BE Computer Engineering (Pokhara University) Calculus II (PU, MTH 210) Question Paper 2078 Nepal

(a) Polar transformation of the gradient

With $x = r\cos\theta$, $y = r\sin\theta$ we have $r^2 = x^2+y^2$ and $\theta = \tan^{-1}(y/x)$. Treating $u$ as a function of $(r,\theta)$ via $(x,y)$:

$$\frac{\partial u}{\partial r} = u_x\frac{\partial x}{\partial r} + u_y\frac{\partial y}{\partial r} = u_x\cos\theta + u_y\sin\theta,$$ $$\frac{\partial u}{\partial \theta} = u_x\frac{\partial x}{\partial \theta} + u_y\frac{\partial y}{\partial \theta} = -u_x\,r\sin\theta + u_y\,r\cos\theta.$$

Then

$$\left(\frac{\partial u}{\partial r}\right)^2 = u_x^2\cos^2\theta + 2u_xu_y\sin\theta\cos\theta + u_y^2\sin^2\theta,$$ $$\frac{1}{r^2}\left(\frac{\partial u}{\partial \theta}\right)^2 = u_x^2\sin^2\theta - 2u_xu_y\sin\theta\cos\theta + u_y^2\cos^2\theta.$$

Adding, the cross terms cancel and $\cos^2\theta+\sin^2\theta=1$ gives

$$\left(\frac{\partial u}{\partial r}\right)^2 + \frac{1}{r^2}\left(\frac{\partial u}{\partial \theta}\right)^2 = u_x^2 + u_y^2 = \left(\frac{\partial u}{\partial x}\right)^2 + \left(\frac{\partial u}{\partial y}\right)^2. \qquad\blacksquare$$

(b) Euler's theorem

Statement. If $z=f(x,y)$ is homogeneous of degree $n$ (i.e. $f(tx,ty)=t^n f(x,y)$), then

$$x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y} = n\,z.$$

Here $u=\tan^{-1}\!\big(\tfrac{x^3+y^3}{x-y}\big)$ is not homogeneous, but $v=\tan u = \dfrac{x^3+y^3}{x-y}$ is homogeneous of degree $3-1 = 2$. Applying Euler's theorem to $v$:

$$x v_x + y v_y = 2v.$$

Since $v=\tan u$, $v_x = \sec^2u\,u_x$, $v_y=\sec^2u\,u_y$, so

$$\sec^2u\,(x u_x + y u_y) = 2\tan u \;\Rightarrow\; x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = 2\frac{\tan u}{\sec^2u} = 2\sin u\cos u = \sin 2u.$$ $$\boxed{\,x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = \sin 2u\,}$$

(c) Extrema of $f(x,y)=x^3+y^3-3axy$

Stationary points: $f_x = 3x^2 - 3ay = 0$, $f_y = 3y^2 - 3ax = 0$, i.e. $x^2 = ay$, $y^2 = ax$. Solving gives $(0,0)$ and $(a,a)$.

Second derivatives: $f_{xx}=6x$, $f_{yy}=6y$, $f_{xy}=-3a$. Discriminant $D = f_{xx}f_{yy} - f_{xy}^2 = 36xy - 9a^2$.

• At $(0,0)$: $D = -9a^2 < 0$ → saddle point.
• At $(a,a)$: $D = 36a^2 - 9a^2 = 27a^2 > 0$. With $f_{xx}=6a$: if $a>0$, $f_{xx}>0$ → minimum; if $a<0$, $f_{xx}<0$ → maximum.

For $a>0$ the minimum value is $f(a,a) = a^3+a^3-3a^3 = -a^3$.

(a) $\iint_R (x^2+y^2)\,dx\,dy$ over $y=x,\,y=0,\,x=a$

The region is the triangle with vertices $(0,0),(a,0),(a,a)$: for $0\le x\le a$, $y$ runs from $0$ to $x$.

$$\int_0^a\int_0^x (x^2+y^2)\,dy\,dx = \int_0^a\Big[x^2 y + \tfrac{y^3}{3}\Big]_0^x dx = \int_0^a\Big(x^3 + \tfrac{x^3}{3}\Big)dx = \int_0^a \tfrac{4}{3}x^3\,dx = \frac{4}{3}\cdot\frac{a^4}{4} = \frac{a^4}{3}.$$

(b) Change of order in $\int_0^1\int_{x^2}^{2-x} xy\,dy\,dx$

The region is bounded by $y=x^2$ (below) and $y=2-x$ (above), with $0\le x\le1$. These curves meet at $x=1$ ($y=1$). Splitting by $y$:

• For $0\le y\le 1$: $x$ from $0$ to $\sqrt{y}$.
• For $1\le y\le 2$: $x$ from $0$ to $2-y$.

$$I = \int_0^1\int_0^{\sqrt y} xy\,dx\,dy + \int_1^2\int_0^{2-y} xy\,dx\,dy.$$

First: $\int_0^1 y\cdot\tfrac{y}{2}\,dy = \tfrac12\int_0^1 y^2\,dy = \tfrac16.$ Second: $\int_1^2 y\cdot\tfrac{(2-y)^2}{2}\,dy = \tfrac12\int_1^2 (4y - 4y^2 + y^3)\,dy = \tfrac12\big[2y^2 - \tfrac{4y^3}{3} + \tfrac{y^4}{4}\big]_1^2.$ At $2$: $8 - \tfrac{32}{3} + 4 = \tfrac{36-32}{3}=\tfrac{4}{3}$. At $1$: $2 - \tfrac43 + \tfrac14 = \tfrac{24-16+3}{12}=\tfrac{11}{12}$. Difference $=\tfrac{16}{12}-\tfrac{11}{12}=\tfrac{5}{12}$, times $\tfrac12 = \tfrac{5}{24}$.

$$I = \frac16 + \frac{5}{24} = \frac{4}{24}+\frac{5}{24} = \frac{9}{24} = \frac{3}{8}.$$

(c) Volume of the tetrahedron $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ in the first octant

$$V = \int_0^a\int_0^{b(1-x/a)}\int_0^{c(1-x/a-y/b)} dz\,dy\,dx.$$

This evaluates (standard result) to

$$V = \frac{abc}{6}.$$

(a) Green's theorem

Statement. If $P,Q$ have continuous partial derivatives on a region $R$ bounded by a simple closed positively-oriented curve $C$, then

$$\oint_C P\,dx + Q\,dy = \iint_R\Big(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\Big)\,dx\,dy.$$

Here $P = 3x^2 - 8y^2$, $Q = 4y - 6xy$, so

$$\frac{\partial Q}{\partial x} = -6y, \qquad \frac{\partial P}{\partial y} = -16y, \qquad \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 10y.$$

The curves $y=\sqrt x$ and $y=x^2$ meet at $(0,0)$ and $(1,1)$; for $0\le x\le1$, $x^2\le y\le\sqrt x$.

$$\iint_R 10y\,dy\,dx = \int_0^1 10\Big[\tfrac{y^2}{2}\Big]_{x^2}^{\sqrt x}dx = \int_0^1 5\,(x - x^4)\,dx = 5\Big(\tfrac12 - \tfrac15\Big) = 5\cdot\tfrac{3}{10} = \frac{3}{2}.$$ $$\boxed{\oint_C P\,dx+Q\,dy = \tfrac32}$$

(b) Verify Stokes' theorem for $\vec F = (2x-y)\hat i - yz^2\hat j - y^2z\,\hat k$

Stokes: $\displaystyle\iint_S (\nabla\times\vec F)\cdot\hat n\,dS = \oint_C \vec F\cdot d\vec r$, with $S$ the upper hemisphere and $C$ the unit circle $x^2+y^2=1$, $z=0$.

Curl.

$$\nabla\times\vec F = \begin{vmatrix}\hat i & \hat j & \hat k\\ \partial_x & \partial_y & \partial_z\\ 2x-y & -yz^2 & -y^2z\end{vmatrix} = (-2yz + 2yz)\hat i - (0-0)\hat j + (0+1)\hat k = \hat k.$$

Surface integral. $\nabla\times\vec F=\hat k$, so $\iint_S \hat k\cdot\hat n\,dS$ equals the area of the projection of $S$ on the $xy$-plane, i.e. the unit disk, $=\pi$.

Line integral. On $C$: $x=\cos t,\,y=\sin t,\,z=0$, so $\vec F=(2\cos t-\sin t)\hat i$, $d\vec r=(-\sin t,\cos t,0)\,dt$.

$$\oint_C\vec F\cdot d\vec r = \int_0^{2\pi}(2\cos t-\sin t)(-\sin t)\,dt = \int_0^{2\pi}(-2\sin t\cos t + \sin^2 t)\,dt = 0 + \pi = \pi.$$

Both sides equal $\pi$, so Stokes' theorem is verified.

(a) Laplace transform and $\mathcal L\{t e^{-2t}\sin 3t\}$

Definition. $\displaystyle \mathcal L\{f(t)\} = F(s) = \int_0^\infty e^{-st} f(t)\,dt$, for $s$ large enough for convergence.

Start from $\mathcal L\{\sin 3t\} = \dfrac{3}{s^2+9}$. By the first shifting theorem $\mathcal L\{e^{-2t}\sin 3t\} = \dfrac{3}{(s+2)^2+9}$. By the multiplication-by-$t$ property $\mathcal L\{t g(t)\} = -G'(s)$:

$$\mathcal L\{t e^{-2t}\sin 3t\} = -\frac{d}{ds}\frac{3}{(s+2)^2+9} = \frac{3\cdot 2(s+2)}{\big[(s+2)^2+9\big]^2} = \frac{6(s+2)}{\big[(s+2)^2+9\big]^2}.$$

(b) IVP $y'' + 4y' + 3y = e^{-t}$, $y(0)=0$, $y'(0)=1$

Let $Y=\mathcal L\{y\}$. Taking transforms:

$$(s^2Y - s\cdot0 - 1) + 4(sY - 0) + 3Y = \frac{1}{s+1}.$$ $$(s^2+4s+3)Y = 1 + \frac{1}{s+1} = \frac{s+2}{s+1}, \qquad s^2+4s+3=(s+1)(s+3).$$ $$Y = \frac{s+2}{(s+1)^2(s+3)}.$$

Partial fractions: $Y = \dfrac{A}{s+1} + \dfrac{B}{(s+1)^2} + \dfrac{C}{s+3}$. Solving: $B=\tfrac12$ (mult. $(s+1)^2$, $s=-1$: $\tfrac{1}{2}$), $C = \dfrac{-3+2}{(-2)^2} = -\tfrac14$ ($s=-3$), and $A=\tfrac14$. Thus

$$y(t) = \tfrac14 e^{-t} + \tfrac12 t e^{-t} - \tfrac14 e^{-3t}.$$

(c) Convolution theorem and $\mathcal L^{-1}\{1/(s^2+a^2)^2\}$

Convolution theorem. $\mathcal L^{-1}\{F(s)G(s)\} = (f*g)(t) = \int_0^t f(\tau)g(t-\tau)\,d\tau$.

Write $\dfrac{1}{(s^2+a^2)^2} = \dfrac{1}{a^2}\cdot\dfrac{a}{s^2+a^2}\cdot\dfrac{a}{s^2+a^2}$, with $\mathcal L^{-1}\{a/(s^2+a^2)\}=\sin at$.

$$\mathcal L^{-1}\Big\{\tfrac{1}{(s^2+a^2)^2}\Big\} = \frac{1}{a^2}\int_0^t \sin a\tau\,\sin a(t-\tau)\,d\tau.$$

Using $\sin A\sin B = \tfrac12[\cos(A-B)-\cos(A+B)]$:

$$= \frac{1}{2a^2}\int_0^t\big[\cos a(2\tau-t) - \cos at\big]d\tau = \frac{1}{2a^2}\Big(\frac{\sin at}{a} - t\cos at\Big).$$ $$\boxed{\mathcal L^{-1}\Big\{\tfrac{1}{(s^2+a^2)^2}\Big\} = \frac{1}{2a^3}\big(\sin at - at\cos at\big)}$$

Solve for $x,y,z$ in terms of $u,v,w$:

$$z = uvw, \quad y = uv - z = uv - uvw = uv(1-w), \quad x = u - (y+z) = u - uv = u(1-v).$$

Jacobian:

$$\frac{\partial(x,y,z)}{\partial(u,v,w)} = \begin{vmatrix} x_u & x_v & x_w\\ y_u & y_v & y_w\\ z_u & z_v & z_w\end{vmatrix} = \begin{vmatrix} 1-v & -u & 0\\ v(1-w) & u(1-w) & -uv\\ vw & uw & uv\end{vmatrix}.$$

Expanding (a standard reduction gives a clean form):

$$J = u^2 v.$$

At $(u,v,w)=(1,1,1)$: $J = (1)^2(1) = 1 \ne 0$, so the transformation is invertible (locally) at that point by the inverse function theorem.

Gradient of $\phi = x^2yz + 4xz^2$

$$\nabla\phi = (2xyz + 4z^2)\hat i + x^2 z\,\hat j + (x^2 y + 8xz)\hat k.$$

At $(1,-2,-1)$: $\,2(1)(-2)(-1)+4(1)=8;\ (1)(-1)=-1;\ (1)(-2)+8(1)(-1)=-10.$

$$\nabla\phi = 8\hat i - \hat j - 10\hat k.$$

Divergence and curl of $\vec F = xz^3\hat i - 2x^2yz\,\hat j + 2yz^4\hat k$

$$\nabla\cdot\vec F = \frac{\partial}{\partial x}(xz^3) + \frac{\partial}{\partial y}(-2x^2yz) + \frac{\partial}{\partial z}(2yz^4) = z^3 - 2x^2 z + 8yz^3.$$

At $(1,-2,-1)$: $(-1)^3 - 2(1)(-1) + 8(-2)(-1)^3 = -1 + 2 + 16 = 17.$

$$\nabla\cdot\vec F = 17.$$ $$\nabla\times\vec F = \begin{vmatrix}\hat i & \hat j & \hat k\\ \partial_x & \partial_y & \partial_z\\ xz^3 & -2x^2yz & 2yz^4\end{vmatrix} = (2z^4 + 2x^2 y)\hat i - (0 - 3xz^2)\hat j + (-4xyz - 0)\hat k.$$

At $(1,-2,-1)$: $\,\hat i: 2(1)+2(1)(-2)=2-4=-2;\ \hat j: 3(1)(1)=3;\ \hat k: -4(1)(-2)(-1)=-8.$

$$\nabla\times\vec F = -2\hat i + 3\hat j - 8\hat k.$$

Given $(x^2+y^2+x)\,dx + xy\,dy = 0$, with $M = x^2+y^2+x$, $N = xy$.

$$M_y = 2y,\quad N_x = y \;\Rightarrow\; \frac{M_y - N_x}{N} = \frac{2y-y}{xy} = \frac{1}{x},$$

which is a function of $x$ alone, so the integrating factor is

$$\mu = e^{\int \frac{1}{x}\,dx} = e^{\ln x} = x.$$

Multiply through by $x$:

$$(x^3 + xy^2 + x^2)\,dx + x^2 y\,dy = 0.$$

Now $M_y = 2xy = N_x$, so it is exact. Find $F$ with $F_x = x^3 + xy^2 + x^2$:

$$F = \frac{x^4}{4} + \frac{x^2 y^2}{2} + \frac{x^3}{3} + g(y).$$

$F_y = x^2 y + g'(y) = x^2 y \Rightarrow g'(y)=0$. Hence the general solution is

$$\boxed{\frac{x^4}{4} + \frac{x^2 y^2}{2} + \frac{x^3}{3} = C.}$$

Equation: $y'' - 4y' + 4y = e^{2x} + \sin 2x$. Auxiliary equation $m^2 - 4m + 4 = (m-2)^2 = 0 \Rightarrow m = 2,2.$

Complementary function: $y_c = (C_1 + C_2 x)e^{2x}.$

Particular integral. Operator $D^2 - 4D + 4 = (D-2)^2$.

For $e^{2x}$: since $2$ is a repeated root, multiply by $x^2$:

$$y_{p1} = \frac{1}{(D-2)^2}e^{2x} = \frac{x^2}{2}e^{2x}.$$

For $\sin 2x$: replace $D^2\to -4$ in $D^2-4D+4$: $-4 - 4D + 4 = -4D$.

$$y_{p2} = \frac{1}{-4D}\sin 2x = -\frac{1}{4}\int\sin 2x\,dx = -\frac{1}{4}\Big(-\frac{\cos 2x}{2}\Big) = \frac{\cos 2x}{8}.$$

General solution:

$$\boxed{y = (C_1 + C_2 x)e^{2x} + \frac{x^2}{2}e^{2x} + \frac{\cos 2x}{8}.}$$

$$\mathcal L^{-1}\Big\{\frac{s+5}{s^2 - 6s + 13}\Big\}.$$

Complete the square: $s^2 - 6s + 13 = (s-3)^2 + 4 = (s-3)^2 + 2^2.$ Write the numerator in terms of $(s-3)$: $s + 5 = (s-3) + 8.$

$$\frac{s+5}{(s-3)^2+4} = \frac{s-3}{(s-3)^2+2^2} + 8\cdot\frac{1}{(s-3)^2+2^2} = \frac{s-3}{(s-3)^2+2^2} + 4\cdot\frac{2}{(s-3)^2+2^2}.$$

Using $\mathcal L^{-1}\{\tfrac{s-a}{(s-a)^2+b^2}\}=e^{at}\cos bt$ and $\mathcal L^{-1}\{\tfrac{b}{(s-a)^2+b^2}\}=e^{at}\sin bt$ with $a=3,\,b=2$:

$$\boxed{\mathcal L^{-1}\Big\{\frac{s+5}{s^2-6s+13}\Big\} = e^{3t}\big(\cos 2t + 4\sin 2t\big).}$$

Fourier series of $f(x)=x^2$ on $(-\pi,\pi)$

$f$ is even, so all $b_n=0$.

$$a_0 = \frac{1}{\pi}\int_{-\pi}^{\pi} x^2\,dx = \frac{2}{\pi}\int_0^\pi x^2\,dx = \frac{2}{\pi}\cdot\frac{\pi^3}{3} = \frac{2\pi^2}{3}.$$ $$a_n = \frac{2}{\pi}\int_0^\pi x^2\cos nx\,dx.$$

Integrating by parts twice: $\int_0^\pi x^2\cos nx\,dx = \dfrac{2\pi}{n^2}\cos n\pi = \dfrac{2\pi(-1)^n}{n^2}.$ Hence

$$a_n = \frac{2}{\pi}\cdot\frac{2\pi(-1)^n}{n^2} = \frac{4(-1)^n}{n^2}.$$

With $a_0/2 = \pi^2/3$:

$$x^2 = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\cos nx.$$

Deduction

Put $x=\pi$ (where the series converges to $\pi^2$): $\cos n\pi=(-1)^n$, so

$$\pi^2 = \frac{\pi^2}{3} + 4\sum_{n=1}^\infty\frac{(-1)^n(-1)^n}{n^2} = \frac{\pi^2}{3} + 4\sum_{n=1}^\infty\frac{1}{n^2}.$$ $$\Rightarrow 4\sum_{n=1}^\infty\frac{1}{n^2} = \pi^2 - \frac{\pi^2}{3} = \frac{2\pi^2}{3} \;\Rightarrow\; \sum_{n=1}^\infty\frac{1}{n^2} = \frac{\pi^2}{6}. \qquad\blacksquare$$

Half-range sine series: $f(x)=\sum_{n=1}^\infty b_n\sin nx$, with

$$b_n = \frac{2}{\pi}\int_0^\pi (\pi - x)\sin nx\,dx.$$

Compute: $\int_0^\pi \pi\sin nx\,dx = \pi\cdot\dfrac{1-\cos n\pi}{n} = \dfrac{\pi[1-(-1)^n]}{n}.$

$\int_0^\pi x\sin nx\,dx = \Big[-\dfrac{x\cos nx}{n}\Big]_0^\pi + \dfrac1n\int_0^\pi\cos nx\,dx = -\dfrac{\pi(-1)^n}{n} + 0 = -\dfrac{\pi(-1)^n}{n}.$

So

$$b_n = \frac{2}{\pi}\left(\frac{\pi[1-(-1)^n]}{n} + \frac{\pi(-1)^n}{n}\right) = \frac{2}{\pi}\cdot\frac{\pi}{n} = \frac{2}{n}.$$

Therefore

$$\boxed{\;\pi - x = \sum_{n=1}^\infty \frac{2}{n}\sin nx = 2\left(\sin x + \frac{\sin 2x}{2} + \frac{\sin 3x}{3} + \cdots\right),\quad 0<x<\pi.\;}$$

Series solution of $y'' + xy = 0$ about $x=0$

Assume $y = \sum_{n=0}^\infty a_n x^n$. Then $y'' = \sum_{n=2}^\infty n(n-1)a_n x^{n-2} = \sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n$, and $xy = \sum_{n=0}^\infty a_n x^{n+1} = \sum_{n=1}^\infty a_{n-1}x^n$.

Substituting and collecting $x^n$:

• $n=0$: $2\cdot1\,a_2 = 0 \Rightarrow a_2 = 0.$
• $n\ge1$: $(n+2)(n+1)a_{n+2} + a_{n-1} = 0.$

Recurrence relation:

$$\boxed{\,a_{n+2} = -\frac{a_{n-1}}{(n+2)(n+1)},\quad n\ge1;\quad a_2=0.\,}$$

The indices split mod 3. With $a_0,a_1$ arbitrary and $a_2=0$ (so $a_2,a_5,a_8,\dots=0$):

From $a_0$: $a_3 = -\dfrac{a_0}{3\cdot2}=-\dfrac{a_0}{6}$, $a_6 = -\dfrac{a_3}{6\cdot5}=\dfrac{a_0}{180}$, ...

From $a_1$: $a_4 = -\dfrac{a_1}{4\cdot3}=-\dfrac{a_1}{12}$, $a_7 = -\dfrac{a_4}{7\cdot6}=\dfrac{a_1}{504}$, ...

Two linearly independent solutions:

$$y_1(x) = 1 - \frac{x^3}{6} + \frac{x^6}{180} - \cdots,$$ $$y_2(x) = x - \frac{x^4}{12} + \frac{x^7}{504} - \cdots.$$

General solution $y = a_0\,y_1(x) + a_1\,y_2(x).$