Engineering Mathematics-I

For $m,n>0$, $\displaystyle B(m,n)=\int_0^1t^{m-1}(1-t)^{n-1}\,dt=\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$.

$\displaystyle f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}(a_n\cos nx+b_n\sin nx)$, where $a_0,a_n,b_n$ are the Fourier coefficients.

$f_x=y^x\ln y$, so $f_{xy}=y^{x-1}(1+x\ln y)$. At $(2,1)$, $\boxed{f_{xy}=1}$.

$\displaystyle L=\int_0^{\pi/2}\sqrt{(-\sin t)^2+(\cos t)^2+\left(\frac2\pi\right)^2}\,dt=\frac\pi2\sqrt{1+\frac4{\pi^2}}=\boxed{\frac{\sqrt{\pi^2+4}}2}$.

$\displaystyle e^x=e^2\sum_{n=0}^{\infty}\frac{(x-2)^n}{n!}$. The coefficient is $\boxed{\frac{e^2}{4!}=\frac{e^2}{24}}$.

A power series $\sum_{n=0}^{\infty}a_n(x-c)^n$ converges at $x$ if its sequence of partial sums has a finite limit.

$\nabla f=(2x,4y,1)$, so at the point it is $(2,4,1)$. The unit direction is $(3/5,-4/5,0)$. Thus $D_{\mathbf a}f=6/5-16/5=\boxed{-2}$.

$\nabla\times\mathbf V=(3y^2-6z)\mathbf i+0\mathbf j+0\mathbf k$. At $(1,1,1)$, $\boxed{\nabla\times\mathbf V=-3\mathbf i}$.

Vorticity is $\nabla\times\mathbf V$. Here $\nabla\times\mathbf V=3x^2\mathbf i+(-6xz-2x)\mathbf k$. At $(1,1,1)$, $\boxed{3\mathbf i-8\mathbf k}$.

In the first quadrant, $x=y^2/4$ and $x=2\sqrt y$ meet at $y=0,4$. Hence $\displaystyle A=\int_0^4(2\sqrt y-y^2/4)\,dy=\boxed{16/3}$.

(i) Using $\int_0^\infty x^{m-1}e^{-x^2}dx=\frac12\Gamma(m/2)$,

$I=\frac12\Gamma\left(\frac52\right)=\boxed{\frac{3\sqrt\pi}{8}}.$

(ii) Using $\int_0^{\pi/2}\sin^p\theta\cos^q\theta\,d\theta=\frac12B((p+1)/2,(q+1)/2)$,

$J=\frac12B\left(\frac72,4\right)=\boxed{\frac{16}{3003}}.$

The curves intersect at $x=\pi/4$. The outer radius is $\cos x$ and inner radius is $\sin x$.

$V=\pi\int_0^{\pi/4}(\cos^2x-\sin^2x)\,dx=\pi\int_0^{\pi/4}\cos2x\,dx=\boxed{\frac\pi2}.$

The $n$th term is $\displaystyle a_n=\frac{(2n-1)(2n)}{(2n+1)^2(2n+2)^2}$. Then $\displaystyle\lim_{n\to\infty}\frac{a_n}{1/n^2}=\frac14$. Since $\sum 1/n^2$ converges, the given series converges by limit comparison.

The coefficients are $a_0=\pi/2$, $a_n=(1-(-1)^n)/(\pi n^2)$ and $b_n=-1/n$. Therefore

$\boxed{f(x)=\frac\pi4+\sum_{n=1}^{\infty}\left[\frac{1-(-1)^n}{\pi n^2}\cos nx-\frac{\sin nx}{n}\right]}.$

Let $D=x^{2014}+\sqrt{x^{2012}+1}$, $A=e^{\sin x}/D$ and $G=A+\cos(xy)$, so $f=\sqrt G$. Then

$G_x=A'-y\sin(xy),\quad G_y=-x\sin(xy),$

$G_{xy}=-\sin(xy)-xy\cos(xy).$

By differentiating $f=G^{1/2}$,

$\boxed{f_{xy}=\frac{-\sin(xy)-xy\cos(xy)}{2\sqrt G}+\frac{x\sin(xy)(A'-y\sin(xy))}{4G^{3/2}}}.$

Here $\displaystyle A'=\frac{e^{\sin x}\cos x}{D}-\frac{e^{\sin x}}{D^2}\left(2014x^{2013}+\frac{1006x^{2011}}{\sqrt{x^{2012}+1}}\right)$.

Since $\nabla\phi=\nabla(xyz)$, the potential is $\phi=xyz$. Along the curve, $xyz=t\cdot t^2\cdot3t^2=3t^5$. Therefore

$\int_C\nabla\phi\cdot d\mathbf r=\phi(3)-\phi(1)=3(3^5-1)=\boxed{726}.$

In polar coordinates, $x=r\cos\theta$, $y=r\sin\theta$. The boundaries give $\pi/6\le\theta\le\pi/3$ and $\csc\theta\le r\le2$.

$A=\frac12\int_{\pi/6}^{\pi/3}(4-\csc^2\theta)\,d\theta=\boxed{\frac\pi3-\frac1{\sqrt3}}.$

For $0<x<2$, the half-range cosine coefficients are $a_0=\pi$, $a_{2m+1}=0$, and $a_{2m}=2[(-1)^m-1]/(m^2\pi)$. Hence

$\boxed{f(x)=\frac\pi2-\frac4\pi\left(\cos\pi x+\frac{\cos3\pi x}{3^2}+\frac{\cos5\pi x}{5^2}+\cdots\right)}.$

By Parseval's identity, $\int_0^2 f^2dx=2a_0^2/4+\sum a_n^2$. Now $\int_0^2f^2dx=2\pi^2/3$ and $\sum a_n^2=\frac{16}{\pi^2}\sum_{m=0}^{\infty}(2m+1)^{-4}$. Thus

$\frac{2\pi^2}{3}=\frac{\pi^2}{2}+\frac{16}{\pi^2}\sum_{m=0}^{\infty}\frac1{(2m+1)^4},\quad\boxed{\sum_{m=0}^{\infty}\frac1{(2m+1)^4}=\frac{\pi^4}{96}}.$

Since $\nabla r^n=n r^{n-2}(x\mathbf i+y\mathbf j+z\mathbf k)$,

$\nabla^2r^n=n\left[3r^{n-2}+(n-2)r^{n-4}(x^2+y^2+z^2)\right]$

$=n[3+n-2]r^{n-2}=\boxed{n(n+1)r^{n-2}}.$

Taking $n=-1$ gives $\boxed{\nabla^2(1/r)=0}$ for $r\ne0$.

Maximize $P=400xyz^2$ subject to $x^2+y^2+z^2=1$. At the positive maximum, symmetry gives $x^2=y^2$. Let $u=x^2=y^2$ and $w=z^2$, so $2u+w=1$ and $xyz^2$ is maximized when $u=1/4$, $w=1/2$. Thus $|x|=|y|=1/2$, $z^2=1/2$ with $xy>0$.

$P_{\max}=400\left(\frac12\cdot\frac12\cdot\frac12\right)=\boxed{50}.$

The field is conservative because its mixed partial derivatives agree. Find a potential:

$\phi_x=y^2\cos x+z^3\implies\phi=y^2\sin x+xz^3+g(y,z).$

Comparing the other components gives $g=-4y+z^2/2$. Therefore $\phi=y^2\sin x+xz^3-4y+z^2/2$.

$W=\phi(Q)-\phi(P)=(7+4\pi)-\left(-\frac72\right)=\boxed{\frac{21}{2}+4\pi}.$

Let $\mathbf F=(y,z,x)$. Then

$\nabla\times\mathbf F=(-1,-1,-1).$

The intersection circle bounds a disk in the plane $x+z=a$. Choose the normal $\mathbf n=(1,0,1)/\sqrt2$. The disk is at distance $a/\sqrt2$ from the origin, so its radius is $a/\sqrt2$ and its area is $\pi a^2/2$.

$\oint_C\mathbf F\cdot d\mathbf r=\iint_S(\nabla\times\mathbf F)\cdot\mathbf n\,dS=(-\sqrt2)\frac{\pi a^2}{2}=\boxed{-\frac{\pi a^2}{\sqrt2}}.$

The sign reverses if the opposite curve orientation is chosen.