Question 1

Using the method of Lagrange's multipliers, find the minimum and maximum distances of the point $P(2, 6, 3)$ from the sphere $x^2 + y^2 + z^2 = 4$.

Accepted Answer

Mathematical Formulation of the Optimization Problem Let $(x, y, z)$ be an arbitrary point on the surface of the sphere of radius $R = 2$ centered at the origin: $$g(x, y, z) = x^2 + y^2 + z^2 - 4 = 0 tag{1}$$ We are given the external point $P(2, 6, 3)$. The Euclidean distance $D$ between $P(2, 6, 3)$ and $(x, y, z)$ is: $$D = sqrt{(x - 2)^2 + (y - 6)^2 + (z - 3)^2}$$ Since the square root is a strictly increasing function for $D ge 0$, the points that extremize $D$ are identical to the points that extremize the square of the distance: $$f(x, y, z) = D^2 = (x - 2)^2 + (y - 6)^2 + (z - 3)^2 tag{2}$$ --- Step 1: Setting up the Lagrange Multiplier Equation We construct the Lagrangian function $L(x, y, z, lambda)$: $$L(x, y, z, lambda) = f(x, y, z) - lambda , g(x, y, z)$$ $$L(x, y, z, lambda) = (x - 2)^2 + (y - 6)^2 + (z - 3)^2 - lambda(x^2 + y^2 + z^2 - 4) tag{3}$$ where $lambda$ is the Lagrange multiplier. The necessary conditions for stationary points are obtained by equating the first partial derivatives of $L$ to zero: $$begin{aligned} frac{partial L}{partial x} &= 2(x - 2) - 2lambda x = 0 implies x - 2 = lambda x implies x(1 - lambda) = 2 tag{4} [6pt] frac{partial L}{partial y} &= 2(y - 6) - 2lambda y = 0 implies y - 6 = lambda y implies y(1 - lambda) = 6 tag{5} [6pt] frac{partial L}{partial z} &= 2(z - 3) - 2lambda z = 0 implies z - 3 = lambda z implies z(1 - lambda) = 3 tag{6} end{aligned}$$ --- Step 2: Parametric Relation for the Coordinates From equations (4), (5), and (6), notice that if $lambda = 1$, we would have $0 = 2$, which is impossible. Hence $lambda eq 1$. Solving for the coordinates: $$x = frac{2}{1 - lambda}, qquad y = frac{6}{1 - lambda}, qquad z = frac{3}{1 - lambda} tag{7}$$ Let $k = frac{1}{1 - lambda}$. Then: $$x = 2k, qquad y = 6k, qquad z = 3k tag{8}$$ This shows that the extreme points $(x, y, z)$ on the sphere must lie along the line joining the origin $O(0, 0, 0)$ to the given point $P(2, 6, 3)$. --- Step 3: Determining the Parameter $k$

Question 2

Prove that a non-commutative group of order $2p$, where $p$ is an odd prime, must have a subgroup of order $p$.

Accepted Answer

Problem Statement & Hypothesis Let $G$ be a finite, non-commutative (non-abelian) group of order: $$|G| = 2p$$ where $p$ is an odd prime ($p ge 3$). We are required to prove that $G$ must contain a subgroup of order $p$. --- Method I: Proof via Cauchy's Theorem and Lagrange's Theorem 1. Existence of an Element of Order $p$: Since $p$ is a prime number and $p$ divides the order of the group $|G| = 2p$: $$p mid |G|$$ By Cauchy's Theorem for Finite Groups, if a prime $p$ divides the order of a finite group $G$, then $G$ contains at least one element $a in G$ of order $p$: $$o(a) = p$$ 2. Construction of the Subgroup: Consider the cyclic subgroup generated by this element $a$: $$H = langle a rangle = {e, a, a^2, a^3, dots, a^{p-1}}$$ By definition of the order of an element: $$|H| = o(a) = p$$ Thus, $H$ is a subgroup of $G$ of order $p$. This completes the direct proof of existence. --- Method II: First-Principles Proof Without Invoking Cauchy's Theorem To provide an exhaustive UPSC-standard proof from foundational axioms, we prove the existence directly from Lagrange's Theorem and the hypothesis of non-commutativity. 1. Permissible Orders of Elements: By Lagrange's Theorem, the order of every element $g in G$ must divide the group order $|G| = 2p$. Since $p$ is prime, the only positive divisors of $2p$ are $1, 2, p, 2p$. Therefore, for every $g in G$: $$o(g) in {1, 2, p, 2p}$$ 2. $G$ Cannot Have an Element of Order $2p$: If $G$ had an element $g$ of order $2p$, then the cyclic subgroup generated by $g$ would have order $2p$: $$|langle g rangle| = 2p = |G| implies G = langle g rangle$$ Every cyclic group is commutative (abelian). However, $G$ is given to be non-commutative. This contradiction proves that $G$ contains no element of order $2p$. 3. $G$ Cannot Have All Non-Identity Elements of Order $2$: Suppose, for contradiction, that $G$ contains no element of order $p$. Then every non-identity element $x in G setminus {e}$ must have order $2$: $$x^2 = e iff x = x^{-1} q

Question 3

A person requires 24, 24 and 20 units of chemicals $A, B$ and $C$ respectively for his garden. Product $P$ contains 2, 4 and 1 units of chemicals $A, B$ and $C$ respectively per jar and product $Q$ contains 2, 1 and 5 units of chemicals $A, B$ and $C$ respectively per jar. If a jar of $P$ costs ₹ 30

Accepted Answer

Linear Programming Problem Formulation Let $x1$ denote the number of jars of product $P$ to be purchased. Let $x2$ denote the number of jars of product $Q$ to be purchased. 1. Objective Function: The cost of each jar of $P$ is ₹ 30 and each jar of $Q$ is ₹ 50. We wish to minimize the total expenditure $Z$: $$text{Minimize } Z = 30x1 + 50x2 tag{1}$$ 2. Constraints: - Chemical $A$ Requirement: Minimum 24 units. $$2x1 + 2x2 ge 24 iff x1 + x2 ge 12 tag{2}$$ - Chemical $B$ Requirement: Minimum 24 units. $$4x1 + x2 ge 24 tag{3}$$ - Chemical $C$ Requirement: Minimum 20 units. $$x1 + 5x2 ge 20 tag{4}$$ - Non-negativity Restrictions: $$x1 ge 0, qquad x2 ge 0 tag{5}$$ --- Step 1: Determination of the Boundary Lines & Feasible Region We plot the boundary lines of the constraints in the first quadrant of the $x1 x2$-plane: 1. Line $L1$: $x1 + x2 = 12$ - Intercepts: $(12, 0)$ on $x1$-axis; $(0, 12)$ on $x2$-axis. 2. Line $L2$: $4x1 + x2 = 24$ - Intercepts: $(6, 0)$ on $x1$-axis; $(0, 24)$ on $x2$-axis. 3. Line $L3$: $x1 + 5x2 = 20$ - Intercepts: $(20, 0)$ on $x1$-axis; $(0, 4)$ on $x2$-axis. Since all constraints are of the "$ge$" type, the feasible region is the unbounded convex polygonal region lying above and to the right of these three lines in the first quadrant. --- Step 2: Determination of Corner (Extreme) Points The boundary of the feasible region is formed by the upper envelope of the constraint lines. The corner points are: 1. Corner Point $K1$: On the $x2$-axis ($x1 = 0$): From $4x1 + x2 ge 24$, setting $x1 = 0$ gives $x2 = 24$. Checking other constraints: - $0 + 24 = 24 ge 12 ; (checkmark)$ - $0 + 5(24) = 120 ge 20 ; (checkmark)$ $$K1 = (0, 24)$$ 2. Corner Point $K2$: Intersection of $L2$ and $L1$: $$begin{cases} 4x1 + x2 = 24 x1 + x2 = 12 end{cases}$$ Subtracting the second equation from the first: $$3x1 = 12 implies x1 = 4$$ Substituting into $x1 + x2 = 12$: $$x2 = 12 - 4 = 8$$ Checking constraint (4): $$x1 + 5x2 = 4 + 5(8) = 4 + 40 = 44 ge 20 quad (checkmark)$$ $$

Question 4

State the sufficient conditions for a function $f(z) = f(x + iy) = u(x, y) + iv(x, y)$ to be analytic in its domain. Hence, show that $f(z) = log z$ is analytic in its domain and find $frac{df}{dz}$.

Accepted Answer

Part 1: Sufficient Conditions for Analyticity A complex-valued function $f(z) = u(x, y) + i v(x, y)$ of a complex variable $z = x + iy$ is said to be analytic (or holomorphic) at a point $z0 = (x0, y0)$ if it is complex differentiable not only at $z0$, but at every point in some open neighborhood of $z0$. It is analytic in a domain $D subseteq mathbb{C}$ if it is analytic at every point of $D$. Theorem (Sufficient Conditions for Differentiability / Analyticity): Let $f(z) = u(x, y) + i v(x, y)$ be defined in an open domain $D subset mathbb{C}$. The function $f(z)$ is analytic in $D$ if: 1. The four first-order partial derivatives: $$frac{partial u}{partial x}, quad frac{partial u}{partial y}, quad frac{partial v}{partial x}, quad frac{partial v}{partial y}$$ exist at every point $(x, y) in D$. 2. All four first-order partial derivatives are continuous throughout $D$. 3. The partial derivatives satisfy the Cauchy-Riemann equations at every point in $D$: $$frac{partial u}{partial x} = frac{partial v}{partial y} tag{CR 1}$$ $$frac{partial u}{partial y} = -frac{partial v}{partial x} tag{CR 2}$$ When these sufficient conditions hold, the derivative $f'(z) = frac{df}{dz}$ exists at each point in $D$ and is given by: $$f'(z) = frac{partial u}{partial x} + i frac{partial v}{partial x} = frac{partial v}{partial y} - i frac{partial u}{partial y}$$ --- Part 2: Analyticity of the Principal Branch of $f(z) = log z$ The complex logarithmic function is multi-valued: $$log z = ln|z| + i arg(z) = ln r + i(theta + 2kpi), quad k in mathbb{Z}$$ To obtain a single-valued analytic function, we consider the principal branch $text{Log}(z)$, defined on the slit complex plane (domain of analyticity): $$D = mathbb{C} setminus {z = x + iy : x le 0, ; y = 0}$$ (the complex plane cut along the non-positive real axis), where the principal argument satisfies $-pi 0$$ $$operatorname{Arg}(z) = theta(x, y)$$ Thus, the real and imaginary parts of $f(z) = u(x, y) + iv(x, y)$ are: $$u(x, y) = lnsqrt{x^2

Question 5

Test the convergence of the series $$sum{n=1}^infty frac{1 cdot 3 cdot 5 cdots (2n - 1)}{2 cdot 4 cdot 6 cdots (2n)} frac{x^{2n+1}}{(2n + 1)}, qquad x 0.$$

Accepted Answer

Statement of the Problem & General Term We are required to test the convergence of the positive-term infinite series: $$sum{n=1}^infty un = sum{n=1}^infty frac{1 cdot 3 cdot 5 cdots (2n - 1)}{2 cdot 4 cdot 6 cdots (2n)} frac{x^{2n+1}}{2n + 1}, qquad x 0 tag{1}$$ The general term $un$ is: $$un = frac{1 cdot 3 cdot 5 cdots (2n - 1)}{2 cdot 4 cdot 6 cdots (2n)} frac{x^{2n+1}}{2n + 1}$$ The next term $u{n+1}$ is: $$u{n+1} = frac{1 cdot 3 cdot 5 cdots (2n - 1)(2n + 1)}{2 cdot 4 cdot 6 cdots (2n)(2n + 2)} frac{x^{2n+3}}{2n + 3}$$ --- Step 1: Application of D'Alembert's Ratio Test We form the ratio of successive terms: $$frac{un}{u{n+1}} = left[ frac{1 cdot 3 cdots (2n - 1)}{2 cdot 4 cdots (2n)} frac{x^{2n+1}}{2n + 1} right] times left[ frac{2 cdot 4 cdots (2n)(2n + 2)}{1 cdot 3 cdots (2n - 1)(2n + 1)} frac{2n + 3}{x^{2n+3}} right]$$ Canceling common product factors: $$frac{un}{u{n+1}} = left( frac{2n + 2}{2n + 1} right) left( frac{2n + 3}{2n + 1} right) frac{1}{x^2}$$ $$frac{un}{u{n+1}} = frac{4n^2 + 10n + 6}{(2n + 1)^2} frac{1}{x^2} = frac{4n^2 + 10n + 6}{4n^2 + 4n + 1} frac{1}{x^2}$$ Taking the limit as $n to infty$: $$L = lim{n to infty} frac{un}{u{n+1}} = lim{n to infty} left( frac{4 + frac{10}{n} + frac{6}{n^2}}{4 + frac{4}{n} + frac{1}{n^2}} right) frac{1}{x^2} = frac{4}{4} cdot frac{1}{x^2} = frac{1}{x^2} tag{2}$$ According to D'Alembert's Ratio Test: 1. If $L 1 iff frac{1}{x^2} 1 iff x^2 1 iff x 1$: the series diverges. 3. If $L = 1 iff x = 1$: the ratio test fails (is inconclusive). --- Step 2: Investigation at the Boundary Case $x = 1$ At $x = 1$, the ratio becomes: $$frac{un}{u{n+1}} = frac{4n^2 + 10n + 6}{4n^2 + 4n + 1}$$ We subtract $1$ to prepare for higher-order tests: $$frac{un}{u{n+1}} - 1 = frac{4n^2 + 10n + 6 - (4n^2 + 4n + 1)}{4n^2 + 4n + 1} = frac{6n + 5}{4n^2 + 4n + 1}$$ Application of Raabe's Test: Raabe's test considers the limit: $$R = lim{n to infty} n left( frac{un}{u{n+1}} - 1 right)$$ Evaluating $R$: $$R = lim{n to infty} n left( frac{6n + 5}{

Question 6

Express the ideal $4mathbb{Z} + 6mathbb{Z}$ as a principal ideal in the integral domain $mathbb{Z}$.

Accepted Answer

Algebraic Context: Ideals in the Ring of Integers Let $(mathbb{Z}, +, cdot)$ denote the ring of integers with usual addition and multiplication. $(mathbb{Z}, +, cdot)$ is a Euclidean Domain (ED), and therefore a Principal Ideal Domain (PID). This means that every ideal $I$ of $mathbb{Z}$ is principal, i.e., generated by a single element: $$I = langle d rangle = dmathbb{Z} = {k d : k in mathbb{Z}}$$ for some non-negative integer $d in mathbb{Z}{ge 0}$. --- Step 1: Definition of the Sum of Ideals Let $I = 4mathbb{Z} = langle 4 rangle$ and $J = 6mathbb{Z} = langle 6 rangle$ be two ideals in $mathbb{Z}$. By definition, the sum of two ideals $I$ and $J$ in a commutative ring $R$ is the set: $$I + J = {x + y : x in I, ; y in J}$$ For $I = 4mathbb{Z}$ and $J = 6mathbb{Z}$: $$4mathbb{Z} + 6mathbb{Z} = {4m + 6n : m, n in mathbb{Z}} tag{1}$$ --- Step 2: Theorem on the Sum of Principal Ideals in a PID Theorem: In a Principal Ideal Domain $R$, the sum of two principal ideals $langle a rangle$ and $langle b rangle$ is the principal ideal generated by the greatest common divisor $gcd(a, b)$: $$langle a rangle + langle b rangle = langle gcd(a, b) rangle$$ Let us prove this rigorously for $mathbb{Z}$: Let $d = gcd(4, 6)$. The divisors of $4$ are ${pm 1, pm 2, pm 4}$, and the divisors of $6$ are ${pm 1, pm 2, pm 3, pm 6}$. The greatest common positive divisor is: $$d = gcd(4, 6) = 2$$ We now demonstrate mutual set inclusion: $4mathbb{Z} + 6mathbb{Z} = 2mathbb{Z}$. 1. Proof that $4mathbb{Z} + 6mathbb{Z} subseteq 2mathbb{Z}$: Let $z in 4mathbb{Z} + 6mathbb{Z}$. Then there exist integers $m, n in mathbb{Z}$ such that: $$z = 4m + 6n$$ Factoring out $2$: $$z = 2(2m + 3n)$$ Since $m, n in mathbb{Z}$, the quantity $k = 2m + 3n$ is an integer ($k in mathbb{Z}$). Therefore: $$z = 2k in 2mathbb{Z}$$ This proves that: $$4mathbb{Z} + 6mathbb{Z} subseteq 2mathbb{Z} tag{2}$$ 2. Proof that $2mathbb{Z} subseteq 4mathbb{Z} + 6mathbb{Z}$: By Bézout's Identity, the greatest common divisor of two integ

Question 7

Let $G$ be a group of order 10 and $G'$ be a group of order 6. Examine whether there exists a homomorphism of $G$ onto $G'$.

Accepted Answer

Theoretical Framework & Group Homomorphism Theorems Let $G$ and $G'$ be two finite groups, and let $phi: G to G'$ be a group homomorphism. The fundamental properties governing group homomorphisms and their structural consequences are: 1. The Kernel of a Homomorphism: $$ker(phi) = {g in G : phi(g) = e{G'}}$$ where $e{G'}$ is the identity element of $G'$. $ker(phi)$ is always a normal subgroup of $G$, denoted $ker(phi) triangleleft G$. 2. The Image of a Homomorphism: $$text{Im}(phi) = phi(G) = {phi(g) : g in G}$$ $text{Im}(phi)$ is always a subgroup of $G'$, denoted $text{Im}(phi) le G'$. 3. First Isomorphism Theorem (Fundamental Theorem of Group Homomorphisms): $$G / ker(phi) cong text{Im}(phi)$$ Consequently, for finite groups, the order of the image is equal to the index of the kernel in $G$: $$|text{Im}(phi)| = [G : ker(phi)] = frac{|G|}{|ker(phi)|} tag{1}$$ 4. Lagrange's Theorem: The order of every subgroup of a finite group divides the order of the group. - Since $ker(phi) le G$, we have $|ker(phi)| mid |G|$. - Since $text{Im}(phi) le G'$, we have $|text{Im}(phi)| mid |G'|$. --- Step 1: Permissible Orders of the Homomorphic Image We are given: $$|G| = 10, qquad |G'| = 6$$ From equation (1): $$|text{Im}(phi)| = frac{10}{|ker(phi)|}$$ This implies that $|text{Im}(phi)|$ must be a positive divisor of $|G| = 10$. The set of positive divisors of $10$ is: $$text{Div}(10) = {1, 2, 5, 10}$$ Therefore: $$|text{Im}(phi)| in {1, 2, 5, 10} tag{2}$$ Simultaneously, since $text{Im}(phi)$ is a subgroup of the codomain $G'$, by Lagrange's Theorem on $G'$: $$|text{Im}(phi)| mid |G'| = 6$$ The set of positive divisors of $6$ is: $$text{Div}(6) = {1, 2, 3, 6}$$ Therefore: $$|text{Im}(phi)| in {1, 2, 3, 6} tag{3}$$ Intersecting the two divisibility constraints (2) and (3): $$|text{Im}(phi)| in {1, 2, 5, 10} cap {1, 2, 3, 6} = {1, 2} tag{4}$$ Thus, any homomorphism from $G$ to $G'$ can have an image of order at most $2$. --- Step 2: Examination of the Onto (Surjective) Condition A h

Question 8

Evaluate $int0^{2pi} frac{cos 2theta}{5 + 4costheta} dtheta$ using contour integration.

Accepted Answer

Method of Contour Integration on the Unit Circle We are required to evaluate the real definite trigonometric integral: $$I = int0^{2pi} frac{cos 2theta}{5 + 4costheta} dtheta tag{1}$$ using the technique of complex contour integration. We transform the integral over the interval $[0, 2pi]$ into a contour integral along the positively oriented unit circle $C = {z in mathbb{C} : |z| = 1}$. --- Step 1: Complex Coordinate Transformation Let $z = e^{itheta}$. As $theta$ ranges from $0$ to $2pi$, $z$ traverses the unit circle $C$ once in the counter-clockwise direction. Differentiating $z = e^{itheta}$: $$dz = i e^{itheta} dtheta = i z , dtheta implies dtheta = frac{dz}{i z} tag{2}$$ Expressing the trigonometric functions in terms of $z$: $$costheta = frac{e^{itheta} + e^{-itheta}}{2} = frac{z + z^{-1}}{2} = frac{z^2 + 1}{2z} tag{3}$$ $$cos 2theta = frac{e^{2itheta} + e^{-2itheta}}{2} = frac{z^2 + z^{-2}}{2} = frac{z^4 + 1}{2z^2} tag{4}$$ --- Step 2: Transforming the Integrand The denominator of the integrand is: $$5 + 4costheta = 5 + 4left( frac{z^2 + 1}{2z} right) = 5 + frac{2(z^2 + 1)}{z} = frac{2z^2 + 5z + 2}{z} tag{5}$$ Factoring the quadratic polynomial in the numerator: $$2z^2 + 5z + 2 = 2z^2 + 4z + z + 2 = 2z(z + 2) + 1(z + 2) = (2z + 1)(z + 2)$$ Now, substituting equations (2), (4), and (5) into integral (1): $$I = ointC frac{frac{z^4 + 1}{2z^2}}{frac{(2z + 1)(z + 2)}{z}} left( frac{dz}{i z} right)$$ $$I = ointC frac{z^4 + 1}{2z^2} cdot frac{z}{(2z + 1)(z + 2)} cdot frac{dz}{i z}$$ $$I = frac{1}{2i} ointC frac{z^4 + 1}{z^2 (2z + 1)(z + 2)} , dz tag{6}$$ Factoring out $2$ from $(2z + 1)$: $$I = frac{1}{4i} ointC frac{z^4 + 1}{z^2 left(z + frac{1}{2}right)(z + 2)} , dz tag{7}$$ Let the integrand function be: $$F(z) = frac{z^4 + 1}{z^2 (2z + 1)(z + 2)}$$ --- Step 3: Location and Classification of Singularities The poles of $F(z)$ are given by the roots of the denominator $z^2 (2z + 1)(z + 2) = 0$: 1. $z = 0$: A pole of order 2 (double pole). $$|z| = 0 1 implies text

Question 9

Prove that $x^2 + 1$ is an irreducible polynomial in $mathbb{Z}3[x]$. Further show that the quotient ring $frac{mathbb{Z}3[x]}{langle x^2 + 1 rangle}$ is a field of 9 elements.

Accepted Answer

Algebraic Setting: Polynomials Over Finite Fields Let $mathbb{Z}3 = {0, 1, 2}$ be the finite field of integers modulo 3 under addition and multiplication modulo 3. Since 3 is a prime, $mathbb{Z}3$ is a field. Let $mathbb{Z}3[x]$ denote the ring of polynomials in one indeterminate $x$ with coefficients in $mathbb{Z}3$. We are given the polynomial: $$p(x) = x^2 + 1 in mathbb{Z}3[x]$$ --- Part 1: Proof that $x^2 + 1$ is Irreducible in $mathbb{Z}3[x]$ Definition: A non-constant polynomial $f(x) in F[x]$ over a field $F$ is irreducible if it cannot be expressed as the product of two polynomials of strictly lower positive degrees in $F[x]$. Root Criterion for Degrees 2 and 3: Let $f(x) in F[x]$ be a polynomial of degree 2 or 3. If $f(x)$ is reducible over $F$, it must have at least one linear factor $(x - alpha)$ in $F[x]$, which by the Factor Theorem is equivalent to $f(x)$ having a root $alpha in F$ such that $f(alpha) = 0$. Consequently, $f(x)$ is irreducible over $F$ if and only if $f(x)$ has no roots in $F$. Here, $deg(p(x)) = 2$. We evaluate $p(x) = x^2 + 1$ at each of the three elements of $mathbb{Z}3$: 1. For $x = 0$: $$p(0) = 0^2 + 1 = 1 otequiv 0 pmod 3$$ 2. For $x = 1$: $$p(1) = 1^2 + 1 = 2 otequiv 0 pmod 3$$ 3. For $x = 2$: $$p(2) = 2^2 + 1 = 4 + 1 = 5 equiv 2 otequiv 0 pmod 3$$ Since $p(x)$ has no root in $mathbb{Z}3$: $$p(x) eq 0 quad forall x in mathbb{Z}3$$ it cannot be factored into linear factors over $mathbb{Z}3$. Therefore, $x^2 + 1$ is an irreducible polynomial in $mathbb{Z}3[x]$. --- Part 2: Proof that the Quotient Ring is a Field Let $I = langle x^2 + 1 rangle$ be the principal ideal generated by $p(x)$ in $mathbb{Z}3[x]$. We consider the quotient ring: $$K = frac{mathbb{Z}3[x]}{langle x^2 + 1 rangle} = {f(x) + I : f(x) in mathbb{Z}3[x]}$$ Theorem: If $F$ is a field and $p(x) in F[x]$ is an irreducible polynomial, then the principal ideal $langle p(x) rangle$ is a maximal ideal in $F[x]$, and the quotient ring $frac{F[x]}{langle p(x) rangle}$ is a f

Question 10

Prove that $u(x, y) = e^x(xcos y - ysin y)$ is harmonic. Find its conjugate harmonic function $v(x, y)$ and express the corresponding analytic function $f(z)$ in terms of $z$.

Accepted Answer

Harmonic Functions & Harmonic Conjugates A real-valued function $u(x, y)$ of two real variables is said to be harmonic in a domain $D subseteq mathbb{R}^2$ if it has continuous second-order partial derivatives and satisfies Laplace's equation: $$ abla^2 u = frac{partial^2 u}{partial x^2} + frac{partial^2 u}{partial y^2} = 0$$ If $u(x, y)$ is harmonic in a simply connected domain $D$, there exists another harmonic function $v(x, y)$, called the harmonic conjugate of $u$, such that $f(z) = u(x, y) + i v(x, y)$ is an analytic function of $z = x + iy$. $u$ and $v$ are connected by the Cauchy-Riemann equations: $$frac{partial u}{partial x} = frac{partial v}{partial y}, qquad frac{partial u}{partial y} = -frac{partial v}{partial x}$$ --- Step 1: Proof that $u(x, y)$ is Harmonic We are given: $$u(x, y) = e^x (xcos y - ysin y) tag{1}$$ 1. Partial Derivatives with Respect to $x$: Treating $y$ as a constant and applying the product rule to $e^x cdot (xcos y - ysin y)$: $$frac{partial u}{partial x} = e^x(xcos y - ysin y) + e^x(cos y) = e^x [ (x + 1)cos y - ysin y ] tag{2}$$ Differentiating again with respect to $x$: $$frac{partial^2 u}{partial x^2} = e^x [ (x + 1)cos y - ysin y ] + e^x(cos y) = e^x [ (x + 2)cos y - ysin y ] tag{3}$$ 2. Partial Derivatives with Respect to $y$: Treating $x$ as a constant: $$frac{partial u}{partial y} = e^x left[ x(-sin y) - (1 cdot sin y + ycos y) right] = -e^x [ (x + 1)sin y + ycos y ] tag{4}$$ Differentiating again with respect to $y$: $$frac{partial^2 u}{partial y^2} = -e^x left[ (x + 1)cos y + (1 cdot cos y - ysin y) right] = -e^x [ (x + 2)cos y - ysin y ] tag{5}$$ 3. Verification of Laplace's Equation: Adding equations (3) and (5): $$ abla^2 u = frac{partial^2 u}{partial x^2} + frac{partial^2 u}{partial y^2} = e^x [ (x + 2)cos y - ysin y ] - e^x [ (x + 2)cos y - ysin y ] = 0$$ Since $u(x, y)$ has continuous partial derivatives of all orders on $mathbb{R}^2$ and satisfies $ abla^2 u = 0$, $u(x, y)$ is harmonic everywhere in $mathbb{C}$. ---

Question 11

Solve the following linear programming problem by Big M method: $$text{Minimize } Z = 2x1 + 3x2$$ subject to $$begin{aligned} x1 + x2 &ge 9 x1 + 2x2 &ge 15 2x1 - 3x2 &le 9 x1, x2 &ge 0 end{aligned}$$ Is the optimal solution unique? Justify your answer.

Accepted Answer

Linear Programming Problem Setup & Standard Form We are given the minimization problem: $$text{Minimize } Z = 2x1 + 3x2 tag{1}$$ subject to: $$begin{aligned} x1 + x2 &ge 9 tag{2} x1 + 2x2 &ge 15 tag{3} 2x1 - 3x2 &le 9 tag{4} x1, x2 &ge 0 tag{5} end{aligned}$$ --- Step 1: Big M Formulation 1. Conversion to Maximization: $$text{Maximize } Z^ = -Z = -2x1 - 3x2$$ 2. Introduction of Slack, Surplus, and Artificial Variables: - In constraint (2) ($ge$ type): Subtract surplus variable $s1 ge 0$ and add artificial variable $A1 ge 0$: $$x1 + x2 - s1 + A1 = 9$$ - In constraint (3) ($ge$ type): Subtract surplus variable $s2 ge 0$ and add artificial variable $A2 ge 0$: $$x1 + 2x2 - s2 + A2 = 15$$ - In constraint (4) ($le$ type): Add slack variable $s3 ge 0$: $$2x1 - 3x2 + s3 = 9$$ 3. Penalized Objective Function: Assigning a large penalty $-M$ ($M gg 0$) to each artificial variable: $$text{Maximize } Z^ = -2x1 - 3x2 + 0s1 + 0s2 + 0s3 - M A1 - M A2$$ --- Step 2: Initial Simplex Table Initial Basic Feasible Solution: $A1 = 9, ; A2 = 15, ; s3 = 9$. Non-basic variables: $x1 = 0, x2 = 0, s1 = 0, s2 = 0$. $$Z^ = -M(9) - M(15) = -24M$$ The net evaluations $zj - cj$: - For $x1$: $(-M)(1) + (-M)(1) + 0(2) - (-2) = -2M + 2$ - For $x2$: $(-M)(1) + (-M)(2) + 0(-3) - (-3) = -3M + 3$ - For $s1$: $(-M)(-1) - 0 = M$ - For $s2$: $(-M)(-1) - 0 = M$ - For $s3$: $0$ - For $A1, A2$: $0$ Since $-3M + 3$ is the most negative entry, $x2$ enters the basis. Minimum ratio test: - Row 1 ($A1$): $9 / 1 = 9$ - Row 2 ($A2$): $15 / 2 = 7.5$ (Minimum!) - Row 3 ($s3$): coefficient is $-3 0$$ - For $s2$: $$z{s2} - c{s2} = c{B} cdot text{col}(s2) - 0 = (-2)(1) + (-3)(-1) + 0(-5) - 0 = -2 + 3 = 1 0$$ Since all $zj - cj ge 0$ for non-basic variables in a maximization problem (or $cj - zj 0$ for original minimization), the current solution is strictly OPTIMAL. --- Step 5: Optimal Solution & Value The optimal values of the decision variables are: $$x1 = 3, qquad x2 = 6$$ Slack variable $s3 = 21$ (meaning $2(3) - 3(6)

Question 12

Prove that the oscillation of a real-valued bounded function $f$ defined on $[a, b]$ is the supremum of the set ${|f(x1) - f(x2)| : x1, x2 in [a, b]}$. [Note: Ensure subscripts $x1, x2$ remain in math mode.]

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Definitions and Mathematical Formulation Let $f: [a, b] to mathbb{R}$ be a real-valued bounded function defined on the closed interval $[a, b]$. Since $f$ is bounded on $[a, b]$, the set of values $f([a, b]) = {f(x) : x in [a, b]}$ is bounded above and below. By the Completeness Axiom of real numbers, the supremum and infimum of $f$ on $[a, b]$ exist as finite real numbers: $$M = sup{x in [a, b]} f(x) = sup {f(x) : x in [a, b]} tag{1}$$ $$m = inf{x in [a, b]} f(x) = inf {f(x) : x in [a, b]} tag{2}$$ Definition of Oscillation: The oscillation of the bounded function $f$ on the interval $[a, b]$, denoted by $omega(f, [a, b])$ or $Omega(f)$, is defined as the difference between its supremum and infimum on that interval: $$omega(f, [a, b]) = M - m tag{3}$$ We are required to prove that: $$omega(f, [a, b]) = sup S$$ where $S$ is the set of absolute differences between values of $f$: $$S = {|f(x1) - f(x2)| : x1, x2 in [a, b]} tag{4}$$ --- Step 1: Proof that $sup S le M - m$ Let $x1, x2$ be any two arbitrary points in $[a, b]$. By definition of supremum $M$ and infimum $m$: $$m le f(x1) le M tag{5}$$ $$m le f(x2) le M tag{6}$$ From (5) and (6): $$f(x1) - f(x2) le M - m tag{7}$$ Similarly, reversing the roles of $x1$ and $x2$: $$f(x2) - f(x1) le M - m implies -(f(x1) - f(x2)) le M - m tag{8}$$ Combining inequalities (7) and (8): $$|f(x1) - f(x2)| le M - m quad forall x1, x2 in [a, b] tag{9}$$ Inequality (9) establishes that the real number $M - m$ is an upper bound for the set $S$. Since the supremum of a set is the least upper bound: $$sup S le M - m tag{10}$$ --- Step 2: Proof that $sup S ge M - m$ Let $epsilon 0$ be an arbitrarily chosen positive real number. 1. Since $M = sup {f(x) : x in [a, b]}$, by the characteristic property of the supremum, $M - frac{epsilon}{2}$ is not an upper bound of $f([a, b])$. Therefore, there exists a point $x1' in [a, b]$ such that: $$f(x1') M - frac{epsilon}{2} tag{11}$$ 2. Since $m = inf {f(x) : x in [a, b]}$, by the characteristic prope

Question 13

Classify the singular point $z = 0$ of the function $f(z) = frac{e^z}{z - sin z}$ and obtain the principal part of its Laurent series expansion.

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Classification of the Singularity at $z = 0$ We are given the complex function: $$f(z) = frac{e^z}{z - sin z} tag{1}$$ Let $h(z) = e^z$ be the numerator and $g(z) = z - sin z$ be the denominator. Both $h(z)$ and $g(z)$ are entire functions (analytic on the entire complex plane $mathbb{C}$). 1. Analysis of the Numerator $h(z) = e^z$: At $z = 0$: $$h(0) = e^0 = 1 eq 0$$ Thus, the numerator does not vanish at $z = 0$. 2. Analysis of the Denominator $g(z) = z - sin z$: Recall the Taylor series expansion of $sin z$ about $z = 0$: $$sin z = z - frac{z^3}{3!} + frac{z^5}{5!} - frac{z^7}{7!} + O(z^9) = z - frac{z^3}{6} + frac{z^5}{120} - frac{z^7}{5040} + O(z^9)$$ Substituting into $g(z)$: $$g(z) = z - left( z - frac{z^3}{6} + frac{z^5}{120} - O(z^7) right) = frac{z^3}{6} - frac{z^5}{120} + O(z^7) tag{2}$$ Factoring out $z^3$: $$g(z) = z^3 left( frac{1}{6} - frac{z^2}{120} + O(z^4) right) tag{3}$$ Notice that: $$g(0) = 0, quad g'(0) = 0, quad g''(0) = 0, quad g'''(0) = 1 eq 0$$ Therefore, $g(z)$ has a zero of order 3 at $z = 0$. 3. Classification of Singularity: Since the numerator $h(z)$ has a non-zero value at $z = 0$ ($h(0) = 1$) and the denominator $g(z)$ has a zero of order 3 at $z = 0$: The ratio $f(z) = frac{h(z)}{g(z)}$ has an isolated pole of order 3 (a triple pole) at $z = 0$. --- Step 2: Derivation of the Laurent Series Expansion We expand $f(z)$ in powers of $z$ valid in the punctured disk $0 < |z| < sqrt{6pi}$: From equation (3): $$f(z) = frac{e^z}{z^3 left( frac{1}{6} - frac{z^2}{120} + O(z^4) right)} = frac{6 e^z}{z^3 left( 1 - frac{z^2}{20} + O(z^4) right)} tag{4}$$ Using the geometric series $(1 - u)^{-1} = 1 + u + u^2 + dots$ for $u = frac{z^2}{20} + O(z^4)$: $$left( 1 - frac{z^2}{20} + O(z^4) right)^{-1} = 1 + frac{z^2}{20} + O(z^4) tag{5}$$ Now write out the series expansion for the numerator $e^z$: $$e^z = 1 + z + frac{z^2}{2!} + frac{z^3}{3!} + frac{z^4}{4!} + O(z^5) = 1 + z + frac{z^2}{2} + frac{z^3}{6} + frac{z^4}{24} + O(z^5) tag{6}$$ Multiplying th

Question 14

A department head has 5 subordinates and 5 jobs to be performed. The time (in hours) that each subordinate will take to perform each job is given in the matrix below: $$begin{array}{cc} & text{Jobs} text{Subordinates} & begin{array}{c|ccccc} & A & B & C & D & E hline I & 4 & 9 & 4 & 12 & 4 II & 15 &

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Assignment Problem Formulation & Hungarian Method Let $c{ij}$ denote the time (in hours) required by subordinate $i$ ($i = I, II, III, IV, V$) to perform job $j$ ($j = A, B, C, D, E$). The given cost matrix is a balanced $5 times 5$ square matrix: $$C = begin{pmatrix} 4 & 9 & 4 & 12 & 4 15 & 11 & 20 & 5 & 8 17 & 7 & 15 & 12 & 18 9 & 13 & 11 & 9 & 14 6 & 11 & 12 & 9 & 14 end{pmatrix}$$ --- Part 1: Unrestricted Case (Hungarian Algorithm) Step 1: Row Reduction Subtract the minimum element of each row from all elements in that row: - Row 1: minimum is $4 implies [4-4, 9-4, 4-4, 12-4, 4-4] = [0, 5, 0, 8, 0]$ - Row 2: minimum is $5 implies [15-5, 11-5, 20-5, 5-5, 8-5] = [10, 6, 15, 0, 3]$ - Row 3: minimum is $7 implies [17-7, 7-7, 15-7, 12-7, 18-7] = [10, 0, 8, 5, 11]$ - Row 4: minimum is $9 implies [9-9, 13-9, 11-9, 9-9, 14-9] = [0, 4, 2, 0, 5]$ - Row 5: minimum is $6 implies [6-6, 11-6, 12-6, 9-6, 14-6] = [0, 5, 6, 3, 8]$ The row-reduced matrix is: $$begin{pmatrix} 0 & 5 & 0 & 8 & 0 10 & 6 & 15 & 0 & 3 10 & 0 & 8 & 5 & 11 0 & 4 & 2 & 0 & 5 0 & 5 & 6 & 3 & 8 end{pmatrix}$$ Step 2: Column Reduction Inspect each column for its minimum element: - Column A has minimum $0$. - Column B has minimum $0$. - Column C has minimum $0$. - Column D has minimum $0$. - Column E has minimum $0$. Since every column already contains at least one zero, the column-reduced matrix is identical to the row-reduced matrix. Step 3: Test for Optimality (Covering Zeros with Minimum Lines) We cover all zeros using a minimum number of horizontal and vertical lines: - Line 1: Row 1 (contains three zeros at A, C, E) - Line 2: Column A (covers remaining zeros in Row 4, Row 5) - Line 3: Column B (covers zero in Row 3) - Line 4: Column D (covers zeros in Row 2, Row 4) Number of lines required to cover all zeros is $4 < 5$ (order of matrix). Hence, the current zero configuration is not optimal. Step 4: Shifting Zeros The uncovered elements are: - Row 2: $15, 3$ - Row 3: $8, 11$ - Row 4: $2, 5$ - Row 5: $6,

Question 15

Solve the system of linear equations $$begin{aligned} 7x1 - x2 + 2x3 &= 11 2x1 + 8x2 - x3 &= 9 x1 - 2x2 + 9x3 &= 7 end{aligned}$$ correct up to 4 significant figures by the Gauss-Seidel iterative method. Take initially guessed solution as $x1 = x2 = x3 = 0$.

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Theoretical Principles of the Gauss-Seidel Method The Gauss-Seidel iterative method is an established numerical technique for solving a system of linear algebraic equations $A x = b$. 1. Diagonal Dominance and Convergence Criterion A square matrix $A = (a{ij}){n times n}$ is strictly diagonally dominant if: $$|a{ii}| sum{j eq i} |a{ij}| quad text{for all } i = 1, 2, dots, n tag{1}$$ By the Scarborough criterion and the Gershgorin circle theorem, strict diagonal dominance provides a sufficient condition for the convergence of the Gauss-Seidel iterative process from any arbitrary initial starting vector $x^{(0)}$. Let us check the given system: $$begin{aligned} 7x1 - x2 + 2x3 &= 11 tag{Row 1} 2x1 + 8x2 - x3 &= 9 tag{Row 2} x1 - 2x2 + 9x3 &= 7 tag{Row 3} end{aligned}$$ - Row 1: $|a{11}| = 7$, while $|a{12}| + |a{13}| = |-1| + |2| = 1 + 2 = 3$. Since $7 3$, Row 1 is strictly diagonally dominant. - Row 2: $|a{22}| = 8$, while $|a{21}| + |a{23}| = |2| + |-1| = 2 + 1 = 3$. Since $8 3$, Row 2 is strictly diagonally dominant. - Row 3: $|a{33}| = 9$, while $|a{31}| + |a{32}| = |1| + |-2| = 1 + 2 = 3$. Since $9 3$, Row 3 is strictly diagonally dominant. Since all rows satisfy the criterion, the system is strictly diagonally dominant, guaranteeing rapid and stable convergence. --- Step 2: Formulation of Iteration Equations Expressing each unknown in terms of the others: $$x1^{(k+1)} = frac{1}{7} left( 11 + x2^{(k)} - 2x3^{(k)} right) tag{2a}$$ $$x2^{(k+1)} = frac{1}{8} left( 9 - 2x1^{(k+1)} + x3^{(k)} right) tag{2b}$$ $$x3^{(k+1)} = frac{1}{9} left( 7 - x1^{(k+1)} + 2x2^{(k+1)} right) tag{2c}$$ Notice that in Gauss-Seidel iteration, the most recently updated components ($x1^{(k+1)}$ in (2b), and both $x1^{(k+1)}, x2^{(k+1)}$ in (2c)) are used immediately within the same iteration cycle. --- Step 3: Iterative Execution We are given the initial starting vector: $$x1^{(0)} = 0, quad x2^{(0)} = 0, quad x3^{(0)} = 0$$ Iteration 1 ($k = 0$): - Compute $x1^{(1)}$: $$x1^{(1)} = frac{1}

Question 16

Find the surface passing through the two lines $z = x = 0$ and $z - 1 = x - y = 0$, and satisfying the partial differential equation $frac{partial^2 z}{partial x^2} - 4frac{partial^2 z}{partial x partial y} + 4frac{partial^2 z}{partial y^2} = 0$.

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Problem Statement & Mathematical Classification We are required to determine the unique integral surface $z = z(x, y)$ that satisfies: 1. The second-order homogeneous linear partial differential equation: $$frac{partial^2 z}{partial x^2} - 4frac{partial^2 z}{partial x partial y} + 4frac{partial^2 z}{partial y^2} = 0 tag{1}$$ 2. The boundary condition that the surface passes through the straight line $L1$: $$z = 0, quad x = 0 tag{2}$$ 3. The boundary condition that the surface passes through the straight line $L2$: $$z - 1 = 0, quad x - y = 0 iff z = 1, quad y = x tag{3}$$ --- Step 1: General Solution of the Homogeneous PDE Let us express equation (1) in differential operator notation. Defining: $$D = frac{partial}{partial x}, quad D' = frac{partial}{partial y}$$ the PDE becomes: $$(D^2 - 4 D D' + 4 D'^2) z = 0 tag{4}$$ Factoring the differential operator: $$(D - 2 D')^2 z = 0 tag{5}$$ The corresponding auxiliary characteristic equation is: $$m^2 - 4m + 4 = 0 iff (m - 2)^2 = 0 tag{6}$$ The auxiliary equation has repeated roots: $$m1 = 2, quad m2 = 2$$ According to the standard theory of linear homogeneous PDEs with constant coefficients, when the auxiliary equation possesses a repeated root $m$, the complementary function contains an arbitrary function $f$ and a second arbitrary function $g$ multiplied by the independent variable $x$: $$z(x, y) = f(y + mx) + x , g(y + mx)$$ With $m = 2$, the general solution is: $$z(x, y) = f(y + 2x) + x , g(y + 2x) tag{7}$$ where $f$ and $g$ are arbitrary, twice-differentiable single-variable functions to be determined from the boundary conditions. --- Step 2: Application of Boundary Condition 1 ($z = 0$ along $x = 0$) The surface must contain the line $L1$ defined by $x = 0, z = 0$ for all values of $y$. Substituting $x = 0$ and $z = 0$ into the general solution (7): $$0 = f(y + 2(0)) + (0) cdot g(y + 2(0))$$ $$0 = f(y) + 0 implies f(y) = 0 quad text{for all } y in mathbb{R} tag{8}$$ Since the functional form of $f$ is identically

Question 17

In a fluid motion, there is a source of strength $2m$ placed at $z = 2$ and two sinks of strength $m$ are placed at $z = 2 + i$ and $z = 2 - i$. Find the streamlines.

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Hydrodynamic Principles and Complex Potential In two-dimensional, irrotational, incompressible fluid dynamics, the flow field is described by a single holomorphic complex potential: $$w(z) = phi(x, y) + i psi(x, y) tag{1}$$ where: - $phi(x, y)$ is the velocity potential, satisfying the Laplace equation $ abla^2 phi = 0$. - $psi(x, y)$ is the stream function, satisfying $ abla^2 psi = 0$. - The velocity components are given by: $$u = frac{partial phi}{partial x} = frac{partial psi}{partial y}, quad v = frac{partial phi}{partial y} = -frac{partial psi}{partial x}$$ $$frac{dw}{dz} = u - iv$$ The streamlines of the flow are the trajectories everywhere tangent to the velocity vector field, defined by the condition that the stream function remains constant: $$psi(x, y) = operatorname{Im}{w(z)} = text{constant} tag{2}$$ --- Singularities in 2D Flow: Sources and Sinks 1. A source of strength $m0$ at point $z = z0$ contributes a complex potential: $$w{text{source}}(z) = -m0 ln(z - z0)$$ 2. A sink of strength $m0$ (which is a negative source) at point $z = z0$ contributes: $$w{text{sink}}(z) = +m0 ln(z - z0)$$ --- Step 1: Construction of the Net Complex Potential We are given the following configuration of singularities: - A source of strength $2m$ located at $z1 = 2$: $$w1(z) = -2m ln(z - 2)$$ - A sink of strength $m$ located at $z2 = 2 + i$: $$w2(z) = +m ln[z - (2 + i)]$$ - A sink of strength $m$ located at $z3 = 2 - i$: $$w3(z) = +m ln[z - (2 - i)]$$ By the principle of linear superposition of potential flows, the total complex potential is: $$w(z) = w1(z) + w2(z) + w3(z)$$ $$w(z) = -2m ln(z - 2) + m ln(z - 2 - i) + m ln(z - 2 + i) tag{3}$$ --- Step 2: Algebraic Simplification of $w(z)$ Factor out $m$: $$w(z) = -m left{ 2ln(z - 2) - ln(z - 2 - i) - ln(z - 2 + i) right}$$ $$w(z) = -m lnleft[ frac{(z - 2)^2}{(z - 2 - i)(z - 2 + i)} right] tag{4}$$ Notice the product in the denominator: $$(z - 2 - i)(z - 2 + i) = (z - 2)^2 - i^2 = (z - 2)^2 + 1$$ Therefore: $$w(z) = -m lnleft

Question 18

A planet of mass $m$ is revolving around the sun of mass $M$. The kinetic energy $T$ and the potential energy $V$ of the planet are given by $T = frac{1}{2}m(dot{r}^2 + r^2dot{theta}^2)$ and $V = G M mleft(frac{1}{2a} - frac{1}{r}right)$, where $(r, theta)$ are the polar coordinates of the planet at

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Theoretical Framework of Hamiltonian Mechanics In classical analytical dynamics, the state of a holonomic conservative system having $n$ degrees of freedom is described in the configuration space by generalized coordinates $q = (q1, dots, qn)$ and generalized velocities $dot{q} = (dot{q}1, dots, dot{q}n)$. The transition to the Hamiltonian phase space $(q, p)$ replaces generalized velocities $dot{q}i$ with the conjugate generalized momenta $pi$, defined by: $$pi = frac{partial L}{partial dot{q}i} tag{1}$$ where $L(q, dot{q}, t) = T - V$ is the Lagrangian of the system. The Hamiltonian $H(q, p, t)$ is defined via the Legendre transformation: $$H(q, p, t) = sum{i=1}^n pi dot{q}i - L(q, dot{q}, t) tag{2}$$ and the system's dynamics are governed by Hamilton's canonical equations of motion: $$dot{q}i = frac{partial H}{partial pi}, quad dot{p}i = -frac{partial H}{partial qi} tag{3}$$ --- Step 1: Formulation of the Lagrangian The motion of the planet of mass $m$ in the plane of its orbit is characterized by two degrees of freedom ($n = 2$): - Generalized coordinates: $q1 = r$, $q2 = theta$. - Generalized velocities: $dot{q}1 = dot{r}$, $dot{q}2 = dot{theta}$. We are given: - Kinetic energy: $$T = frac{1}{2}m left( dot{r}^2 + r^2 dot{theta}^2 right) tag{4}$$ - Potential energy: $$V(r) = G M m left( frac{1}{2a} - frac{1}{r} right) tag{5}$$ (Note: The term $frac{GMm}{2a}$ represents a constant energy offset associated with the total orbital binding energy $E = -frac{GMm}{2a}$, while $-frac{GMm}{r}$ is the standard Newton gravitational potential). The Lagrangian $L = T - V$ is: $$L(r, theta, dot{r}, dot{theta}) = frac{1}{2}m left( dot{r}^2 + r^2 dot{theta}^2 right) - G M m left( frac{1}{2a} - frac{1}{r} right) tag{6}$$ --- Step 2: Determination of Generalized Conjugate Momenta Differentiating $L$ with respect to the generalized velocities: 1. Radial conjugate momentum $pr$: $$pr = frac{partial L}{partial dot{r}} = frac{partial T}{partial dot{r}} = m dot{r} tag{7a}$$ Inverting

Question 19

Evaluate, using the binary arithmetic, the following numbers in their given system: (i) $(634.235)8 - (132.223)8$ (ii) $(7text{AB}.432){16} - (5text{CA}.text{D}61){16}$

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Comprehensive Guide to Base Arithmetic via Binary Representation The problem requires us to perform subtraction between two numbers in base 8 (octal) and two numbers in base 16 (hexadecimal) by employing binary arithmetic, and then expressing the final result in the original respective number systems. The fundamental relationship connecting these positional systems is powers of 2: - Base 8 (Octal): $8 = 2^3$, so each octal digit maps uniquely and directly to a group of $3$ binary bits. - Base 16 (Hexadecimal): $16 = 2^4$, so each hexadecimal digit maps uniquely and directly to a group of $4$ binary bits. --- Part (i): Evaluation of $(634.235)8 - (132.223)8$ Step 1: Conversion of Octal Numbers to Binary Using the 3-bit equivalent for each octal digit: $$0 leftrightarrow 000, quad 1 leftrightarrow 001, quad 2 leftrightarrow 010, quad 3 leftrightarrow 011$$ $$4 leftrightarrow 100, quad 5 leftrightarrow 101, quad 6 leftrightarrow 110, quad 7 leftrightarrow 111$$ 1. Minuend $A = (634.235)8$: $$begin{array}{ccccccc} 6 & 3 & 4 & . & 2 & 3 & 5 downarrow & downarrow & downarrow & & downarrow & downarrow & downarrow 110 & 011 & 100 & . & 010 & 011 & 101 end{array}$$ $$A = (110011100.010011101)2$$ 2. Subtrahend $B = (132.223)8$: $$begin{array}{ccccccc} 1 & 3 & 2 & . & 2 & 2 & 3 downarrow & downarrow & downarrow & & downarrow & downarrow & downarrow 001 & 011 & 010 & . & 010 & 010 & 011 end{array}$$ $$B = (001011010.010010011)2$$ --- Step 2: Binary Subtraction ($A - B$) We perform column-by-column binary subtraction: $$0 - 0 = 0, quad 1 - 0 = 1, quad 1 - 1 = 0, quad 0 - 1 = 1 text{ (with a borrow of } 1 text{ from next higher bit)}$$ Let us align the two binary numbers: $$begin{array}{r@{quad}l} & 110011100 . 0100111012 - & 001011010 . 0100100112 hline end{array}$$ Let us compute the bits from right to left: - Fractional part: - Bit $-9$: $1 - 1 = 0$ - Bit $-8$: $0 - 1 implies$ borrow 1 from bit $-7$ (which becomes 0): $(2)2 - 1 = 1$ - Bit $-7$: $0 - 0 = 0$ - Bit $-6$: $1 - 1 =

Question 20

Given $frac{dy}{dx} = frac{y^2 - x}{y^2 + x}$ with initial condition $y = 1$ at $x = 0$. Find the value of $y$ for $x = 0.4$ by Euler's method, correct to 4 decimal places, taking step length $h = 0.1$.

Accepted Answer

Mathematical Formulation and Method Overview We are given the initial value problem (IVP): $$frac{dy}{dx} = f(x, y) = frac{y^2 - x}{y^2 + x}, quad y(x0) = y0 tag{1}$$ with initial values: $$x0 = 0.0, quad y0 = 1.0$$ and constant step size: $$h = 0.1$$ We are required to compute the numerical approximation of $y(x)$ at $x = 0.4$ using Euler's forward method, retaining $4$ decimal places in the final result. --- The Classical Euler Iteration Scheme Euler's method is the fundamental explicit single-step numerical integration method for first-order ODEs. Given the mesh points: $$xn = x0 + n h, quad n = 0, 1, 2, dots$$ the successive approximations $yn approx y(xn)$ are generated by the recurrence relation: $$y{n+1} = yn + h cdot f(xn, yn) tag{2}$$ where the slope at each stage is: $$kn = f(xn, yn) = frac{yn^2 - xn}{yn^2 + xn} tag{3}$$ To achieve $x = 0.4$ starting from $x0 = 0.0$ with $h = 0.1$, we require exactly $4$ integration steps: $$x0 = 0.0 xrightarrow{text{Step 1}} x1 = 0.1 xrightarrow{text{Step 2}} x2 = 0.2 xrightarrow{text{Step 3}} x3 = 0.3 xrightarrow{text{Step 4}} x4 = 0.4$$ --- Step-by-Step Numerical Computation Step 1: Evaluation at $x1 = 0.1$ - Given: $x0 = 0.0$, $y0 = 1.0$. - Evaluate the slope function $f(x0, y0)$: $$f(x0, y0) = frac{y0^2 - x0}{y0^2 + x0} = frac{1.0^2 - 0.0}{1.0^2 + 0.0} = frac{1.0}{1.0} = 1.000000$$ - Apply Euler's formula: $$y1 = y0 + h cdot f(x0, y0) = 1.0 + (0.1)(1.000000) = 1.0 + 0.100000 = 1.100000$$ Thus, at $x1 = 0.1$: $$mathbf{y1 = 1.100000}$$ --- Step 2: Evaluation at $x2 = 0.2$ - Given: $x1 = 0.1$, $y1 = 1.100000$. - Compute $y1^2$: $$y1^2 = (1.1)^2 = 1.210000$$ - Evaluate the slope function $f(x1, y1)$: $$f(x1, y1) = frac{y1^2 - x1}{y1^2 + x1} = frac{1.210000 - 0.1}{1.210000 + 0.1} = frac{1.110000}{1.310000} approx 0.847328$$ - Apply Euler's formula: $$y2 = y1 + h cdot f(x1, y1) = 1.100000 + (0.1)(0.847328) = 1.100000 + 0.084733 = 1.184733$$ Thus, at $x2 = 0.2$: $$mathbf{y2 = 1.184733}$$ --- Step 3: Evaluation at $x3 = 0.3$

Question 21

By eliminating the arbitrary functions $f$ and $g$ from $z = f(x^2 - y) + g(x^2 + y)$, form partial differential equation.

Accepted Answer

Problem Formulation and Theoretical Foundation We are given the relation: $$z = f(x^2 - y) + g(x^2 + y) tag{1}$$ where $f$ and $g$ are arbitrary, twice-differentiable single-variable functions, and $z$ is the dependent variable of two independent variables $x$ and $y$. In the theory of partial differential equations, a relation containing two arbitrary functions $f$ and $g$ generally gives rise to a partial differential equation of second order upon the systematic elimination of $f, g$ and their successive derivatives. Let us introduce the intermediate characteristic variables: $$u = x^2 - y tag{2a}$$ $$v = x^2 + y tag{2b}$$ Then the given function (1) is expressed compactly as: $$z = f(u) + g(v) tag{3}$$ --- Step 1: First-Order Partial Derivatives Let us compute the first-order partial derivatives of $z$ with respect to $x$ and $y$ using the multivariable chain rule. 1. Differentiating with respect to $x$: $$p = frac{partial z}{partial x} = frac{d f}{d u} frac{partial u}{partial x} + frac{d g}{d v} frac{partial v}{partial x}$$ From (2a) and (2b): $$frac{partial u}{partial x} = 2x, quad frac{partial v}{partial x} = 2x$$ Substituting these into the expression for $p$: $$p = frac{partial z}{partial x} = 2x f'(u) + 2x g'(v) = 2x left[ f'(u) + g'(v) right] tag{4}$$ Dividing by $2x$ (assuming $x eq 0$): $$frac{p}{2x} = frac{1}{2x} frac{partial z}{partial x} = f'(u) + g'(v) tag{5}$$ 2. Differentiating with respect to $y$: $$q = frac{partial z}{partial y} = frac{d f}{d u} frac{partial u}{partial y} + frac{d g}{d v} frac{partial v}{partial y}$$ From (2a) and (2b): $$frac{partial u}{partial y} = -1, quad frac{partial v}{partial y} = +1$$ Thus: $$q = frac{partial z}{partial y} = -f'(u) + g'(v) tag{6}$$ --- Step 2: Second-Order Partial Derivatives We now calculate the second-order partial derivatives: $$r = frac{partial^2 z}{partial x^2}, quad s = frac{partial^2 z}{partial x partial y}, quad t = frac{partial^2 z}{partial y^2}$$ 1. Computing $r = frac{partial^2 z}{partial x^2}$

Question 22

A mechanical system with 2 degrees of freedom has the Lagrangian $$L = frac{1}{2}m(dot{x}^2 + dot{y}^2) - frac{1}{2}m(w1^2 x^2 + w2^2 y^2) + kxy$$ where $m, w1, w2, k$ are constants. Find the parameter $theta$ so that under the transformation $$x = q1costheta - q2sintheta, quad y = q1sintheta + q2co

Accepted Answer

Problem Formulation & Coordinate Decoupling We consider a classical conservative mechanical system possessing two degrees of freedom with generalized coordinates $(x, y)$ and the Lagrangian: $$L(x, y, dot{x}, dot{y}) = T - V = frac{1}{2}m(dot{x}^2 + dot{y}^2) - frac{1}{2}m(w1^2 x^2 + w2^2 y^2) + kxy tag{1}$$ Here: - The kinetic energy is: $$T = frac{1}{2}m(dot{x}^2 + dot{y}^2) tag{2}$$ - The potential energy is: $$V = frac{1}{2}m(w1^2 x^2 + w2^2 y^2) - kxy tag{3}$$ We are given an orthogonal rotation transformation by an angle $theta$: $$begin{pmatrix} x y end{pmatrix} = begin{pmatrix} costheta & -sintheta sintheta & costheta end{pmatrix} begin{pmatrix} q1 q2 end{pmatrix} tag{4}$$ Explicitly: $$x = q1 costheta - q2 sintheta tag{5a}$$ $$y = q1 sintheta + q2 costheta tag{5b}$$ We are tasked with: 1. Finding the rotation angle $theta$ such that the cross-coupling term $q1 q2$ vanishes from the transformed Lagrangian $L(q1, q2, dot{q}1, dot{q}2)$. 2. Deriving the decoupled equations of motion for $q1$ and $q2$ expressed independently of the parameter $theta$. --- Step 1: Transformation of Kinetic Energy Differentiating equations (5a) and (5b) with respect to time $t$ (noting that $theta$ is a constant rotation angle): $$dot{x} = dot{q}1 costheta - dot{q}2 sintheta$$ $$dot{y} = dot{q}1 sintheta + dot{q}2 costheta$$ Evaluating the sum of squares of velocities: $$dot{x}^2 + dot{y}^2 = (dot{q}1 costheta - dot{q}2 sintheta)^2 + (dot{q}1 sintheta + dot{q}2 costheta)^2$$ $$= dot{q}1^2 (cos^2theta + sin^2theta) + dot{q}2^2 (sin^2theta + cos^2theta) + 2dot{q}1dot{q}2(-costhetasintheta + sinthetacostheta)$$ $$dot{x}^2 + dot{y}^2 = dot{q}1^2 + dot{q}2^2 tag{6}$$ Since orthogonal transformations preserve the Euclidean vector norm, the kinetic energy transforms into an uncoupled, isotropic quadratic form: $$T = frac{1}{2}m(dot{q}1^2 + dot{q}2^2) tag{7}$$ Notice that $T$ naturally contains no cross terms $dot{q}1dot{q}2$. --- Step 2: Transformation of Potential Energy and Elimination

Question 23

(i) Find the conjunctive normal form (CNF) of the following Boolean function : $$f(x, y, z, t) = x cdot y cdot z + bar{x} cdot y cdot (t + bar{z})$$ (ii) Express the Boolean function $$f(x, y, z) = x + (bar{x} cdot bar{y} + bar{x} cdot z) + z$$ in disjunctive normal form (DNF) and construct the trut

Accepted Answer

Mathematical Foundations of Boolean Normal Forms In Boolean algebra and propositional logic: 1. Conjunctive Normal Form (CNF / Product of Sums): A Boolean expression represented as a conjunction (logical AND / product $prod$) of clauses, where each clause is a disjunction (logical OR / sum $sum$) of literals: $$f = prod{i} Mi quad text{or} quad f = (l{11} + l{12} + dots)(l{21} + l{22} + dots)dots$$ The canonical CNF is the product of all maxterms where the function evaluates to $0$. 2. Disjunctive Normal Form (DNF / Sum of Products): A Boolean expression represented as a disjunction (logical OR / sum $sum$) of terms, where each term is a conjunction (logical AND / product $prod$) of literals: $$f = sumj mj quad text{or} quad f = (l{11} cdot l{12} dots) + (l{21} cdot l{22} dots) + dots$$ The canonical DNF is the sum of all minterms where the function evaluates to $1$. --- Part (i): CNF of $f(x, y, z, t) = x cdot y cdot z + bar{x} cdot y cdot (t + bar{z})$ We are given the 4-variable Boolean function: $$f(x, y, z, t) = x y z + bar{x} y (t + bar{z}) tag{1}$$ Step 1: Algebraic Simplification Factor out $y$ from both terms: $$f = y left[ x z + bar{x}(t + bar{z}) right] = y left( x z + bar{x} t + bar{x} bar{z} right) tag{2}$$ Notice that if $y = 0$, $f = 0$ irrespective of the values of $x, z, t$. If $y = 1$: $$f = x z + bar{x} t + bar{x} bar{z}$$ - Case $x = 1$: $f = 1 cdot z + 0 = z$. Thus $f = 1$ when $z = 1$ (for $t in {0, 1}$). Minterms: $(x, y, z, t) in {(1, 1, 1, 0), (1, 1, 1, 1)} implies m{14}, m{15}$. - Case $x = 0$: $f = 0 + 1 cdot t + 1 cdot bar{z} = t + bar{z}$. Thus $f = 1$ if $t = 1$ or $z = 0$. Combinations with $x = 0, y = 1$: - $(0, 1, 0, 0) implies m4$ ($f = 1$) - $(0, 1, 0, 1) implies m5$ ($f = 1$) - $(0, 1, 1, 1) implies m7$ ($f = 1$) - $(0, 1, 1, 0) implies m6$ gives $f = 0 + 0 = 0$. Step 2: Identification of Maxterms (Where $f = 0$) The function evaluates to $1$ only for the $5$ minterms: $$text{ON-set} = sum m(4, 5, 7, 14, 15)$$ Therefore, the funct

Question 24

A perfectly rough ball is at rest within a hollow cylindrical roller. The roller is drawn along a level path with uniform velocity $V$. Let $a$ and $b$ be the radii of the ball and the roller respectively. If $V^2 frac{27}{7}g(b - a)$, then show that the ball will roll completely round the inside of

Accepted Answer

Physical Setup & Dynamics in a Translating Reference Frame We consider a solid uniform spherical ball of radius $a$ and mass $M$ initially at rest at the bottom of a hollow cylindrical roller of inner radius $b$. The roller is drawn horizontally along a level path at a constant, uniform velocity $V$. The inside surface of the roller is perfectly rough, ensuring that the ball rolls without slipping at all times. Level Path O (Cylinder Axis) Uniform Velocity V a &952; Apex (&952; = &960;) N (Normal Force) Mg Radius b Condition for Complete Loop: V&178; > (27/7) g (b - a) Figure: Geometry and critical forces of a rough sphere rolling inside a hollow cylinder translating at uniform speed $V$. --- Step 1: Selection of Reference Frame and Initial Conditions Let us adopt the reference frame attached to the axis of the translating cylindrical roller: - Since the roller moves with a uniform (constant) velocity $V$ along a straight level path: $$frac{dvec{V}}{dt} = vec{0}$$ - Hence, the reference frame translating with the roller is an inertial frame. No fictitious forces (such as d'Alembert inertial forces) act in this frame. In this moving frame of reference: 1. The hollow cylindrical roller of radius $b$ is stationary. 2. At time $t = 0^-$, the ball was at rest in the laboratory frame. 3. When the roller is drawn forward with velocity $V hat{i}$, the ball's center at the lowest position relative to the roller has a horizontal velocity of magnitude $v0 = V$ in the opposite direction. 4. Because the surface is perfectly rough, pure rolling is established, so the initial speed of the center of mass of the ball at the lowest equilibrium point ($theta = 0$) is: $$v0 = V tag{1}$$ --- Step 2: Kinematics of Pure Rolling Inside a Cylinder Let: - $c = b - a$ be the distance from the axis of the cylinder $O$ to the center of the spherical ball $C$. - $theta$ be the angular displacement of the center of the ball measured from the downward vertical. - As the ball moves along the cyl

Question 25

Solve the partial differential equation $$a^2 frac{partial^2 u}{partial x^2} = frac{partial^2 u}{partial t^2}, quad 0 0$$ subject to the conditions $$u(0, t) = 0, quad u(L, t) = 0, quad t 0$$ $$u(x, 0) = x, quad left(frac{partial u}{partial t}right){t = 0} = 1, quad 0 < x < L$$

Accepted Answer

Mathematical Formulation & Physical Context We are required to solve the one-dimensional homogeneous wave equation describing transverse vibrations of a flexible stretched string of length $L$ with fixed boundaries and wave speed $c = a$: $$frac{partial^2 u}{partial t^2} = a^2 frac{partial^2 u}{partial x^2}, quad 0 0 tag{1}$$ The boundary conditions are: $$u(0, t) = 0 tag{2a}$$ $$u(L, t) = 0 tag{2b}$$ for all $t 0$. The initial conditions are: - Initial displacement: $$u(x, 0) = f(x) = x, quad 0 < x < L tag{3a}$$ - Initial velocity: $$left( frac{partial u}{partial t} right){t = 0} = g(x) = 1, quad 0 < x < L tag{3b}$$ --- Step 1: Separation of Variables Assume a product solution of the form: $$u(x, t) = X(x) cdot T(t) tag{4}$$ Differentiating and substituting into (1): $$X(x) T''(t) = a^2 X''(x) T(t)$$ Dividing both sides by $a^2 X(x) T(t)$: $$frac{T''(t)}{a^2 T(t)} = frac{X''(x)}{X(x)} = -lambda^2 tag{5}$$ where $-lambda^2$ is the separation constant (chosen negative to produce oscillatory harmonic motion). This yields two coupled ordinary differential equations: 1. Spatial Equation: $$X''(x) + lambda^2 X(x) = 0 tag{6}$$ 2. Temporal Equation: $$T''(t) + (lambda a)^2 T(t) = 0 tag{7}$$ --- Step 2: Spatial Sturm-Liouville Boundary Value Problem The spatial ODE (6) has general solution: $$X(x) = C1 cos(lambda x) + C2 sin(lambda x)$$ Applying the boundary conditions: 1. $u(0, t) = X(0) T(t) = 0 implies X(0) = 0$: $$X(0) = C1 (1) + C2 (0) = 0 implies C1 = 0$$ Thus $X(x) = C2 sin(lambda x)$. 2. $u(L, t) = X(L) T(t) = 0 implies X(L) = 0$: $$C2 sin(lambda L) = 0$$ For non-trivial solutions ($C2 eq 0$): $$sin(lambda L) = 0 implies lambdan L = n pi implies lambdan = frac{n pi}{L}, quad n = 1, 2, 3, dots tag{8}$$ The eigenfunctions are: $$Xn(x) = sinleft( frac{n pi x}{L} right) tag{9}$$ --- Step 3: Temporal Equation and Superposition For each eigenvalue $lambdan = frac{n pi}{L}$, define the circular frequency: $$omegan = lambdan a = frac{n pi a}{L} tag{10}$$ Equation (7) become

Question 26

Reduce the partial differential equation $$frac{partial^2 z}{partial y^2} - frac{partial^2 z}{partial x partial y} + frac{partial z}{partial x} - frac{partial z}{partial y}left(1 + frac{1}{x}right) + frac{z}{x} = 0$$ to canonical form.

Accepted Answer

Classification of Second-Order Linear Partial Differential Equations We are given the second-order linear PDE: $$frac{partial^2 z}{partial y^2} - frac{partial^2 z}{partial x partial y} + frac{partial z}{partial x} - left(1 + frac{1}{x}right)frac{partial z}{partial y} + frac{z}{x} = 0 tag{1}$$ Let us write the general second-order linear PDE in two independent variables $(x, y)$: $$R frac{partial^2 z}{partial x^2} + S frac{partial^2 z}{partial x partial y} + T frac{partial^2 z}{partial y^2} + P frac{partial z}{partial x} + Q frac{partial z}{partial y} + Z z = 0 tag{2}$$ Matching coefficients of equation (1): - $R = 0$ - $S = -1$ - $T = 1$ - $P = 1$ - $Q = -left(1 + frac{1}{x}right)$ - $Z = frac{1}{x}$ --- Step 1: Determination of PDE Type (Discriminant) The discriminant is: $$Delta = S^2 - 4 R T = (-1)^2 - 4(0)(1) = 1 - 0 = 1 0 tag{3}$$ Since $Delta = 1 0$ strictly for all $(x, y)$ in the domain ($x eq 0$), the given partial differential equation is hyperbolic everywhere. --- Step 2: Characteristic Curves and Coordinates The characteristic differential equations for a second-order PDE are given by: $$R left(frac{dy}{dx}right)^2 - S left(frac{dy}{dx}right) + T = 0$$ Since $R = 0$, it is advantageous to write the characteristic equation in terms of $frac{dx}{dy}$: $$T left(frac{dx}{dy}right)^2 - S left(frac{dx}{dy}right) + R = 0 tag{4}$$ Substitute $T = 1, S = -1, R = 0$: $$1 cdot left(frac{dx}{dy}right)^2 - (-1) left(frac{dx}{dy}right) + 0 = 0$$ $$left(frac{dx}{dy}right)^2 + frac{dx}{dy} = 0 iff frac{dx}{dy}left( frac{dx}{dy} + 1 right) = 0 tag{5}$$ This yields two distinct families of characteristic ODEs: 1. First family: $$frac{dx}{dy} = 0 implies dx = 0 implies x = c1$$ Thus, we define the first canonical coordinate: $$xi = x tag{6a}$$ 2. Second family: $$frac{dx}{dy} + 1 = 0 implies frac{dx}{dy} = -1 implies dx + dy = 0 implies x + y = c2$$ Thus, we define the second canonical coordinate: $$eta = x + y tag{6b}$$ The Jacobian of the transformation $(xi, eta) leftri

Question 27

Compute a root of the equation $log{10}(2x + 1) - x^2 + 3 = 0$, in the interval $[0, 3]$, by Regula-Falsi method, correct to 6 decimal places.

Accepted Answer

Theoretical Principles of the Regula-Falsi (False Position) Method The Regula-Falsi method (or method of false position) is an ancient, robust bracketed numerical method for determining real roots of non-linear algebraic and transcendental equations $f(x) = 0$. Given a continuous real function $f(x)$ on an interval $[a, b]$ such that: $$f(a) cdot f(b) 0$, the root lies in $[c, b]$, so set $a leftarrow c$. --- Step 1: Function Definition & Initial Bracket Verification We are given: $$f(x) = log{10}(2x + 1) - x^2 + 3 = 0 tag{3}$$ on the initial interval: $$[a0, b0] = [0, 3]$$ Let us evaluate $f(x)$ at the interval endpoints: 1. At $a0 = 0$: $$f(0) = log{10}(2(0) + 1) - 0^2 + 3 = log{10}(1) + 3 = 0 + 3 = +3.00000000 0$$ 2. At $b0 = 3$: $$f(3) = log{10}(2(3) + 1) - 3^2 + 3 = log{10}(7) - 9 + 3 = 0.84509804 - 6 = -5.15490196 0$$ Since $f(c1) 0$, set $a leftarrow c1 = 1.10363068$. --- Iteration 2: - $a = 1.10363068$, $f(a) = +2.28813367$ - $b = 3.00000000$, $f(b) = -5.15490196$ - Compute $c2$: $$c2 = frac{(1.10363068)(-5.15490196) - (3)(2.28813367)}{-5.15490196 - 2.28813367} = frac{-5.68910811 - 6.86440101}{-7.44303563} = frac{-12.55350912}{-7.44303563} approx 1.68661143$$ - Evaluate $f(c2)$: $$f(1.68661143) = log{10}(4.37322286) - (1.68661143)^2 + 3 = 0.64080133 - 2.84465811 + 3 = +0.79614322 0$$ Since $f(c2) 0$, set $a leftarrow c2 = 1.68661143$. --- Iteration 3: - $a = 1.68661143$, $f(a) = +0.79614322$ - $b = 3.00000000$, $f(b) = -5.15490196$ - Compute $c3$: $$c3 = frac{(1.68661143)(-5.15490196) - (3)(0.79614322)}{-5.15490196 - 0.79614322} = frac{-8.69431718 - 2.38842966}{-5.95104518} approx 1.86231934$$ - Evaluate $f(c3)$: $$f(1.86231934) = log{10}(4.72463868) - (1.86231934)^2 + 3 = 0.67436848 - 3.46823332 + 3 = +0.20613516 0$$ Since $f(c3) 0$, set $a leftarrow c3 = 1.86231934$. --- Iteration 4: - $a = 1.86231934$, $f(a) = +0.20613516$ - $b = 3.00000000$, $f(b) = -5.15490196$ - Compute $c4$: $$c4 = frac{(1.86231934)(-5.15490196) - (3)(0.20613516)}{-5.15490196 - 0.2061

Question 28

Determine under what conditions the velocity field $u = c(x^2 - y^2), v = -2cxy, w = 0$ is a solution to the Navier-Stokes momentum equations. Assuming that the conditions are met, determine the resulting pressure distribution, when $z$ is up and the external body forces are $Bx = 0, By = 0, Bz = -g

Accepted Answer

Hydrodynamic Principles & The Navier-Stokes Equations The motion of an incompressible Newtonian fluid of constant density $rho$ and dynamic viscosity $mu$ (kinematic viscosity $ u = mu/rho$) subject to external body forces $vec{B} = (Bx, By, Bz)$ is governed by: 1. The Continuity Equation (Conservation of Mass): $$ abla cdot vec{V} = frac{partial u}{partial x} + frac{partial v}{partial y} + frac{partial w}{partial z} = 0 tag{1}$$ 2. The Navier-Stokes Momentum Equations: $$rho left[ frac{partial vec{V}}{partial t} + (vec{V} cdot abla)vec{V} right] = - abla p + rho vec{B} + mu abla^2 vec{V} tag{2}$$ where $vec{V} = u hat{i} + v hat{j} + w hat{k}$ is the velocity vector field and $p(x, y, z)$ is the scalar pressure field. We are given the velocity field: $$u = c(x^2 - y^2), quad v = -2cxy, quad w = 0 tag{3}$$ where $c$ is a given real constant, and external body forces: $$Bx = 0, quad By = 0, quad Bz = -g tag{4}$$ --- Step 1: Verification of the Incompressibility Condition Let us evaluate the divergence of the velocity field: $$frac{partial u}{partial x} = frac{partial}{partial x}left[ c(x^2 - y^2) right] = 2cx tag{5a}$$ $$frac{partial v}{partial y} = frac{partial}{partial y}left[ -2cxy right] = -2cx tag{5b}$$ $$frac{partial w}{partial z} = 0 tag{5c}$$ Summing these partial derivatives: $$ abla cdot vec{V} = frac{partial u}{partial x} + frac{partial v}{partial y} + frac{partial w}{partial z} = 2cx - 2cx + 0 = 0 tag{6}$$ Since the divergence vanishes identically everywhere, the fluid flow is kinematically possible and strictly incompressible for any real value of $c$. --- Step 2: Analysis of Viscous Dissipation and Laplacian $ abla^2 vec{V}$ Let us compute the Laplacian of each velocity component: 1. For $u = c(x^2 - y^2)$: $$frac{partial^2 u}{partial x^2} = frac{partial}{partial x}(2cx) = 2c$$ $$frac{partial^2 u}{partial y^2} = frac{partial}{partial y}(-2cy) = -2c$$ $$frac{partial^2 u}{partial z^2} = 0$$ $$ abla^2 u = frac{partial^2 u}{partial x^2} + frac{partial^2 u}{

Maths Optional 2023 Paper II

28 Questions Available

Maths Optional 2023 Paper II

28 Questions Available

Question 1: Using the method of Lagrange's multipliers, find the minimum and maximum distances of the point $P(2, 6, 3)$ from the sphere $x^2 + y^2 + z^2 = 4$.

Question 2: Prove that a non-commutative group of order $2p$, where $p$ is an odd prime, must have a subgroup of order $p$.

Question 3: A person requires 24, 24 and 20 units of chemicals $A, B$ and $C$ respectively for his garden. Product $P$ contains 2, 4 and 1 units of chemicals $A,

Question 4: State the sufficient conditions for a function $f(z) = f(x + iy) = u(x, y) + iv(x, y)$ to be analytic in its domain. Hence, show that $f(z) = log z$ i

Question 5: Test the convergence of the series $$sum{n=1}^infty frac{1 cdot 3 cdot 5 cdots (2n - 1)}{2 cdot 4 cdot 6 cdots (2n)} frac{x^{2n+1}}{(2n + 1)}, qquad x

Question 6: Express the ideal $4mathbb{Z} + 6mathbb{Z}$ as a principal ideal in the integral domain $mathbb{Z}$.

Question 7: Let $G$ be a group of order 10 and $G'$ be a group of order 6. Examine whether there exists a homomorphism of $G$ onto $G'$.

Question 8: Evaluate $int0^{2pi} frac{cos 2theta}{5 + 4costheta} dtheta$ using contour integration.

Question 9: Prove that $x^2 + 1$ is an irreducible polynomial in $mathbb{Z}3[x]$. Further show that the quotient ring $frac{mathbb{Z}3[x]}{langle x^2 + 1 rangle}$

Question 10: Prove that $u(x, y) = e^x(xcos y - ysin y)$ is harmonic. Find its conjugate harmonic function $v(x, y)$ and express the corresponding analytic functio

Question 11: Solve the following linear programming problem by Big M method: $$text{Minimize } Z = 2x1 + 3x2$$ subject to $$begin{aligned} x1 + x2 &ge 9 x1 + 2x2 &

Question 12: Prove that the oscillation of a real-valued bounded function $f$ defined on $[a, b]$ is the supremum of the set ${|f(x1) - f(x2)| : x1, x2 in [a, b]}$

Question 13: Classify the singular point $z = 0$ of the function $f(z) = frac{e^z}{z - sin z}$ and obtain the principal part of its Laurent series expansion.

Question 14: A department head has 5 subordinates and 5 jobs to be performed. The time (in hours) that each subordinate will take to perform each job is given in t

Question 15: Solve the system of linear equations $$begin{aligned} 7x1 - x2 + 2x3 &= 11 2x1 + 8x2 - x3 &= 9 x1 - 2x2 + 9x3 &= 7 end{aligned}$$ correct up to 4 sign

Question 16: Find the surface passing through the two lines $z = x = 0$ and $z - 1 = x - y = 0$, and satisfying the partial differential equation $frac{partial^2 z

Question 17: In a fluid motion, there is a source of strength $2m$ placed at $z = 2$ and two sinks of strength $m$ are placed at $z = 2 + i$ and $z = 2 - i$. Find

Question 18: A planet of mass $m$ is revolving around the sun of mass $M$. The kinetic energy $T$ and the potential energy $V$ of the planet are given by $T = frac

Question 19: Evaluate, using the binary arithmetic, the following numbers in their given system: (i) $(634.235)8 - (132.223)8$ (ii) $(7text{AB}.432){16} - (5text{C

Question 20: Given $frac{dy}{dx} = frac{y^2 - x}{y^2 + x}$ with initial condition $y = 1$ at $x = 0$. Find the value of $y$ for $x = 0.4$ by Euler's method, correc

Question 21: By eliminating the arbitrary functions $f$ and $g$ from $z = f(x^2 - y) + g(x^2 + y)$, form partial differential equation.

Question 22: A mechanical system with 2 degrees of freedom has the Lagrangian $$L = frac{1}{2}m(dot{x}^2 + dot{y}^2) - frac{1}{2}m(w1^2 x^2 + w2^2 y^2) + kxy$$ whe

Question 23: (i) Find the conjunctive normal form (CNF) of the following Boolean function : $$f(x, y, z, t) = x cdot y cdot z + bar{x} cdot y cdot (t + bar{z})$$ (

Question 24: A perfectly rough ball is at rest within a hollow cylindrical roller. The roller is drawn along a level path with uniform velocity $V$. Let $a$ and $b

Question 25: Solve the partial differential equation $$a^2 frac{partial^2 u}{partial x^2} = frac{partial^2 u}{partial t^2}, quad 0 0$$ subject to the conditions $$

Question 26: Reduce the partial differential equation $$frac{partial^2 z}{partial y^2} - frac{partial^2 z}{partial x partial y} + frac{partial z}{partial x} - frac

Question 27: Compute a root of the equation $log{10}(2x + 1) - x^2 + 3 = 0$, in the interval $[0, 3]$, by Regula-Falsi method, correct to 6 decimal places.

Question 28: Determine under what conditions the velocity field $u = c(x^2 - y^2), v = -2cxy, w = 0$ is a solution to the Navier-Stokes momentum equations. Assumin