BE Computer Engineering (IOE, TU) Engineering Mathematics I (IOE, SH 401) Question Paper 2079 Nepal

(a) Euler's theorem and application

Statement. If $u = f(x,y)$ is a homogeneous function of degree $n$ in $x$ and $y$, then

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

Proof. A function homogeneous of degree $n$ can be written as $u = x^n\,g\!\left(\dfrac{y}{x}\right)$. Put $t = y/x$. Then

$$\frac{\partial u}{\partial x} = n x^{n-1} g(t) + x^n g'(t)\left(-\frac{y}{x^2}\right) = n x^{n-1}g(t) - x^{n-2}y\,g'(t),$$ $$\frac{\partial u}{\partial y} = x^n g'(t)\cdot\frac{1}{x} = x^{n-1}g'(t).$$

Hence

$$x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = n x^n g(t) - x^{n-1}y\,g'(t) + x^{n-1}y\,g'(t) = n x^n g(t) = n u. \qquad\blacksquare$$

Application. Let $u = \tan^{-1}\!\left(\dfrac{x^3+y^3}{x-y}\right)$. Put $z = \tan u = \dfrac{x^3+y^3}{x-y}$.

Here $z$ is homogeneous of degree $3-1 = 2$. By Euler's theorem,

$$x z_x + y z_y = 2z.$$

Since $z = \tan u$, $z_x = \sec^2 u\,u_x$ and $z_y = \sec^2 u\,u_y$, so

$$\sec^2 u\,(x u_x + y u_y) = 2\tan u.$$

Therefore

$$x u_x + y u_y = \frac{2\tan u}{\sec^2 u} = 2\sin u\cos u = \sin 2u. \qquad\blacksquare$$

(b) Change to polar coordinates

With $x = r\cos\theta,\; y = r\sin\theta$ and $z = f(x,y)$, by the chain rule

$$\frac{\partial z}{\partial r} = z_x\cos\theta + z_y\sin\theta,\qquad \frac{\partial z}{\partial \theta} = -z_x\,r\sin\theta + z_y\,r\cos\theta.$$

Then

$$\left(\frac{\partial z}{\partial r}\right)^2 = z_x^2\cos^2\theta + 2z_xz_y\cos\theta\sin\theta + z_y^2\sin^2\theta,$$ $$\frac{1}{r^2}\left(\frac{\partial z}{\partial \theta}\right)^2 = z_x^2\sin^2\theta - 2z_xz_y\cos\theta\sin\theta + z_y^2\cos^2\theta.$$

Adding,

$$\left(\frac{\partial z}{\partial r}\right)^2 + \frac{1}{r^2}\left(\frac{\partial z}{\partial \theta}\right)^2 = z_x^2(\cos^2\theta+\sin^2\theta) + z_y^2(\sin^2\theta+\cos^2\theta) = z_x^2 + z_y^2 = \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2. \qquad\blacksquare$$

(a) Asymptotes of $x^3 + 2x^2y - xy^2 - 2y^3 + 4y^2 + 2xy + y - 1 = 0$

Oblique asymptotes. The third-degree terms give $\phi_3(m)$ by putting $x=1,\,y=m$ in the highest-degree part:

$$\phi_3(m) = 1 + 2m - m^2 - 2m^3 = -(2m^3 + m^2 - 2m - 1) = -(2m+1)(m^2-1) = -(2m+1)(m-1)(m+1).$$

Roots: $m = 1,\; -1,\; -\tfrac12$. For each slope $c$ is found from

$$c = -\frac{\phi_2(m)}{\phi_3'(m)},$$

where $\phi_2(m) = 4m^2 + 2m$ (from the degree-2 terms $4y^2+2xy$, with $x=1,y=m$) and $\phi_3'(m) = 2 - 2m - 6m^2$.

• $m=1$: $\phi_2 = 6,\ \phi_3' = 2-2-6 = -6,\ c = -6/(-6)=1$. Asymptote $y = x + 1$.
• $m=-1$: $\phi_2 = 4-2 = 2,\ \phi_3' = 2+2-6 = -2,\ c = -2/(-2) = 1$. Asymptote $y = -x + 1$.
• $m=-\tfrac12$: $\phi_2 = 1-1 = 0,\ \phi_3' = 2+1-\tfrac32 = \tfrac32,\ c = 0$. Asymptote $y = -\tfrac12 x$, i.e. $2y + x = 0$.

Asymptotes: $\boxed{y = x+1,\quad y = -x+1,\quad x + 2y = 0.}$

(b) Radius of curvature

Derivation. Curvature $\kappa = \dfrac{d\psi}{ds}$ where $\tan\psi = y_1 = dy/dx$ and $ds = \sqrt{1+y_1^2}\,dx$. Differentiating $\psi = \tan^{-1}y_1$,

$$\frac{d\psi}{dx} = \frac{y_2}{1+y_1^2}.$$

Hence

$$\kappa = \frac{d\psi}{ds} = \frac{d\psi/dx}{ds/dx} = \frac{y_2/(1+y_1^2)}{\sqrt{1+y_1^2}} = \frac{y_2}{(1+y_1^2)^{3/2}},$$

so

$$\rho = \frac{1}{\kappa} = \frac{(1+y_1^2)^{3/2}}{y_2}.$$

Application. $y = x^3 - 6x^2 + 3x + 1$, so $y_1 = 3x^2 - 12x + 3,\ y_2 = 6x - 12$. At $(1,-1)$: $y_1 = 3 - 12 + 3 = -6,\ y_2 = 6 - 12 = -6$.

$$\rho = \frac{(1+36)^{3/2}}{-6} = \frac{37^{3/2}}{-6}.$$

Taking magnitude, $\rho = \dfrac{37\sqrt{37}}{6} \approx \dfrac{225.1}{6} \approx 37.5$ units.

(a) Reduction formula for $I_n = \int_0^{\pi/2}\sin^n x\,dx$

Write $I_n = \int_0^{\pi/2}\sin^{n-1}x\,\sin x\,dx$ and integrate by parts with $u=\sin^{n-1}x,\ dv=\sin x\,dx$:

$$I_n = \big[-\sin^{n-1}x\cos x\big]_0^{\pi/2} + (n-1)\int_0^{\pi/2}\sin^{n-2}x\cos^2x\,dx.$$

The boundary term vanishes. Using $\cos^2x = 1-\sin^2x$:

$$I_n = (n-1)(I_{n-2} - I_n) \;\Rightarrow\; I_n = \frac{n-1}{n}\,I_{n-2}.$$

Evaluation of $\int_0^{\pi/2}\sin^6x\,dx$. With $I_0 = \pi/2$:

$$I_6 = \frac{5}{6}\cdot\frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2} = \frac{15}{48}\cdot\frac{\pi}{2} = \frac{5\pi}{32}.$$

(b) Reduction formula for $I_{m,n} = \int \cos^m x\,\sin^n x\,dx$

Write $I_{m,n} = \int \cos^{m-1}x\,(\sin^n x\cos x)\,dx$. Integrate by parts with $u = \cos^{m-1}x,\ dv = \sin^n x\cos x\,dx$ so $v = \dfrac{\sin^{n+1}x}{n+1}$:

$$I_{m,n} = \frac{\cos^{m-1}x\sin^{n+1}x}{n+1} + \frac{m-1}{n+1}\int \cos^{m-2}x\sin^{n+2}x\,dx.$$

Replace $\sin^{n+2}x = \sin^n x(1-\cos^2x)$:

$$I_{m,n} = \frac{\cos^{m-1}x\sin^{n+1}x}{n+1} + \frac{m-1}{n+1}\big(I_{m-2,n} - I_{m,n}\big).$$

Collecting $I_{m,n}$:

$$I_{m,n} = \frac{\cos^{m-1}x\,\sin^{n+1}x}{m+n} + \frac{m-1}{m+n}\,I_{m-2,n}.$$

Evaluation of $\int_0^{\pi/2}\cos^4x\sin^2x\,dx$. Over $[0,\pi/2]$ the bracket term vanishes, so the definite form gives

$$I_{m,n} = \frac{m-1}{m+n}\,I_{m-2,n}.$$

• $I_{4,2} = \dfrac{3}{6}I_{2,2} = \dfrac12 I_{2,2}$.
• $I_{2,2} = \dfrac{1}{4}I_{0,2}$.
• $I_{0,2} = \int_0^{\pi/2}\sin^2x\,dx = \dfrac{\pi}{4}$.

Hence $I_{2,2} = \dfrac14\cdot\dfrac{\pi}{4} = \dfrac{\pi}{16}$ and

$$\int_0^{\pi/2}\cos^4x\sin^2x\,dx = \frac12\cdot\frac{\pi}{16} = \frac{\pi}{32}.$$

Leibnitz's theorem

If $u$ and $v$ are functions of $x$ possessing derivatives up to order $n$, then the $n$th derivative of the product is

$$(uv)_n = \sum_{r=0}^{n}\binom{n}{r} u_{n-r}\,v_r = u_n v + \binom{n}{1}u_{n-1}v_1 + \binom{n}{2}u_{n-2}v_2 + \cdots + u\,v_n.$$

Recurrence for $y = \sin^{-1}x$

$y_1 = \dfrac{1}{\sqrt{1-x^2}}\Rightarrow (1-x^2)y_1^2 = 1$. Differentiating: $(1-x^2)2y_1y_2 - 2x y_1^2 = 0$, i.e.

$$(1-x^2)y_2 - x y_1 = 0.$$

Differentiate this $n$ times by Leibnitz's theorem. For the term $(1-x^2)y_2$ take $u=y_2,\,v=(1-x^2)$ (whose derivatives are $-2x$ and $-2$, higher ones zero):

$$(1-x^2)y_{n+2} + n(-2x)y_{n+1} + \frac{n(n-1)}{2}(-2)y_n.$$

For $-xy_1$: $-\big(x y_{n+1} + n\,y_n\big)$. Adding,

$$(1-x^2)y_{n+2} - 2nx\,y_{n+1} - n(n-1)y_n - x y_{n+1} - n y_n = 0,$$ $$\boxed{(1-x^2)y_{n+2} - (2n+1)x\,y_{n+1} - n^2 y_n = 0.}$$

Value of $(y_n)_0$ at $x=0$

Put $x=0$: $(y_{n+2})_0 = n^2 (y_n)_0$. With $y(0)=0,\ y_1(0)=1,\ y_2(0)=0$:

• $n$ even: $(y_n)_0 = 0$.
• $n$ odd: $(y_n)_0 = (n-2)^2(n-4)^2\cdots 3^2\cdot 1^2 = \big[(n-2)(n-4)\cdots 3\cdot 1\big]^2.$

Thus $(y_n)_0 = 0$ for even $n$, and $(y_n)_0 = \{(n-2)!!\}^2$ for odd $n$ (with $(y_1)_0 = 1$).

(a) Convergence of $\int_1^{\infty}\dfrac{dx}{x^p}$

For $p \neq 1$,

$$\int_1^{R}x^{-p}\,dx = \left[\frac{x^{1-p}}{1-p}\right]_1^{R} = \frac{R^{1-p}-1}{1-p}.$$

As $R\to\infty$: if $p>1$, $R^{1-p}\to 0$ and the integral $\to \dfrac{1}{p-1}$ (converges). If $p<1$, $R^{1-p}\to\infty$ (diverges). For $p=1$, $\int_1^R \frac{dx}{x} = \ln R \to \infty$ (diverges).

Conclusion: the integral converges iff $p>1$, with value $\dfrac{1}{p-1}$; it diverges for $p\le 1$.

(b) Evaluate $\int_0^1 \dfrac{dx}{\sqrt{1-x^2}}$

The integrand is unbounded at $x=1$, so it is an improper integral of the second kind:

$$\int_0^1\frac{dx}{\sqrt{1-x^2}} = \lim_{b\to 1^-}\int_0^b\frac{dx}{\sqrt{1-x^2}} = \lim_{b\to 1^-}\big[\sin^{-1}x\big]_0^b = \lim_{b\to1^-}\sin^{-1}b = \sin^{-1}1 = \frac{\pi}{2}.$$

The limit exists and is finite, so the integral converges to $\dfrac{\pi}{2}$.

(a) Exact equation

$M = x^2 - 4xy - 2y^2,\quad N = y^2 - 4xy - 2x^2$. Test exactness:

$$M_y = -4x - 4y,\qquad N_x = -4y - 4x.$$

Since $M_y = N_x$, the equation is exact. The solution is $F(x,y)=C$ where

$$F = \int M\,dx = \frac{x^3}{3} - 2x^2 y - 2xy^2 + g(y).$$

Then $F_y = -2x^2 - 4xy + g'(y)$ must equal $N = y^2 - 4xy - 2x^2$, so $g'(y) = y^2$, giving $g(y) = \dfrac{y^3}{3}$.

Solution: $\dfrac{x^3}{3} - 2x^2 y - 2xy^2 + \dfrac{y^3}{3} = C$, i.e. $\;x^3 + y^3 - 6x^2y - 6xy^2 = C_1.$

(b) Linear equation $\dfrac{dy}{dx} + \dfrac{y}{x} = x^2$

Here $P = 1/x$, integrating factor $\mu = e^{\int dx/x} = e^{\ln x} = x$. Multiply through:

$$\frac{d}{dx}(xy) = x\cdot x^2 = x^3.$$

Integrate: $xy = \dfrac{x^4}{4} + C$, so

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

Volume of revolution about the $x$-axis

Region bounded by $y^2 = 4x$ and $x = 4$, revolved about the $x$-axis. Using disks of radius $y$ with $y^2 = 4x$:

$$V = \int_0^4 \pi y^2\,dx = \int_0^4 \pi (4x)\,dx = 4\pi\left[\frac{x^2}{2}\right]_0^4 = 4\pi\cdot 8 = 32\pi \text{ cubic units}.$$

Surface area generated by the arc $y^2 = 4x$

For the parabola, $y = 2\sqrt{x}$, $\dfrac{dy}{dx} = \dfrac{1}{\sqrt{x}}$, so

$$1 + \left(\frac{dy}{dx}\right)^2 = 1 + \frac{1}{x} = \frac{x+1}{x}.$$

Surface of revolution about the $x$-axis:

$$S = \int_0^4 2\pi y\sqrt{1+y_1^2}\,dx = \int_0^4 2\pi (2\sqrt{x})\sqrt{\frac{x+1}{x}}\,dx = 4\pi\int_0^4 \sqrt{x+1}\,dx.$$ $$= 4\pi\left[\frac{2}{3}(x+1)^{3/2}\right]_0^4 = \frac{8\pi}{3}\big[5^{3/2} - 1\big] = \frac{8\pi}{3}(5\sqrt5 - 1)\text{ square units}.$$

Numerically $S \approx \dfrac{8\pi}{3}(11.180 - 1) \approx 85.3$ square units.

Reduce $3x^2 + 2xy + 3y^2 - 16y + 23 = 0$

Remove the $xy$ term. For $ax^2 + 2hxy + by^2+\dots$ with $a=3,\,h=1,\,b=3$, rotate by angle $\theta$ where $\tan 2\theta = \dfrac{2h}{a-b} = \dfrac{2}{0} = \infty$, so $2\theta = 90^\circ,\ \theta = 45^\circ$.

The new quadratic coefficients are the eigenvalues of $\begin{pmatrix}3&1\\1&3\end{pmatrix}$: $\lambda = 3\pm1 = 4,\,2$. So the rotated quadratic part is $4X^2 + 2Y^2$.

With $x = \dfrac{X-Y}{\sqrt2},\ y = \dfrac{X+Y}{\sqrt2}$, the linear term $-16y = -\dfrac{16}{\sqrt2}(X+Y) = -8\sqrt2\,(X+Y)$. The equation becomes

$$4X^2 + 2Y^2 - 8\sqrt2\,X - 8\sqrt2\,Y + 23 = 0.$$

Complete the square.

$$4\big(X^2 - 2\sqrt2 X\big) + 2\big(Y^2 - 4\sqrt2 Y\big) + 23 = 0,$$ $$4(X-\sqrt2)^2 - 8 + 2(Y-2\sqrt2)^2 - 16 + 23 = 0,$$ $$4(X-\sqrt2)^2 + 2(Y-2\sqrt2)^2 = 1.$$

With $X' = X-\sqrt2,\ Y' = Y-2\sqrt2$:

$$\frac{X'^2}{1/4} + \frac{Y'^2}{1/2} = 1.$$

Standard form: $\dfrac{X'^2}{(1/2)^2} + \dfrac{Y'^2}{(1/\sqrt2)^2} = 1.$

Type: an ellipse. Semi-axes: $\dfrac{1}{\sqrt2} \approx 0.707$ (along $Y'$, semi-major) and $\dfrac{1}{2} = 0.5$ (along $X'$, semi-minor).

(a) Angle of intersection of $r = a(1+\cos\theta)$ and $r = b(1-\cos\theta)$

Use $\tan\phi = r\dfrac{d\theta}{dr}$.

Curve 1: $r = a(1+\cos\theta)$, $\dfrac{dr}{d\theta} = -a\sin\theta$.

$$\tan\phi_1 = \frac{a(1+\cos\theta)}{-a\sin\theta} = \frac{2\cos^2(\theta/2)}{-2\sin(\theta/2)\cos(\theta/2)} = -\cot\frac{\theta}{2} = \tan\!\left(\frac{\pi}{2}+\frac{\theta}{2}\right).$$

So $\phi_1 = \dfrac{\pi}{2} + \dfrac{\theta}{2}$.

Curve 2: $r = b(1-\cos\theta)$, $\dfrac{dr}{d\theta} = b\sin\theta$.

$$\tan\phi_2 = \frac{b(1-\cos\theta)}{b\sin\theta} = \frac{2\sin^2(\theta/2)}{2\sin(\theta/2)\cos(\theta/2)} = \tan\frac{\theta}{2}.$$

So $\phi_2 = \dfrac{\theta}{2}$.

The angle of intersection is

$$\phi_1 - \phi_2 = \frac{\pi}{2}.$$

The two cardioids intersect orthogonally (cut at right angles) at every common point.

(b) Pedal equation of $r = a(1+\cos\theta)$

The pedal relation is $\dfrac{1}{p^2} = \dfrac{1}{r^2} + \dfrac{1}{r^4}\left(\dfrac{dr}{d\theta}\right)^2$, or simply $p = r\sin\phi$.

From part (a), $\phi = \dfrac{\pi}{2} + \dfrac{\theta}{2}$, so $\sin\phi = \cos\dfrac{\theta}{2}$ and $p = r\cos\dfrac{\theta}{2}$.

Thus $p^2 = r^2\cos^2\dfrac{\theta}{2} = r^2\cdot\dfrac{1+\cos\theta}{2}$. Since $1+\cos\theta = \dfrac{r}{a}$,

$$p^2 = r^2\cdot\frac{r}{2a} = \frac{r^3}{2a}\quad\Rightarrow\quad \boxed{2a\,p^2 = r^3.}$$

Maxima/minima of $f(x,y) = x^3 + y^3 - 3axy$

Stationary points.

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

So $y = x^2/a$ and $x = y^2/a$. Substituting: $x = \dfrac{(x^2/a)^2}{a} = \dfrac{x^4}{a^3}$, giving $x^4 = a^3 x$, i.e. $x(x^3 - a^3)=0$. Hence $x = 0$ or $x = a$.

Stationary points: $(0,0)$ and $(a,a)$.

Second derivatives.

$$r = f_{xx} = 6x,\quad t = f_{yy} = 6y,\quad s = f_{xy} = -3a.$$

Discriminant $D = rt - s^2 = 36xy - 9a^2$.

At $(0,0)$: $D = 0 - 9a^2 = -9a^2 < 0$ $\Rightarrow$ saddle point (no extremum).

At $(a,a)$: $D = 36a^2 - 9a^2 = 27a^2 > 0$, and $r = 6a$.

• If $a > 0$: $r > 0 \Rightarrow$ minimum, with $f(a,a) = a^3 + a^3 - 3a^3 = -a^3$.
• If $a < 0$: $r < 0 \Rightarrow$ maximum, with $f(a,a) = -a^3$.

Conclusion: $(0,0)$ is a saddle; $(a,a)$ gives an extremum value $-a^3$ (minimum for $a>0$, maximum for $a<0$).

Orthogonal trajectories of $x^2 + y^2 = 2cx$

Form the differential equation of the family. Differentiate $x^2 + y^2 = 2cx$:

$$2x + 2y y' = 2c.$$

Eliminate $c$ using $c = \dfrac{x^2+y^2}{2x}$:

$$2x + 2yy' = \frac{x^2+y^2}{x}\;\Rightarrow\; 2x^2 + 2xyy' = x^2 + y^2\;\Rightarrow\; y' = \frac{y^2 - x^2}{2xy}.$$

Replace $y'$ by $-1/y'$ for orthogonal trajectories:

$$-\frac{1}{y'} = \frac{y^2-x^2}{2xy}\;\Rightarrow\; \frac{dy}{dx} = \frac{2xy}{x^2 - y^2}.$$

Solve (homogeneous). Put $y = vx$, $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$:

$$v + x\frac{dv}{dx} = \frac{2v}{1 - v^2}\;\Rightarrow\; x\frac{dv}{dx} = \frac{2v - v(1-v^2)}{1-v^2} = \frac{v + v^3}{1-v^2} = \frac{v(1+v^2)}{1-v^2}.$$

Separate:

$$\frac{1-v^2}{v(1+v^2)}\,dv = \frac{dx}{x}.$$

Partial fractions: $\dfrac{1-v^2}{v(1+v^2)} = \dfrac{1}{v} - \dfrac{2v}{1+v^2}$. Integrate:

$$\ln v - \ln(1+v^2) = \ln x + \ln k\;\Rightarrow\; \frac{v}{1+v^2} = kx.$$

Back-substitute $v = y/x$: $\dfrac{y/x}{1 + y^2/x^2} = kx$, i.e. $\dfrac{xy}{x^2+y^2} = kx$, giving

$$x^2 + y^2 = \frac{y}{k} = 2c'y,$$

so the orthogonal trajectories are the family

$$\boxed{x^2 + y^2 = 2c' y}$$

i.e. the family of circles through the origin with centres on the $y$-axis (the original family being circles through the origin with centres on the $x$-axis).