BE Computer Engineering (Pokhara University) Calculus II (PU, MTH 210) Question Paper 2079 Nepal

(a) Polar transformation identity

With $x = r\cos\theta$, $y = r\sin\theta$ we have $r^2 = x^2+y^2$, $\tan\theta = y/x$.

By the chain rule,

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

Squaring and combining:

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

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

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

(b) Extrema of $f(x,y)=x^3+y^3-3axy$, $a>0$

Stationary points:

$$f_x = 3x^2 - 3ay = 0 \Rightarrow y = x^2/a, \qquad f_y = 3y^2 - 3ax = 0 \Rightarrow x = y^2/a.$$

Substituting: $x = x^4/a^3 \Rightarrow x(x^3 - a^3)=0 \Rightarrow x=0$ or $x=a$. This gives the critical points $(0,0)$ 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 $(0,0)$: $D = 0 - 9a^2 = -9a^2 < 0 \Rightarrow$ saddle point.
• At $(a,a)$: $D = 36a^2 - 9a^2 = 27a^2 > 0$ and $f_{xx}=6a>0 \Rightarrow$ relative minimum, with $f(a,a)=a^3+a^3-3a\cdot a\cdot a = -a^3$.

Conclusion: $f$ has a local minimum value $-a^3$ at $(a,a)$ and a saddle point at $(0,0)$; there is no local maximum.

(a) $\displaystyle \int_0^1 \int_{x^2}^{\sqrt{x}} xy\,dy\,dx$

Region: between $y=x^2$ (lower) and $y=\sqrt{x}$ (upper) for $0\le x\le 1$; the two curves meet at $(0,0)$ and $(1,1)$, enclosing a leaf-shaped region in the first quadrant.

Inner integral:

$$\int_{x^2}^{\sqrt{x}} xy\,dy = x\left[\frac{y^2}{2}\right]_{x^2}^{\sqrt{x}} = \frac{x}{2}\left(x - x^4\right) = \frac{x^2 - x^5}{2}.$$

Then

$$\int_0^1 \frac{x^2 - x^5}{2}\,dx = \frac12\left[\frac{x^3}{3} - \frac{x^6}{6}\right]_0^1 = \frac12\left(\frac13 - \frac16\right) = \frac12\cdot\frac16 = \boxed{\frac{1}{12}}.$$

(b) Change of order: $\displaystyle \int_0^a \int_y^a \frac{x}{x^2+y^2}\,dx\,dy$

Region: $0\le y\le a$, $y\le x\le a$, i.e. the triangle $0\le x\le a$, $0\le y\le x$. Reversing:

$$\int_0^a \int_0^x \frac{x}{x^2+y^2}\,dy\,dx = \int_0^a x\left[\frac1x\tan^{-1}\frac{y}{x}\right]_0^x dx = \int_0^a \tan^{-1}(1)\,dx = \frac{\pi}{4}\int_0^a dx = \boxed{\frac{\pi a}{4}}.$$

(c) Volume by triple integration

The solid lies over the disk $x^2+y^2\le 4$ between $z=0$ and $z=3-x$. Since on the disk $|x|\le 2 < 3$, $3-x>0$ throughout, so

$$V = \iint_{x^2+y^2\le 4}\int_0^{3-x} dz\,dA = \iint_{x^2+y^2\le 4}(3-x)\,dA.$$

The term $\iint x\,dA = 0$ by symmetry, and $\iint 3\,dA = 3\times(\text{area of disk}) = 3\cdot\pi(2)^2 = 12\pi$.

$$\boxed{V = 12\pi \text{ cubic units}}.$$

(a) $y'' - 4y' + 4y = e^{2x} + \sin 2x$

Complementary function. Auxiliary equation $m^2 - 4m + 4 = (m-2)^2 = 0 \Rightarrow m=2,2$.

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

Particular integral. $y_p = y_{p1} + y_{p2}$ with $P.I. = \dfrac{1}{(D-2)^2}(e^{2x}+\sin 2x)$.

For $e^{2x}$: root $2$ is a double root, so multiply by $x^2$:

$$y_{p1} = \frac{x^2}{2!}e^{2x} = \frac{x^2 e^{2x}}{2}.$$

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

$$y_{p2} = \frac{1}{-4D}\sin 2x = -\frac{1}{4}\int \sin 2x\,dx = -\frac14\left(-\frac{\cos 2x}{2}\right) = \frac{\cos 2x}{8}.$$

General solution:

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

(b) $(D^2+1)y = \sec x$ by variation of parameters

C.F.: $m^2+1=0\Rightarrow y_c = C_1\cos x + C_2\sin x$, so $y_1=\cos x,\ y_2=\sin x$, Wronskian $W = \cos x\cos x - \sin x(-\sin x) = 1$.

$$u = -\int \frac{y_2\,R}{W}dx = -\int \sin x\sec x\,dx = -\int \tan x\,dx = \ln|\cos x|,$$ $$v = \int \frac{y_1\,R}{W}dx = \int \cos x\sec x\,dx = \int dx = x.$$

Particular integral: $y_p = u y_1 + v y_2 = \cos x\ln|\cos x| + x\sin x.$

General solution:

$$\boxed{y = C_1\cos x + C_2\sin x + \cos x\,\ln|\cos x| + x\sin x.}$$

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

Convolution theorem. If $L^{-1}\{F(s)\}=f(t)$ and $L^{-1}\{G(s)\}=g(t)$, then

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

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

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

Using $\sin A\sin B = \tfrac12[\cos(A-B)-\cos(A+B)]$ with $A=a\tau,\ B=a(t-\tau)$:

$$=\frac{1}{2a^2}\int_0^t\big[\cos a(2\tau - t) - \cos at\big]d\tau = \frac{1}{2a^2}\left[\frac{\sin a(2\tau-t)}{2a} - \tau\cos at\right]_0^t.$$

At $\tau=t$: $\frac{\sin at}{2a} - t\cos at$; at $\tau=0$: $\frac{\sin(-at)}{2a} = -\frac{\sin at}{2a}$. Difference $= \frac{\sin at}{a} - t\cos at$. Hence

$$\boxed{L^{-1}\left\{\frac{1}{(s^2+a^2)^2}\right\} = \frac{1}{2a^3}\big(\sin at - at\cos at\big).}$$

(b) $y''+4y=\sin t$, $y(0)=0,\ y'(0)=0$

Taking Laplace transforms with $L\{y\}=Y$:

$$(s^2 Y - 0 - 0) + 4Y = \frac{1}{s^2+1} \Rightarrow Y = \frac{1}{(s^2+1)(s^2+4)}.$$

Partial fractions: $\dfrac{1}{(s^2+1)(s^2+4)} = \dfrac{1}{3}\left(\dfrac{1}{s^2+1} - \dfrac{1}{s^2+4}\right).$

Inverting,

$$y(t) = \frac13\left(\sin t - \frac{\sin 2t}{2}\right) = \boxed{\frac{1}{3}\sin t - \frac{1}{6}\sin 2t.}$$

Given $x+y+z=u$, $y+z=uv$, $z=uvw$. Solve explicitly for $x,y,z$:

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

The Jacobian is

$$\frac{\partial(x,y,z)}{\partial(u,v,w)} = \begin{vmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} & \dfrac{\partial x}{\partial w} \\[4pt] \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} & \dfrac{\partial y}{\partial w} \\[4pt] \dfrac{\partial z}{\partial u} & \dfrac{\partial z}{\partial v} & \dfrac{\partial z}{\partial w} \end{vmatrix} = \begin{vmatrix} 1-v & -u & 0 \\ v(1-w) & u(1-w) & -uv \\ vw & uw & uv \end{vmatrix}.$$

Expanding along the first row:

$$= (1-v)\big[u(1-w)\cdot uv - (-uv)(uw)\big] - (-u)\big[v(1-w)\cdot uv - (-uv)(vw)\big]$$ $$= (1-v)\big[u^2v(1-w) + u^2vw\big] + u\big[uv^2(1-w) + uv^2 w\big]$$ $$= (1-v)(u^2 v) + u(uv^2) = u^2v(1-v) + u^2v^2 = u^2 v.$$ $$\boxed{\dfrac{\partial(x,y,z)}{\partial(u,v,w)} = u^2 v.}$$

(a) Directional derivative

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

At $(1,-2,-1)$:

$$2xyz+4z^2 = 2(1)(-2)(-1)+4(1) = 4+4 = 8,$$ $$x^2 z = (1)(-1) = -1,\qquad x^2 y + 8xz = (1)(-2)+8(1)(-1) = -2-8 = -10.$$

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

Unit vector along $2\hat i - \hat j - 2\hat k$: magnitude $\sqrt{4+1+4}=3$, so $\hat a = \tfrac13(2\hat i - \hat j - 2\hat k)$.

$$D_{\hat a}\phi = \nabla\phi\cdot\hat a = \frac{1}{3}\big(8\cdot2 + (-1)(-1) + (-10)(-2)\big) = \frac{16+1+20}{3} = \boxed{\frac{37}{3}}.$$

(b) Irrotational field

$\vec F = (y^2\cos x + z^3)\hat i + (2y\sin x - 4)\hat j + (3xz^2 + 2)\hat k$. Compute $\nabla\times\vec F$:

$$(\nabla\times\vec F)_x = \frac{\partial}{\partial y}(3xz^2+2) - \frac{\partial}{\partial z}(2y\sin x-4) = 0 - 0 = 0,$$ $$(\nabla\times\vec F)_y = \frac{\partial}{\partial z}(y^2\cos x+z^3) - \frac{\partial}{\partial x}(3xz^2+2) = 3z^2 - 3z^2 = 0,$$ $$(\nabla\times\vec F)_z = \frac{\partial}{\partial x}(2y\sin x-4) - \frac{\partial}{\partial y}(y^2\cos x+z^3) = 2y\cos x - 2y\cos x = 0.$$

Since $\nabla\times\vec F = \vec 0$, the field is irrotational. $\blacksquare$

Green's theorem: $\displaystyle\oint_C (M\,dx + N\,dy) = \iint_R\left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right)dA$, with $M = 3x^2 - 8y^2$, $N = 4y - 6xy$.

Double integral (RHS)

$$\frac{\partial N}{\partial x} = -6y, \qquad \frac{\partial M}{\partial y} = -16y \Rightarrow \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = -6y + 16y = 10y.$$

Region $R$: triangle bounded by $x=0,\ y=0,\ x+y=1$, so $0\le x\le 1$, $0\le y\le 1-x$.

$$\iint_R 10y\,dA = \int_0^1\int_0^{1-x} 10y\,dy\,dx = \int_0^1 5(1-x)^2 dx = 5\left[-\frac{(1-x)^3}{3}\right]_0^1 = \frac{5}{3}.$$

Line integral (LHS)

Traverse $C$ counter-clockwise: $C_1$ ($y=0$, $x:0\to1$), $C_2$ ($x+y=1$, $x:1\to0$), $C_3$ ($x=0$, $y:1\to0$).

$C_1$ ($y=0,\ dy=0$): $\int_0^1 3x^2\,dx = 1.$

$C_2$ ($y=1-x,\ dy=-dx,\ x:1\to0$): $M\,dx + N\,dy = (3x^2 - 8(1-x)^2)dx + (4(1-x) - 6x(1-x))(-dx)$. $= [3x^2 - 8(1-x)^2 - 4(1-x) + 6x(1-x)]dx$. Expanding: $3x^2 - 8(1-2x+x^2) - 4 + 4x + 6x - 6x^2 = 3x^2 - 8 + 16x - 8x^2 - 4 + 10x - 6x^2 = -11x^2 + 26x - 12.$ $\int_1^0(-11x^2+26x-12)dx = -\int_0^1(-11x^2+26x-12)dx = -\left(-\tfrac{11}{3} + 13 - 12\right) = -\left(\tfrac{-11+3}{3}\right) = \tfrac{8}{3}.$

$C_3$ ($x=0,\ dx=0,\ y:1\to0$): $N = 4y$, $\int_1^0 4y\,dy = -2.$

Sum: $1 + \tfrac{8}{3} - 2 = \tfrac{8}{3} - 1 = \tfrac{5}{3}.$

Conclusion

Line integral $= \dfrac{5}{3} =$ double integral. Green's theorem is verified. $\blacksquare$

By Stokes' theorem, $\displaystyle\oint_C \vec F\cdot d\vec r = \iint_S (\nabla\times\vec F)\cdot\hat n\,dS$. The boundary $C$ of the upper hemisphere is the unit circle $x^2+y^2=1$, $z=0$, so it suffices to use the flat disk $S: x^2+y^2\le 1,\ z=0$ with $\hat n = \hat k$.

Curl: $\vec F = (2x-y)\hat i - yz^2\hat j - y^2 z\,\hat k$.

$$\nabla\times\vec F = \left(\frac{\partial(-y^2 z)}{\partial y} - \frac{\partial(-yz^2)}{\partial z}\right)\hat i + \left(\frac{\partial(2x-y)}{\partial z} - \frac{\partial(-y^2 z)}{\partial x}\right)\hat j + \left(\frac{\partial(-yz^2)}{\partial x} - \frac{\partial(2x-y)}{\partial y}\right)\hat k.$$ $$= (-2yz + 2yz)\hat i + (0 - 0)\hat j + (0 - (-1))\hat k = \hat k.$$

Flux through the disk:

$$\iint_S (\nabla\times\vec F)\cdot\hat k\,dS = \iint_{x^2+y^2\le1} 1\,dA = \text{area of unit disk} = \pi.$$ $$\boxed{\oint_C \vec F\cdot d\vec r = \pi.}$$

(a) $\dfrac{dy}{dx} + y\cot x = \cos x$

Linear equation, $P=\cot x$, integrating factor $\mu = e^{\int\cot x\,dx} = e^{\ln\sin x} = \sin x$.

$$\frac{d}{dx}(y\sin x) = \cos x\sin x = \tfrac12\sin 2x.$$

Integrating: $y\sin x = -\tfrac14\cos 2x + C$, so

$$\boxed{y = \frac{C - \tfrac14\cos 2x}{\sin x}}\quad\left(\text{equivalently } y\sin x = \frac{\sin^2 x}{2} + C'\right).$$

(b) Exact equation

$M = 2xy + y - \tan y$, $N = x^2 - x\tan^2 y + \sec^2 y$. Check exactness: $M_y = 2x + 1 - \sec^2 y$ and $N_x = 2x - \tan^2 y$. Since $\sec^2 y = 1 + \tan^2 y$, $M_y = 2x + 1 - (1+\tan^2 y) = 2x - \tan^2 y = N_x$. Exact.

Integrate $M$ w.r.t. $x$ (treating $y$ constant):

$$F = \int(2xy + y - \tan y)\,dx = x^2 y + xy - x\tan y + g(y).$$

Differentiate w.r.t. $y$: $F_y = x^2 + x - x\sec^2 y + g'(y)$. Set equal to $N = x^2 - x\tan^2 y + \sec^2 y = x^2 - x(\sec^2 y - 1) + \sec^2 y = x^2 + x - x\sec^2 y + \sec^2 y$. Hence $g'(y) = \sec^2 y \Rightarrow g(y) = \tan y$.

Solution:

$$\boxed{x^2 y + xy - x\tan y + \tan y = C.}$$

Since $f(x)=x^2$ is even on $(-\pi,\pi)$, all $b_n = 0$. The Fourier series is $f(x) = \dfrac{a_0}{2} + \sum_{n=1}^\infty a_n\cos nx$.

Coefficient $a_0$:

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

Coefficient $a_n$:

$$a_n = \frac{2}{\pi}\int_0^\pi x^2\cos nx\,dx.$$

Integrating by parts twice, $\int_0^\pi x^2\cos nx\,dx = \dfrac{2\pi\cos n\pi}{n^2} = \dfrac{2\pi(-1)^n}{n^2}$ (the other boundary terms vanish). Thus

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

Fourier series:

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

Deduction. Put $x=\pi$ (a point where $f$ is continuous, $f(\pi)=\pi^2$); $\cos n\pi = (-1)^n$ so $(-1)^n\cos n\pi = 1$:

$$\pi^2 = \frac{\pi^2}{3} + 4\sum_{n=1}^\infty \frac{1}{n^2} \Rightarrow 4\sum_{n=1}^\infty\frac{1}{n^2} = \frac{2\pi^2}{3} \Rightarrow \sum_{n=1}^\infty\frac{1}{n^2} = \frac{\pi^2}{6}. \quad\blacksquare$$

$x=0$ is an ordinary point. Assume $y = \sum_{n=0}^\infty a_n x^n$, so $y' = \sum_{n=1}^\infty n a_n x^{n-1}$, $y'' = \sum_{n=2}^\infty n(n-1)a_n x^{n-2}$.

Substitute into $y'' + xy' + y = 0$:

$$\sum_{n=2}^\infty n(n-1)a_n x^{n-2} + \sum_{n=1}^\infty n a_n x^n + \sum_{n=0}^\infty a_n x^n = 0.$$

Shift the first sum ($n\to n+2$); note $\sum n a_n x^n + \sum a_n x^n = \sum_{n=0}^\infty (n+1)a_n x^n$:

$$\sum_{n=0}^\infty\Big[(n+2)(n+1)a_{n+2} + (n+1)a_n\Big]x^n = 0.$$

Recurrence relation:

$$a_{n+2} = -\frac{(n+1)a_n}{(n+2)(n+1)} = -\frac{a_n}{n+2}, \qquad n\ge 0.$$

Even terms (from $a_0$): $a_2 = -\dfrac{a_0}{2}$, $a_4 = -\dfrac{a_2}{4} = \dfrac{a_0}{2\cdot4}$, $a_6 = -\dfrac{a_4}{6} = -\dfrac{a_0}{2\cdot4\cdot6}$, giving

$$y_1 = a_0\left(1 - \frac{x^2}{2} + \frac{x^4}{2\cdot4} - \frac{x^6}{2\cdot4\cdot6} + \cdots\right) = a_0\sum_{k=0}^\infty \frac{(-1)^k x^{2k}}{2^k k!} = a_0\,e^{-x^2/2}.$$

Odd terms (from $a_1$): $a_3 = -\dfrac{a_1}{3}$, $a_5 = -\dfrac{a_3}{5} = \dfrac{a_1}{3\cdot5}$, $a_7 = -\dfrac{a_5}{7} = -\dfrac{a_1}{3\cdot5\cdot7}$, giving

$$y_2 = a_1\left(x - \frac{x^3}{3} + \frac{x^5}{3\cdot5} - \frac{x^7}{3\cdot5\cdot7} + \cdots\right).$$

General solution:

$$\boxed{y = a_0\left(1 - \frac{x^2}{2} + \frac{x^4}{8} - \cdots\right) + a_1\left(x - \frac{x^3}{3} + \frac{x^5}{15} - \cdots\right),}$$

where $a_0, a_1$ are arbitrary constants.

(a) $L\{t\,e^{-2t}\sin 3t\}$

Start from $L\{\sin 3t\} = \dfrac{3}{s^2+9}$. By the first shifting theorem ($s\to s+2$):

$$L\{e^{-2t}\sin 3t\} = \frac{3}{(s+2)^2 + 9}.$$

Multiplication by $t$ gives $L\{t f(t)\} = -\dfrac{d}{ds}F(s)$:

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

(b) Unit-step representation and transform

Write $f(t)$ using $u(t-a)$. Building up from pieces:

$$f(t) = t\,[u(t) - u(t-2)] + (4-t)[u(t-2) - u(t-4)].$$

Collecting jumps in slope/value:

$$f(t) = t - 2(t-2)\,u(t-2) + (t-4)\,u(t-4).$$

(Check: for $0<t<2$, $f=t$; for $2<t<4$, $f = t - 2(t-2) = 4-t$; for $t>4$, $f = t - 2(t-2) + (t-4) = 0$.)

Using $L\{(t-a)u(t-a)\} = \dfrac{e^{-as}}{s^2}$ and $L\{t\} = \dfrac{1}{s^2}$:

$$L\{f(t)\} = \frac{1}{s^2} - \frac{2e^{-2s}}{s^2} + \frac{e^{-4s}}{s^2} = \boxed{\frac{1 - 2e^{-2s} + e^{-4s}}{s^2} = \frac{(1 - e^{-2s})^2}{s^2}.}$$