BE Computer Engineering (IOE, TU) Engineering Mathematics II (IOE, SH 451) Question Paper 2078 Nepal

(a) Change of order of integration

Given $I=\displaystyle\int_{0}^{a}\int_{y}^{a}\frac{x}{x^{2}+y^{2}}\,dx\,dy$.

Region. For each $y\in[0,a]$, $x$ runs from $x=y$ to $x=a$. This is the triangle $0\le y\le x\le a$.

Reverse order. For each $x\in[0,a]$, $y$ runs from $0$ to $x$:

$$I=\int_{0}^{a}\int_{0}^{x}\frac{x}{x^{2}+y^{2}}\,dy\,dx.$$

Inner integral (treat $x$ as constant):

$$\int_{0}^{x}\frac{x}{x^{2}+y^{2}}\,dy=x\cdot\frac{1}{x}\left[\tan^{-1}\frac{y}{x}\right]_{0}^{x}=\tan^{-1}(1)-\tan^{-1}(0)=\frac{\pi}{4}.$$

Outer integral:

$$I=\int_{0}^{a}\frac{\pi}{4}\,dx=\frac{\pi}{4}\,a=\boxed{\dfrac{\pi a}{4}}.$$

(b) Volume in polar coordinates

The solid is bounded below by $z=0$, above by the plane $z=4-y$, over the disc $x^{2}+y^{2}\le 4$. Height $=z_{\text{top}}-z_{\text{bottom}}=4-y$.

$$V=\iint_{x^{2}+y^{2}\le4}(4-y)\,dA.$$

Put $x=r\cos\theta,\;y=r\sin\theta,\;dA=r\,dr\,d\theta$, with $0\le r\le 2,\;0\le\theta\le2\pi$:

$$V=\int_{0}^{2\pi}\int_{0}^{2}(4-r\sin\theta)\,r\,dr\,d\theta=\int_{0}^{2\pi}\int_{0}^{2}(4r-r^{2}\sin\theta)\,dr\,d\theta.$$

Inner: $\big[2r^{2}-\tfrac{r^{3}}{3}\sin\theta\big]_{0}^{2}=8-\tfrac{8}{3}\sin\theta.$

$$V=\int_{0}^{2\pi}\Big(8-\tfrac{8}{3}\sin\theta\Big)d\theta=8(2\pi)-\tfrac{8}{3}\big[-\cos\theta\big]_{0}^{2\pi}=16\pi-0=\boxed{16\pi}.$$

(a) $y''-3y'+2y=x\,e^{3x}$

Complementary function. Auxiliary equation $m^{2}-3m+2=0\Rightarrow(m-1)(m-2)=0\Rightarrow m=1,2$.

$$y_c=C_1e^{x}+C_2e^{2x}.$$

Particular integral. Since $3$ is not a root, try $y_p=(ax+b)e^{3x}$.

$$y_p'=(3ax+3b+a)e^{3x},\quad y_p''=(9ax+9b+6a)e^{3x}.$$

Substitute into $y''-3y'+2y$:

$$e^{3x}\big[(9ax+9b+6a)-3(3ax+3b+a)+2(ax+b)\big]=e^{3x}\big[2ax+(2b+3a)\big]=x\,e^{3x}.$$

Match: $2a=1\Rightarrow a=\tfrac12$; $2b+3a=0\Rightarrow b=-\tfrac34$.

$$y_p=\left(\tfrac{1}{2}x-\tfrac{3}{4}\right)e^{3x}.$$

General solution.

$$\boxed{\,y=C_1e^{x}+C_2e^{2x}+\left(\tfrac{1}{2}x-\tfrac{3}{4}\right)e^{3x}.}$$

(b) Variation of parameters for $y''+y=\sec x$

CF: $m^2+1=0\Rightarrow m=\pm i$, so $y_1=\cos x,\;y_2=\sin x$. Wronskian

$$W=\begin{vmatrix}\cos x & \sin x\\ -\sin x & \cos x\end{vmatrix}=1.$$

With $R(x)=\sec x$:

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

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

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

(a) Definitions and $\nabla\times(\nabla\phi)=\vec0$

Gradient of a scalar field $\phi$: $\nabla\phi=\dfrac{\partial\phi}{\partial x}\hat i+\dfrac{\partial\phi}{\partial y}\hat j+\dfrac{\partial\phi}{\partial z}\hat k$ — a vector pointing in the direction of greatest increase of $\phi$.

Divergence of $\vec F=F_1\hat i+F_2\hat j+F_3\hat k$: $\nabla\cdot\vec F=\dfrac{\partial F_1}{\partial x}+\dfrac{\partial F_2}{\partial y}+\dfrac{\partial F_3}{\partial z}$ — a scalar measuring net outward flux per unit volume.

Curl of $\vec F$: $\nabla\times\vec F=\begin{vmatrix}\hat i&\hat j&\hat k\\ \partial_x&\partial_y&\partial_z\\ F_1&F_2&F_3\end{vmatrix}$ — a vector measuring local rotation.

Proof $\nabla\times(\nabla\phi)=\vec0$. The $\hat i$-component is

$$\frac{\partial}{\partial y}\Big(\frac{\partial\phi}{\partial z}\Big)-\frac{\partial}{\partial z}\Big(\frac{\partial\phi}{\partial y}\Big)=\phi_{zy}-\phi_{yz}=0,$$

since mixed partials are equal for a twice-differentiable $\phi$ (Clairaut). Similarly the $\hat j$- and $\hat k$-components vanish, so $\nabla\times(\nabla\phi)=\vec0$.

(b) $\vec F=(x^2-yz)\hat i+(y^2-zx)\hat j+(z^2-xy)\hat k$

Divergence: $\nabla\cdot\vec F=2x+2y+2z=2(x+y+z).$

Curl:

• $\hat i$: $\partial_y(z^2-xy)-\partial_z(y^2-zx)=-x-(-x)=0.$
• $\hat j$: $\partial_z(x^2-yz)-\partial_x(z^2-xy)=-y-(-y)=0.$
• $\hat k$: $\partial_x(y^2-zx)-\partial_y(x^2-yz)=-z-(-z)=0.$

So $\operatorname{curl}\vec F=\vec0$, hence $\vec F$ is irrotational.

Scalar potential $\phi$ with $\nabla\phi=\vec F$:

$$\phi_x=x^2-yz\Rightarrow\phi=\tfrac{x^3}{3}-xyz+f(y,z).$$

$\phi_y=-xz+f_y=y^2-zx\Rightarrow f_y=y^2\Rightarrow f=\tfrac{y^3}{3}+g(z).$ $\phi_z=-xy+g'(z)=z^2-xy\Rightarrow g'=z^2\Rightarrow g=\tfrac{z^3}{3}.$

$$\boxed{\;\phi=\tfrac{1}{3}(x^3+y^3+z^3)-xyz+C.\;}$$

(a) Cayley–Hamilton theorem

Statement. Every square matrix satisfies its own characteristic equation: if $p(\lambda)=\det(A-\lambda I)=0$ is the characteristic equation, then $p(A)=0$.

For $A=\begin{pmatrix}1&2\\2&-1\end{pmatrix}$: $\det(A-\lambda I)=(1-\lambda)(-1-\lambda)-4=\lambda^2-5.$ Characteristic equation: $\lambda^2-5=0$.

Verification. $A^2=\begin{pmatrix}1&2\\2&-1\end{pmatrix}\begin{pmatrix}1&2\\2&-1\end{pmatrix}=\begin{pmatrix}5&0\\0&5\end{pmatrix}=5I.$ Thus $A^2-5I=0$ ✓ (theorem verified).

Inverse. From $A^2-5I=0\Rightarrow A^2=5I\Rightarrow A\cdot A=5I\Rightarrow A^{-1}=\tfrac{1}{5}A.$

$$\boxed{A^{-1}=\frac{1}{5}\begin{pmatrix}1&2\\2&-1\end{pmatrix}.}$$

(b) Eigenvalues and eigenvectors of $A=\begin{pmatrix}2&0&1\\0&2&0\\1&0&2\end{pmatrix}$

Characteristic equation.

$$\det(A-\lambda I)=(2-\lambda)\big[(2-\lambda)^2\big]-1\big[(2-\lambda)\big]\cdot?$$

Expand along the middle (sparse) structure. Since the matrix is block-like in the $x,z$ plane with $y$ decoupled:

$$\det(A-\lambda I)=(2-\lambda)\Big[(2-\lambda)^2-1\Big].$$

So $(2-\lambda)\big[(2-\lambda)^2-1\big]=0\Rightarrow 2-\lambda=0$ or $(2-\lambda)^2=1$. Eigenvalues: $\lambda=2,\;\lambda=3,\;\lambda=1.$

Eigenvectors.

• $\lambda=1$: $(A-I)x=0\Rightarrow\begin{pmatrix}1&0&1\\0&1&0\\1&0&1\end{pmatrix}x=0\Rightarrow x_1+x_3=0,\;x_2=0.$ Eigenvector $(1,0,-1)^T.$
• $\lambda=3$: $(A-3I)x=0\Rightarrow\begin{pmatrix}-1&0&1\\0&-1&0\\1&0&-1\end{pmatrix}x=0\Rightarrow x_1=x_3,\;x_2=0.$ Eigenvector $(1,0,1)^T.$
• $\lambda=2$: $(A-2I)x=0\Rightarrow\begin{pmatrix}0&0&1\\0&0&0\\1&0&0\end{pmatrix}x=0\Rightarrow x_1=0,\;x_3=0,\;x_2$ free. Eigenvector $(0,1,0)^T.$

$$\boxed{\lambda_1=1\to(1,0,-1),\quad\lambda_2=2\to(0,1,0),\quad\lambda_3=3\to(1,0,1).}$$

By De Moivre's theorem, $\cos4\theta+i\sin4\theta=(\cos\theta+i\sin\theta)^4.$ Expanding the right side by the binomial theorem (write $c=\cos\theta,\;s=\sin\theta$):

$$(c+is)^4=c^4+4c^3(is)+6c^2(is)^2+4c(is)^3+(is)^4=c^4-6c^2s^2+s^4+i(4c^3s-4cs^3).$$

Equating real parts:

$$\cos4\theta=c^4-6c^2s^2+s^4.$$

Replace $s^2=1-c^2$: $s^4=(1-c^2)^2=1-2c^2+c^4$ and $6c^2s^2=6c^2-6c^4$:

$$\cos4\theta=c^4-(6c^2-6c^4)+(1-2c^2+c^4)=8c^4-8c^2+1.$$ $$\boxed{\cos4\theta=8\cos^4\theta-8\cos^2\theta+1.}$$

Write $1+i$ in polar form: $r=\sqrt{1^2+1^2}=\sqrt2,\;\arg=\dfrac{\pi}{4}.$ So

$$1+i=\sqrt2\Big(\cos\tfrac{\pi}{4}+i\sin\tfrac{\pi}{4}\Big).$$

By De Moivre, the cube roots are

$$(1+i)^{1/3}=2^{1/6}\left[\cos\frac{\tfrac{\pi}{4}+2k\pi}{3}+i\sin\frac{\tfrac{\pi}{4}+2k\pi}{3}\right],\quad k=0,1,2.$$

The three arguments are $\dfrac{\pi}{12},\;\dfrac{\pi}{12}+\dfrac{2\pi}{3}=\dfrac{3\pi}{4},\;\dfrac{\pi}{12}+\dfrac{4\pi}{3}=\dfrac{17\pi}{12}.$

All have modulus $2^{1/6}\approx1.122$. Numerically:

• $k=0:\;2^{1/6}(\cos15^\circ+i\sin15^\circ)\approx1.084+0.291i.$
• $k=1:\;2^{1/6}(\cos135^\circ+i\sin135^\circ)\approx-0.794+0.794i.$
• $k=2:\;2^{1/6}(\cos255^\circ+i\sin255^\circ)\approx-0.291-1.084i.$

Argand diagram. The three roots lie on a circle of radius $2^{1/6}$ centred at the origin, equally spaced $120^\circ$ apart, at angles $15^\circ,\,135^\circ,\,255^\circ$ — i.e. vertices of an equilateral triangle.

Use the ratio test on $\sum a_n$ with $a_n=\dfrac{n^2}{2^n}$ (positive terms).

$$\frac{a_{n+1}}{a_n}=\frac{(n+1)^2}{2^{n+1}}\cdot\frac{2^n}{n^2}=\frac{1}{2}\left(\frac{n+1}{n}\right)^2=\frac{1}{2}\left(1+\frac{1}{n}\right)^2.$$

Taking the limit:

$$L=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\frac{1}{2}\cdot1=\frac{1}{2}<1.$$

Since $L<1$, by the ratio (d'Alembert) test the series $\displaystyle\sum_{n=1}^{\infty}\frac{n^2}{2^n}$ converges.

Let $a_n=\dfrac{(x-2)^n}{n\,3^n}.$ Apply the ratio test for the radius:

$$\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{(x-2)^{n+1}}{(n+1)3^{n+1}}\cdot\frac{n\,3^n}{(x-2)^n}\right|=\frac{|x-2|}{3}\cdot\frac{n}{n+1}\xrightarrow{n\to\infty}\frac{|x-2|}{3}.$$

Convergence requires $\dfrac{|x-2|}{3}<1\Rightarrow|x-2|<3.$ So the radius of convergence is $R=3$, giving the open interval $-1<x<5.$

Endpoints.

• $x=5$: series becomes $\sum\dfrac{3^n}{n3^n}=\sum\dfrac{1}{n}$, the harmonic series — diverges.
• $x=-1$: series becomes $\sum\dfrac{(-3)^n}{n3^n}=\sum\dfrac{(-1)^n}{n}$ — converges (alternating harmonic).

$$\boxed{R=3,\qquad\text{interval of convergence }[-1,5).}$$

Line 1: point $A=(1,2,3)$, direction $\vec b_1=(2,3,4)$. Line 2: point $C=(2,4,5)$, direction $\vec b_2=(3,4,5)$.

Cross product $\vec b_1\times\vec b_2$:

$$\begin{vmatrix}\hat i&\hat j&\hat k\\2&3&4\\3&4&5\end{vmatrix}=\hat i(15-16)-\hat j(10-12)+\hat k(8-9)=(-1,\,2,\,-1).$$

$|\vec b_1\times\vec b_2|=\sqrt{1+4+1}=\sqrt6.$

Connecting vector $\vec{AC}=C-A=(1,2,2).$

Shortest distance:

$$d=\frac{|(\vec{AC})\cdot(\vec b_1\times\vec b_2)|}{|\vec b_1\times\vec b_2|}=\frac{|(1)(-1)+(2)(2)+(2)(-1)|}{\sqrt6}=\frac{|-1+4-2|}{\sqrt6}=\frac{1}{\sqrt6}.$$ $$\boxed{d=\dfrac{1}{\sqrt6}\approx0.408\text{ units}.}$$

The required plane is perpendicular to the line of intersection of the two given planes, so the line's direction $\vec d$ is the normal of the required plane.

The line of intersection is along $\vec n_1\times\vec n_2$, where $\vec n_1=(1,2,3),\;\vec n_2=(2,-1,1).$

$$\vec d=\vec n_1\times\vec n_2=\begin{vmatrix}\hat i&\hat j&\hat k\\1&2&3\\2&-1&1\end{vmatrix}=\hat i(2+3)-\hat j(1-6)+\hat k(-1-4)=(5,\,5,\,-5).$$

Divide by $5$: normal direction $(1,1,-1).$

Plane through $(1,-2,3)$ with normal $(1,1,-1)$:

$$1(x-1)+1(y+2)-1(z-3)=0\Rightarrow x+y-z+4=0.$$ $$\boxed{x+y-z+4=0.}$$

First check the vectors close to form a triangle: $\vec a-\vec b+?$. Note $\vec b+\vec c=(1+3,\,-3-4,\,-5-4)=(4,-7,-9)$, not $\vec a$, so they are taken as three sides; we test the angles via scalar (dot) products. A triangle is right-angled iff one pair of side-vectors is perpendicular (dot product $=0$).

Compute the dot products:

$$\vec a\cdot\vec b=(2)(1)+(-1)(-3)+(1)(-5)=2+3-5=0.$$

Since $\vec a\cdot\vec b=0$, the sides $\vec a$ and $\vec b$ are perpendicular, so the triangle they bound has a right angle between those two sides.

(For completeness: $\vec a\cdot\vec c=6+4-4=6\neq0$ and $\vec b\cdot\vec c=3+12+20=35\neq0$.) The single zero dot product $\vec a\cdot\vec b=0$ establishes that the vectors form the sides of a right-angled triangle, right-angled between $\vec a$ and $\vec b$.

Innermost integral over $z$ from $0$ to $1-x-y$. Let $s=1-x-y$.

$$\int_{0}^{s}(x+y+z)\,dz=(x+y)z+\tfrac{z^2}{2}\Big|_0^{s}=(x+y)s+\tfrac{s^2}{2}.$$

With $x+y=1-s$: $=(1-s)s+\tfrac{s^2}{2}=s-\tfrac{s^2}{2}.$ So integrand in $y$ is $\tfrac{1}{2}\big[2s-s^2\big]=\tfrac12\big[1-(1-s)^2\big]$. Using $s=1-x-y$:

$$\int_{0}^{1-x}\Big[s-\tfrac{s^2}{2}\Big]dy,\quad s=1-x-y.$$

Substitute $u=1-x-y$ ($du=-dy$; as $y:0\to1-x,\;u:1-x\to0$):

$$\int_{0}^{1-x}\Big(u-\tfrac{u^2}{2}\Big)dy=\int_{0}^{1-x}\Big(u-\tfrac{u^2}{2}\Big)du=\frac{(1-x)^2}{2}-\frac{(1-x)^3}{6}.$$

Let $t=1-x$. Outer integral over $x:0\to1$ (so $t:1\to0$, $dt=-dx$):

$$\int_{0}^{1}\Big(\tfrac{t^2}{2}-\tfrac{t^3}{6}\Big)dt=\frac{1}{2}\cdot\frac{1}{3}-\frac{1}{6}\cdot\frac{1}{4}=\frac{1}{6}-\frac{1}{24}=\frac{4-1}{24}=\frac{3}{24}=\frac{1}{8}.$$ $$\boxed{\;\frac{1}{8}\;}$$

System:

$$x+y+z=6,\quad x+2y+3z=10,\quad x+2y+\lambda z=\mu.$$

Subtract equation 2 from equation 3: $(\lambda-3)z=\mu-10.$ The coefficient matrix determinant:

$$\det\begin{pmatrix}1&1&1\\1&2&3\\1&2&\lambda\end{pmatrix}=1(2\lambda-6)-1(\lambda-3)+1(2-2)=2\lambda-6-\lambda+3=\lambda-3.$$

(ii) Unique solution: determinant $\neq0\Rightarrow\lambda\neq3$ (for any $\mu$). Rank of $A$ = rank of augmented = 3.

When $\lambda=3$, determinant $=0$ and $(\lambda-3)z=\mu-10$ becomes $0=\mu-10$:

(i) No solution: $\lambda=3$ and $\mu\neq10$ (inconsistent, $0=\mu-10\neq0$).

(iii) Infinitely many solutions: $\lambda=3$ and $\mu=10$ (the third equation duplicates the second; one free parameter).

This is a Cauchy–Euler (equidimensional) equation. Try $y=x^m.$ Then $y'=mx^{m-1},\;y''=m(m-1)x^{m-2}.$ Substituting into $x^2y''-2xy'+2y=0$:

$$x^2\,m(m-1)x^{m-2}-2x\,m\,x^{m-1}+2x^m=0\Rightarrow\big[m(m-1)-2m+2\big]x^m=0.$$

Indicial equation: $m^2-3m+2=0\Rightarrow(m-1)(m-2)=0\Rightarrow m=1,\,2.$

Distinct real roots, so the general solution is

$$\boxed{\,y=C_1x+C_2x^{2}.}$$

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

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

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

• $\partial_x=2(1)(-2)(-1)+4(-1)^2=4+4=8.$
• $\partial_y=(1)^2(-1)=-1.$
• $\partial_z=(1)^2(-2)+8(1)(-1)=-2-8=-10.$

So $\nabla\phi=(8,\,-1,\,-10).$

Unit vector in direction $\vec a=2\hat i-\hat j-2\hat k$: $|\vec a|=\sqrt{4+1+4}=3$, $\hat a=\tfrac13(2,-1,-2).$

Directional derivative:

$$D_{\hat a}\phi=\nabla\phi\cdot\hat a=\frac{1}{3}\big[(8)(2)+(-1)(-1)+(-10)(-2)\big]=\frac{16+1+20}{3}=\frac{37}{3}.$$ $$\boxed{D_{\hat a}\phi=\dfrac{37}{3}\approx12.33.}$$