac{1}{R} = frac{2}{R}$$ Comparing the two terms: $$frac{1}{h} = frac{1.6}{R} frac{1}{R} = frac{1}{rho1} + frac{1}{rho2}$$ Since $frac{1}{h} frac{1}{rho1} + frac{1}{rho2}$, any slight roll tilts the Center of Gravity upwards, generating

Question 22

6. (b) A particle executes simple harmonic motion such that in two of its positions, velocities are $u$ and $v$, and the two corresponding accelerations are $f1$ and $f2$. For what value(s) of $k$, the distance between the two positions is $k(v^2 - u^2)$? Show also that the amplitude of the motion i

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Theory of Simple Harmonic Motion (SHM) In Simple Harmonic Motion (SHM) about the origin $O$ with amplitude $a$ and angular frequency $omega$: 1. The displacement of the particle is $x(t)$. 2. The acceleration is proportional to the displacement directed towards the center: $$f = -omega^2 x$$ Taking the magnitudes of accelerations at positions $x1$ and $x2$: $$f1 = omega^2 x1 implies x1 = frac{f1}{omega^2}$$ $$f2 = omega^2 x2 implies x2 = frac{f2}{omega^2}$$ 3. The velocity-displacement relation is given by the conservation of energy / standard SHM equation: $$V^2 = omega^2 (a^2 - x^2)$$ At the two given positions $x1$ and $x2$: $$u^2 = omega^2 (a^2 - x1^2) quad text{--- (1)}$$ $$v^2 = omega^2 (a^2 - x2^2) quad text{--- (2)}$$ --- Step 1: Finding $k$ such that the Distance is $k(v^2 - u^2)$ Subtracting Equation (1) from Equation (2): $$v^2 - u^2 = omega^2 (a^2 - x2^2) - omega^2 (a^2 - x1^2) = omega^2 (x1^2 - x2^2)$$ Factoring the difference of squares: $$v^2 - u^2 = omega^2 (x1 - x2)(x1 + x2) quad text{--- (3)}$$ Let $d = x1 - x2$ be the distance between the two positions (assuming without loss of generality $x1 x2$). Now express $(x1 + x2)$ in terms of the accelerations: $$x1 + x2 = frac{f1}{omega^2} + frac{f2}{omega^2} = frac{f1 + f2}{omega^2}$$ Substitute this into Equation (3): $$v^2 - u^2 = omega^2 d left( frac{f1 + f2}{omega^2} right) = d (f1 + f2)$$ Solving for the distance $d$: $$d = frac{v^2 - u^2}{f1 + f2}$$ Comparing with the given expression $d = k(v^2 - u^2)$: $$k = frac{1}{f1 + f2}$$ If the positions are on opposite sides of the center, $x1 - x2$ involves $f1 - f2$, so in general: $$k = frac{1}{f1 pm f2}$$ --- Step 2: Deriving the Expression for the Amplitude of Motion From the acceleration equations: $$x1 = frac{f1}{omega^2}, quad x2 = frac{f2}{omega^2}$$ Squaring: $$x1^2 = frac{f1^2}{omega^4}, quad x2^2 = frac{f2^2}{omega^4}$$ Substitute these into Equations (1) and (2): $$u^2 = omega^2 left( a^2 - frac{f1^2}{omega^4} right) = omega^2 a^2 - frac{f1^2}

Question 23

8. (a) Using Laplace transform, solve the initial value problem $y'' + 2y' + 5y = delta(t - 2)$, $y(0) = 0$, $y'(0) = 0$, where $delta(t - 2)$ denotes the Dirac delta function.

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Theory of Laplace Transforms Involving the Dirac Delta Function The Dirac delta function $delta(t - a)$ represents an idealized unit impulse applied at the instant $t = a$. Its fundamental property under integration is the sifting property: $$int{0}^{infty} f(t)delta(t - a),dt = f(a) quad (text{for } a 0)$$ Consequently, the Laplace transform of the shifted Dirac delta function is: $$mathcal{L}{delta(t - a)} = int0^infty e^{-st} delta(t - a),dt = e^{-as}$$ Recall also the Second Shifting Theorem (Translation in the $t$-domain): $$mathcal{L}^{-1}{e^{-as} F(s)} = f(t - a) u(t - a)$$ where $u(t - a)$ is the Heaviside unit step function: $$u(t - a) = begin{cases} 0, & t 2$, the system undergoes damped sinusoidal oscillations governed by natural frequency $omega = 2$ and damping decay rate $e^{-(t-2)}$. Final Answer The solution to the initial value problem is: $$y(t) = frac{1}{2} e^{-(t - 2)} sin(2t - 4) , u(t - 2)$$

Question 24

7. (b) A heavy particle hanging vertically from a fixed point by a light inextensible string of length $l$ starts to move with initial velocity $u$ in a circle so as to make a complete revolution in a vertical plane. Show that the sum of tensions at the ends of any diameter is constant.

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Physical Setup and Dynamic Equations Let a particle of mass $m$ be suspended from a fixed point $O$ by a light, inextensible string of length $l$. The particle is projected horizontally from its lowest vertical position $A$ with an initial speed $u$. Let: - $theta$ be the angle that the string makes with the downward vertical $OA$ at time $t$. - $v(theta)$ be the speed of the particle when the string makes an angle $theta$ with the vertical. - $T(theta)$ be the tension in the string at that instant. The only external forces acting on the particle are: 1. The tension $vec{T}$ directed radially inward along the string towards $O$. 2. The downward gravitational force $mvec{g}$. --- Step 1: Conservation of Mechanical Energy Since the string is inextensible and frictionless, the tension force does no work. Mechanical energy is conserved. Choosing the lowest position $A$ as the reference datum for gravitational potential energy: - Potential energy at angle $theta$: $$V(theta) = mgl(1 - costheta)$$ - Kinetic energy at angle $theta$: $$K(theta) = frac{1}{2}mv^2$$ - Total mechanical energy at initial state ($ heta = 0, v = u$): $$E0 = frac{1}{2}mu^2$$ By conservation of energy: $$frac{1}{2}mv^2 + mgl(1 - costheta) = frac{1}{2}mu^2$$ Multiplying by $frac{2}{m}$: $$v^2 = u^2 - 2gl(1 - costheta) quad text{--- (1)}$$ --- Step 2: Radial Equation of Motion The forces acting along the inward normal (radially towards $O$) are: - The inward tension $T$, - The radial component of gravity acting away from the center: $mgcostheta$. Applying Newton's second law in the radial direction: $$T - mgcostheta = frac{mv^2}{l}$$ $$T = mgcostheta + frac{mv^2}{l} quad text{--- (2)}$$ Substituting $v^2$ from Equation (1) into Equation (2): $$T(theta) = mgcostheta + frac{m}{l}left[ u^2 - 2gl(1 - costheta) right]$$ $$T(theta) = mgcostheta + frac{mu^2}{l} - 2mg + 2mgcostheta$$ $$T(theta) = frac{mu^2}{l} - 2mg + 3mgcostheta quad text{--- (3)}$$ --- O A (u at θ = 0) P (θ) T(θ) P' (θ + π) T(θ + π) θ T(θ)

Question 25

8. (b) Using Gauss divergence theorem, evaluate the integral $$iintS (y^2hat{i} + xz^3hat{j} + (z - 1)^2hat{k}) cdot hat{n},dS$$ over the region bounded by the cylinder $x^2 + y^2 = 16$ and the planes $z = 1$ and $z = 5$.

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Statement of Gauss Divergence Theorem Let $V$ be a bounded, closed region in $mathbb{R}^3$ with a piecewise-smooth boundary surface $S = partial V$. Let $hat{n}$ denote the outward unit normal vector to $S$. If $vec{F}$ is a vector field with continuous partial derivatives throughout $V$, then: $$iint{S} vec{F} cdot hat{n} , dS = iiint{V} ( abla cdot vec{F}) , dV$$ --- Step 1: Identification of the Vector Field and Region The given vector field is: $$vec{F} = F1hat{i} + F2hat{j} + F3hat{k} = y^2hat{i} + xz^3hat{j} + (z - 1)^2hat{k}$$ The closed bounding surface $S$ encloses the solid cylindrical region $V$ defined by: $$V = { (x, y, z) in mathbb{R}^3 : x^2 + y^2 le 16, ; 1 le z le 5 }$$ This is a right circular cylinder of radius $R = 4$, centered on the $z$-axis, extending vertically from $z = 1$ to $z = 5$. --- Step 2: Calculating the Divergence $ abla cdot vec{F}$ Compute the divergence of $vec{F}$: $$ abla cdot vec{F} = frac{partial F1}{partial x} + frac{partial F2}{partial y} + frac{partial F3}{partial z}$$ $$frac{partial F1}{partial x} = frac{partial}{partial x}(y^2) = 0$$ $$frac{partial F2}{partial y} = frac{partial}{partial y}(xz^3) = 0$$ $$frac{partial F3}{partial z} = frac{partial}{partial z}left((z - 1)^2right) = 2(z - 1)$$ Summing these terms: $$ abla cdot vec{F} = 0 + 0 + 2(z - 1) = 2(z - 1)$$ --- Step 3: Setting Up the Triple Integral in Cylindrical Coordinates By Gauss's Divergence Theorem: $$I = iintS vec{F} cdot hat{n},dS = iiintV 2(z - 1) , dV$$ Transforming to cylindrical coordinates $(r, theta, z)$: $$x = rcostheta, quad y = rsintheta, quad z = z$$ The volume element is: $$dV = r , dr , dtheta , dz$$ The limits of integration for the solid cylinder $V$ are: $$0 le r le 4, quad 0 le theta le 2pi, quad 1 le z le 5$$ Hence: $$I = int{0}^{2pi} left( int{0}^{4} r , dr right) dtheta cdot left( int{1}^{5} 2(z - 1) , dz right)$$ --- Step 4: Evaluating the Iterated Integrals Notice that the integrals over $(r, theta)$ and $z$ decouple completely: 1. Integ

Question 26

8. (c) A particle moves with a central acceleration $mu left(frac{3}{r^3} + frac{d^2}{r^5}right)$ being projected from a distance $d$ at an angle $45^circ$ with a velocity equal to that in a circle at the same distance. Prove that the time it takes to reach the centre of force is $frac{d^2}{sqrt{2mu

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Theory of Central Orbits and Polar Dynamics Consider a particle of unit mass moving under the influence of an attractive central force $P(r)$ directed towards the origin $O$: $$P(r) = mu left( frac{3}{r^3} + frac{d^2}{r^5} right)$$ In polar coordinates $(r, theta)$: 1. The areal velocity is conserved: $$h = r^2 frac{dtheta}{dt} = v0 p0$$ where $v0$ is the projection speed and $p0$ is the perpendicular distance from the center of force to the initial velocity vector. 2. The radial equation of energy for central motion is: $$frac{1}{2}left( left(frac{dr}{dt}right)^2 + r^2left(frac{dtheta}{dt}right)^2 right) - int P(r),dr = E$$ $$left(frac{dr}{dt}right)^2 + frac{h^2}{r^2} = 2 int P(r),dr + C$$ --- Step 1: Projection Velocity and Areal Constant $h$ For a circular orbit of radius $d$ under central acceleration $P(r)$, the normal acceleration $frac{vc^2}{d}$ must equal the central force: $$frac{vc^2}{d} = P(d) = mu left( frac{3}{d^3} + frac{d^2}{d^5} right) = mu left( frac{3}{d^3} + frac{1}{d^3} right) = frac{4mu}{d^3}$$ Multiplying by $d$: $$vc^2 = frac{4mu}{d^2} implies vc = frac{2sqrt{mu}}{d}$$ The particle is projected from initial distance $r0 = d$ with speed $v0 = vc = frac{2sqrt{mu}}{d}$ at an angle $phi = 45^circ$ to the radius vector. The perpendicular distance from the center to the velocity vector is: $$p = d sin(45^circ) = frac{d}{sqrt{2}}$$ Thus, the constant of areal velocity $h$ is: $$h = v0 p = left( frac{2sqrt{mu}}{d} right) left( frac{d}{sqrt{2}} right) = frac{2sqrt{mu}}{sqrt{2}} = sqrt{2mu}$$ $$h^2 = 2mu$$ --- Step 2: Energy Integral and Radial Velocity $frac{dr}{dt}$ Compute the potential function $V(r) = -int P(r),dr$: $$int P(r),dr = mu int left( 3r^{-3} + d^2 r^{-5} right) dr = mu left( -frac{3}{2r^2} - frac{d^2}{4r^4} right) = -frac{mu}{4r^4}(6r^2 + d^2)$$ The conservation of energy equation is: $$v^2 = left(frac{dr}{dt}right)^2 + frac{h^2}{r^2} = C - 2 left( frac{3mu}{2r^2} + frac{mu d^2}{4r^4} right) = C - frac{3mu}{r^2} - frac{mu d^2}{2r^4}$$ At

Question 27

6. (c) (i) Find the second solution of the differential equation $xy'' + (x - 1)y' - y = 0$ using $u(x) = -e^{-x}$ as one of the solutions.

Accepted Answer

Theory of Reduction of Order (Method of D'Alembert) Consider a homogeneous linear second-order differential equation written in standard normal form: $$y'' + P(x)y' + Q(x)y = 0$$ If one non-trivial solution $y1(x) = u(x)$ is already known, a second linearly independent solution $y2(x)$ can be constructed by setting: $$y(x) = u(x) v(x)$$ where $v(x)$ is an unknown function to be determined. Differentiating $y(x)$: $$y' = u'v + uv'$$ $$y'' = u''v + 2u'v' + uv''$$ Substituting into the standard differential equation: $$(u''v + 2u'v' + uv'') + P(uv' + u'v) + Q(uv) = 0$$ $$v(u'' + Pu' + Qu) + u v'' + (2u' + Pu)v' = 0$$ Since $u(x)$ is a known solution, $u'' + Pu' + Qu = 0$. The equation reduces to a first-order equation for $w = v'$: $$u w' + (2u' + Pu)w = 0$$ Separating variables yields the classical Abel-Liouville reduction formula: $$v'(x) = frac{e^{-int P(x),dx}}{[u(x)]^2} implies v(x) = int frac{e^{-int P(x),dx}}{[u(x)]^2},dx$$ and the second linearly independent solution is given by $y2(x) = u(x)v(x)$. --- Step 1: Verification of the Given Solution $u(x) = -e^{-x}$ The given equation is: $$xy'' + (x - 1)y' - y = 0 quad text{--- (1)}$$ Let $u(x) = -e^{-x}$. Its derivatives are: $$u'(x) = e^{-x}$$ $$u''(x) = -e^{-x}$$ Substituting into the left-hand side of Equation (1): $$text{LHS} = x(-e^{-x}) + (x - 1)(e^{-x}) - (-e^{-x})$$ $$= e^{-x} [-x + (x - 1) + 1] = e^{-x} [0] = 0$$ Thus, $u(x) = -e^{-x}$ is indeed a valid solution. --- Step 2: Applying Reduction of Order Divide Equation (1) by $x$ (for $x eq 0$) to put it in standard form: $$y'' + left(1 - frac{1}{x}right)y' - frac{1}{x}y = 0$$ Here: $$P(x) = 1 - frac{1}{x}$$ Compute the integrating factor: $$int P(x),dx = int left(1 - frac{1}{x}right)dx = x - ln|x|$$ $$e^{-int P(x),dx} = e^{-(x - ln x)} = e^{-x} e^{ln x} = x e^{-x} quad (text{for } x 0)$$ Now compute the denominator $[u(x)]^2$: $$[u(x)]^2 = (-e^{-x})^2 = e^{-2x}$$ Substitute into the formula for $v'(x)$: $$v'(x) = frac{e^{-int P(x),dx}}{[u(x)]^2} = frac{x

Question 28

6. (c) (ii) Find the general solution of the differential equation $x^2 y'' - 2xy' + 2y = x^3 sin x$ by the method of variation of parameters.

Accepted Answer

Theory of Variation of Parameters Consider a non-homogeneous second-order linear differential equation in standard form: $$y'' + P(x)y' + Q(x)y = R(x)$$ Let $y1(x)$ and $y2(x)$ be two linearly independent solutions of the associated homogeneous equation $y'' + P(x)y' + Q(x)y = 0$, so that the complementary function is: $$yc(x) = c1 y1(x) + c2 y2(x)$$ Their Wronskian is: $$W(x) = W(y1, y2) = y1 y2' - y1' y2$$ By the Method of Variation of Parameters, a particular integral $yp(x)$ is sought in the form: $$yp(x) = u1(x) y1(x) + u2(x) y2(x)$$ where the unknown functions $u1(x)$ and $u2(x)$ satisfy the system: $$begin{cases} u1' y1 + u2' y2 = 0 u1' y1' + u2' y2' = R(x) end{cases}$$ Solving this system via Cramer's rule yields: $$u1'(x) = -frac{y2(x) R(x)}{W(x)}, quad u2'(x) = frac{y1(x) R(x)}{W(x)}$$ --- Step 1: Solving the Associated Homogeneous Equation The given equation is: $$x^2 y'' - 2xy' + 2y = x^3 sin x quad text{--- (1)}$$ The homogeneous equation is the Cauchy-Euler equation: $$x^2 y'' - 2xy' + 2y = 0$$ Assuming a trial solution of the form $y = x^m$: $$y' = m x^{m-1}, quad y'' = m(m - 1)x^{m-2}$$ Substituting into the homogeneous equation: $$x^2 [m(m - 1)x^{m-2}] - 2x [m x^{m-1}] + 2x^m = 0$$ $$[m(m - 1) - 2m + 2] x^m = 0$$ $$m^2 - 3m + 2 = 0$$ $$(m - 1)(m - 2) = 0 implies m1 = 1, ; m2 = 2$$ Thus, two linearly independent solutions of the homogeneous equation are: $$y1(x) = x, quad y2(x) = x^2$$ The complementary function is: $$yc(x) = c1 x + c2 x^2$$ --- Step 2: Putting the ODE in Standard Normal Form and Computing Wronskian Divide Equation (1) by the leading coefficient $x^2$: $$y'' - frac{2}{x}y' + frac{2}{x^2}y = x sin x$$ The right-hand side in standard form is: $$R(x) = x sin x$$ Compute the Wronskian of $y1$ and $y2$: $$W(x) = begin{vmatrix} y1 & y2 y1' & y2' end{vmatrix} = begin{vmatrix} x & x^2 1 & 2x end{vmatrix} = x(2x) - 1(x^2) = 2x^2 - x^2 = x^2$$ For $x eq 0$, $W(x) = x^2 eq 0$. --- Step 3: Determining $u1(x)$ and $u2(x)$ 1. Determining $u1(x)$:

Question 29

7. (a) State uniqueness theorem for the existence of unique solution of the initial value problem $frac{dy}{dx} = f(x, y)$, $y(x0) = y0$ in the rectangular region $R : |x - x0| le a, |y - y0| le b$. Test the existence and uniqueness of the solution of the initial value problem $frac{dy}{dx} = 2sqrt{

Accepted Answer

Statement of Picard-Lindelöf Existence and Uniqueness Theorem Let $R$ denote the closed rectangular domain in the $xy$-plane centered at $(x0, y0)$: $$R = { (x, y) in mathbb{R}^2 : |x - x0| le a, ; |y - y0| le b }$$ where $a, b 0$. Consider the Initial Value Problem (IVP): $$frac{dy}{dx} = f(x, y), quad y(x0) = y0$$ Theorem: 1. Existence (Peano): If $f(x, y)$ is continuous on $R$, then there exists at least one solution $y(x)$ to the IVP on the interval $|x - x0| le h$, where: $$h = minleft(a, frac{b}{M}right), quad M = max{(x, y) in R} |f(x, y)|$$ 2. Uniqueness (Picard-Lindelöf): If, in addition to continuity, $f(x, y)$ satisfies a Lipschitz condition with respect to $y$ on $R$: $$|f(x, y1) - f(x, y2)| le L |y1 - y2| quad forall (x, y1), (x, y2) in R$$ for some positive constant $L$ (which is guaranteed if $frac{partial f}{partial y}$ exists and is continuous/bounded on $R$), then the solution $y(x)$ is unique on the interval $|x - x0| le h$. --- Step 1: Testing Existence and Uniqueness for $frac{dy}{dx} = 2sqrt{y}$, $y(1) = 0$ Here $x0 = 1, y0 = 0$, and the function is: $$f(x, y) = 2sqrt{y}$$ Notice that $f(x, y)$ is only defined for real values when $y ge 0$. Consider a semi-rectangle $R = { (x, y) : |x - 1| le a, ; 0 le y le b }$. 1. Existence: $f(x, y) = 2sqrt{y}$ is continuous for all $y ge 0$ on $R$. Therefore, by Peano's existence theorem, a solution to the IVP exists. 2. Uniqueness: Examine the partial derivative with respect to $y$: $$frac{partial f}{partial y} = frac{partial}{partial y}(2y^{1/2}) = frac{1}{sqrt{y}}$$ As $y to 0^+$: $$lim{y to 0^+} frac{partial f}{partial y} = lim{y to 0^+} frac{1}{sqrt{y}} = +infty$$ Because $frac{partial f}{partial y}$ is unbounded in any neighborhood containing $y0 = 0$, the Lipschitz condition fails at $y = 0$. Specifically, taking $y1 = y 0$ and $y2 = 0$: $$frac{|f(x, y) - f(x, 0)|}{|y - 0|} = frac{2sqrt{y}}{y} = frac{2}{sqrt{y}} to infty quad text{as } y to 0^+$$ No finite Lipschitz constant $L$ exists. Thus, uniquen

Question 30

7. (c) State Stokes' theorem and verify it for the vector field $vec{F} = xyhat{i} + yzhat{j} + zxhat{k}$ over the surface $S$, which is the upwardly oriented part of the cylinder $z = 1 - x^2$, for $0 le x le 1$, $-2 le y le 2$.

Accepted Answer

Statement of Stokes' Theorem Let $S$ be a piecewise-smooth, oriented surface in $mathbb{R}^3$ bounded by a simple, closed, piecewise-smooth boundary curve $C = partial S$ with positive orientation (governed by the right-hand rule with respect to the normal vector $hat{n}$ of $S$). Let $vec{F}$ be a vector field whose components have continuous first-order partial derivatives in an open region containing $S$. Then: $$oint{C} vec{F} cdot dvec{r} = iint{S} ( abla times vec{F}) cdot hat{n} , dS$$ --- Step 1: Surface Description and Normal Vector The given surface is: $$S: z = 1 - x^2, quad 0 le x le 1, quad -2 le y le 2$$ Representing $S$ as the level surface $g(x, y, z) = z + x^2 - 1 = 0$, an upward-pointing normal is: $$ abla g = frac{partial g}{partial x}hat{i} + frac{partial g}{partial y}hat{j} + frac{partial g}{partial z}hat{k} = 2xhat{i} + 0hat{j} + hat{k} = 2xhat{i} + hat{k}$$ Since the $hat{k}$-component is $+1 0$, this normal vector points upward, matching the specified orientation. The vector surface element is: $$hat{n} , dS = (2xhat{i} + hat{k}) , dx , dy$$ --- Step 2: Evaluating the Surface Integral $iintS ( abla times vec{F}) cdot hat{n} , dS$ Compute the curl of $vec{F} = xyhat{i} + yzhat{j} + zxhat{k}$: $$ abla times vec{F} = begin{vmatrix} hat{i} & hat{j} & hat{k} [1ex] frac{partial}{partial x} & frac{partial}{partial y} & frac{partial}{partial z} [1ex] xy & yz & zx end{vmatrix}$$ $$ abla times vec{F} = hat{i}left(frac{partial(zx)}{partial y} - frac{partial(yz)}{partial z}right) - hat{j}left(frac{partial(zx)}{partial x} - frac{partial(xy)}{partial z}right) + hat{k}left(frac{partial(yz)}{partial x} - frac{partial(xy)}{partial y}right)$$ $$ abla times vec{F} = hat{i}(0 - y) - hat{j}(z - 0) + hat{k}(0 - x) = -yhat{i} - zhat{j} - xhat{k}$$ Now compute the integrand: $$( abla times vec{F}) cdot hat{n} , dS = (-yhat{i} - zhat{j} - xhat{k}) cdot (2xhat{i} + hat{k}) , dx , dy = (-2xy - x) , dx , dy$$ Integrate over the region $D: 0 le x le 1, -2 le y le 2$: $$i

Maths Optional 2024 Paper I

30 Questions Available

Maths Optional 2024 Paper I

30 Questions Available

Question 1: 2. (a) Consider a linear operator $T$ on $mathbb{R}^3$ over $mathbb{R}$ defined by $T(x, y, z) = (2x, 4x - y, 2x + 3y - z)$. Is $T$ invertible? If yes

Question 2: 2. (b) If $u = (x + y)/(1 - xy)$ and $v = tan^{-1} x + tan^{-1} y$, then find $frac{partial(u, v)}{partial(x, y)}$. Are $u$ and $v$ functionally relat

Question 3: 1. (e) Find the equation of the right circular cylinder which passes through the circle $x^2 + y^2 + z^2 = 9$, $x - y + z = 3$.

Question 4: 2. (c) Find the image of the line $x = 3 - 6t, y = 2t, z = 3 + 2t$ in the plane $3x + 4y - 5z + 26 = 0$.

Question 5: 1. (a) Let $H$ be a subspace of $mathbb{R}^4$ spanned by the vectors $v1 = (1, -2, 5, -3)$, $v2 = (2, 3, 1, -4)$, $v3 = (3, 8, -3, -5)$. Then find a b

Question 6: 1. (b) Let $T : mathbb{R}^3 to mathbb{R}^3$ be a linear operator and $B = {v1, v2, v3}$ be a basis of $mathbb{R}^3$ over $mathbb{R}$. Suppose that $T

Question 7: 1. (c) Discuss the continuity of the function $$f(x) = begin{cases} frac{1}{1 - e^{-1/x}}, & x eq 0 0, & x = 0 end{cases}$$ for all values of $x$.

Question 8: 1. (d) Expand $ln(x)$ in powers of $(x - 1)$ by Taylor's theorem and hence find the value of $ln(1.1)$ correct up to four decimal places.

Question 9: 4. (a) Let $A = begin{pmatrix} 3 & 2 & 4 2 & 0 & 2 4 & 2 & 3 end{pmatrix}$ be a $3 times 3$ matrix. Find the eigenvalues and the corresponding eigenve

Question 10: 3. (a) Let $V = M{2times 2}(mathbb{R})$ denote a vector space over the field of real numbers. Find the matrix of the linear mapping $phi : V to V$ giv

Question 11: 4. (b) Using double integration, find the area lying inside the cardioid $r = a(1 + costheta)$ and outside the circle $r = a$.

Question 12: 3. (b) Find the volume of the greatest cylinder which can be inscribed in a cone of height $h$ and semi-vertical angle $alpha$.

Question 13: 3. (c) Find the vertex of the cone $4x^2 - y^2 + 2z^2 + 2xy - 3yz + 12x - 11y + 6z + 4 = 0$.

Question 14: 4. (c) Find the equation of the sphere which touches the plane $3x + 2y - z + 2 = 0$ at the point $(1, -2, 1)$ and cuts orthogonally the sphere $x^2 +

Question 15: 5. (a) Find the orthogonal trajectories of the family of curves $r = c(sectheta + tantheta)$, where $c$ is a parameter.

Question 16: 5. (b) Solve the integral equation $y(t) = cos t + int0^t y(x)cos(t - x),dx$ using Laplace transform.

Question 17: 5. (c) At any time $t$ (in seconds), the coterminous edges of a variable parallelepiped are represented by the vectors $vec{alpha} = that{i} + (t + 1)

Question 18: 5. (e) (i) Let $C$ be a plane curve $vec{r}(t) = f(t)hat{i} + g(t)hat{j}$, where $f$ and $g$ have second-order derivatives. Show that the curvature at

Question 19: 5. (e) (ii) Show that the principal normals at two consecutive points of a curve do not intersect unless torsion $tau$ is zero.

Question 20: 6. (a) A regular tetrahedron, formed of six light rods, each of length $l$, rests on a smooth horizontal plane. A ring of weight $W$ and radius $r$ is

Question 21: 5. (d) A solid hemisphere rests in equilibrium on a solid sphere of equal radius. Determine the stability of the equilibrium in the two situations—(i)

Question 22: 6. (b) A particle executes simple harmonic motion such that in two of its positions, velocities are $u$ and $v$, and the two corresponding acceleratio

Question 23: 8. (a) Using Laplace transform, solve the initial value problem $y'' + 2y' + 5y = delta(t - 2)$, $y(0) = 0$, $y'(0) = 0$, where $delta(t - 2)$ denotes

Question 24: 7. (b) A heavy particle hanging vertically from a fixed point by a light inextensible string of length $l$ starts to move with initial velocity $u$ in

Question 25: 8. (b) Using Gauss divergence theorem, evaluate the integral $$iintS (y^2hat{i} + xz^3hat{j} + (z - 1)^2hat{k}) cdot hat{n},dS$$ over the region bound

Question 26: 8. (c) A particle moves with a central acceleration $mu left(frac{3}{r^3} + frac{d^2}{r^5}right)$ being projected from a distance $d$ at an angle $45^

Question 27: 6. (c) (i) Find the second solution of the differential equation $xy'' + (x - 1)y' - y = 0$ using $u(x) = -e^{-x}$ as one of the solutions.

Question 28: 6. (c) (ii) Find the general solution of the differential equation $x^2 y'' - 2xy' + 2y = x^3 sin x$ by the method of variation of parameters.

Question 29: 7. (a) State uniqueness theorem for the existence of unique solution of the initial value problem $frac{dy}{dx} = f(x, y)$, $y(x0) = y0$ in the rectan

Question 30: 7. (c) State Stokes' theorem and verify it for the vector field $vec{F} = xyhat{i} + yzhat{j} + zxhat{k}$ over the surface $S$, which is the upwardly