BE Computer Engineering (IOE, TU) Engineering Mathematics I (IOE, SH 401) Question Paper 2078 Nepal

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

Slopes of the asymptotes. Put $y = mx$ in the highest-degree (3rd) terms; the leading coefficient $\phi_3(m)$ is obtained from $x^3(1 + 2m - m^2 - 2m^3)$:

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

Factorise: $2m^3 + m^2 - 2m - 1 = m^2(2m+1) - (2m+1) = (2m+1)(m^2-1) = (2m+1)(m-1)(m+1).$

So $\phi_3(m)=0 \Rightarrow m = 1,\ -1,\ -\tfrac12$. All roots are real and distinct, so there are three oblique asymptotes $y = mx + c$.

Intercepts $c$. Using $c = -\dfrac{\phi_2(m)}{\phi_3'(m)}$, where $\phi_2(m)$ comes from the 2nd-degree terms $4y^2 + 2xy$, i.e. $x^2(4m^2 + 2m)$, so $\phi_2(m) = 4m^2 + 2m$, and $\phi_3'(m) = -(6m^2 + 2m - 2)$.

• $m=1$: $\phi_2=6$, $\phi_3'(1) = -(6+2-2) = -6$, so $c = -6/(-6)\cdot(-1)$. Carefully, $c = -\phi_2/\phi_3'= -6/(-6) = 1.$ Asymptote: $y = x + 1$.
• $m=-1$: $\phi_2 = 4-2 = 2$, $\phi_3'(-1) = -(6-2-2) = -2$, so $c = -2/(-2) = 1.$ Asymptote: $y = -x + 1$.
• $m=-\tfrac12$: $\phi_2 = 4\cdot\tfrac14 - 1 = 0$, $\phi_3'(-\tfrac12) = -(6\cdot\tfrac14 - 1 - 2) = -(1.5 - 3) = 1.5$, so $c = -0/1.5 = 0.$ Asymptote: $y = -\tfrac12 x$, i.e. $2y + x = 0$.

The three asymptotes are:

$$\boxed{\,y = x + 1,\qquad y = -x + 1,\qquad x + 2y = 0\,}$$

(b) Radius of curvature at the origin (Newton's method)

Given $y - x = x^2 + 2xy + y^2$. The curve passes through the origin. Near $O$ the lowest-degree terms give the tangent: $y - x = 0$, so $y = x$ is the tangent at $O$ — it is not along a coordinate axis. We use the general Newton formula $\rho = \lim\limits_{(x,y)\to(0,0)} \dfrac{(x^2+y^2)^{3/2} \cdot \tfrac{1}{2}\ \text{form}}{\cdots}$ — more directly use $p=\lim \dfrac{x^2}{2y}$-type only when the tangent is an axis. Here, since the tangent is $y=x$, rotate so the tangent becomes an axis, or compute $\rho$ from derivatives.

Differentiate implicitly. At $O$, $y_1 = \dfrac{dy}{dx}=1$ (slope of tangent $y=x$). Differentiating $y - x = x^2 + 2xy + y^2$:

$$y_1 - 1 = 2x + 2y + 2xy_1 + 2yy_1.$$

At $O\,(0,0)$: $y_1 - 1 = 0 \Rightarrow y_1 = 1.$ Differentiate again:

$$y_2 = 2 + 2y_1 + 2y_1 + 2xy_2 + 2y_1^2 + 2yy_2.$$

At $O$ with $y_1=1$: $y_2 = 2 + 2 + 2 + 0 + 2 + 0 = 8.$

Radius of curvature:

$$\rho = \frac{(1+y_1^2)^{3/2}}{|y_2|} = \frac{(1+1)^{3/2}}{8} = \frac{2\sqrt2}{8} = \frac{\sqrt2}{4}.$$ $$\boxed{\rho = \dfrac{\sqrt2}{4}\ \text{units}}$$

(a) Euler's theorem for $u = \sin^{-1}\!\left(\dfrac{x^2+y^2}{x+y}\right)$

Let $z = \sin u = \dfrac{x^2+y^2}{x+y}$. Then $z$ is a homogeneous function: replacing $(x,y)\to(tx,ty)$ gives $z(tx,ty) = \dfrac{t^2(x^2+y^2)}{t(x+y)} = t\,z(x,y)$, so $z$ is homogeneous of degree $n = 1$.

By Euler's theorem, $x z_x + y z_y = n z = z$. Since $z = \sin u$, $z_x = \cos u\, u_x$ and $z_y = \cos u\, u_y$:

$$x\cos u\,u_x + y\cos u\,u_y = \sin u.$$

Dividing by $\cos u$:

$$\boxed{\,x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = \frac{\sin u}{\cos u} = \tan u\,}$$

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

Stationary points.

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

So $x^2 = ay$ and $y^2 = ax$. From these, $x^4 = a^2 y^2 = a^2(ax) = a^3 x \Rightarrow x(x^3 - a^3)=0$, giving $x=0$ or $x=a$. Correspondingly the stationary points are $(0,0)$ and $(a,a)$.

Second-order test. $r = f_{xx} = 6x$, $t = f_{yy} = 6y$, $s = f_{xy} = -3a$.

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

• At $(0,0)$: $rt - s^2 = 0 - 9a^2 = -9a^2 < 0$ → saddle point (no extremum).
• At $(a,a)$: $rt - s^2 = 36a^2 - 9a^2 = 27a^2 > 0$ and $r = 6a > 0$ (since $a>0$) → minimum.

Minimum value:

$$f(a,a) = a^3 + a^3 - 3a\cdot a\cdot a = 2a^3 - 3a^3 = -a^3.$$ $$\boxed{\text{Minimum value } = -a^3 \text{ at } (a,a);\ (0,0)\text{ is a saddle point.}}$$

(a) Area of one loop of $r = a\cos 2\theta$

For one loop, $r$ goes from $0$ to $0$ as $\cos 2\theta$ goes from $0$ to $0$; this happens for $2\theta \in [-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$, i.e. $\theta \in [-\tfrac{\pi}{4}, \tfrac{\pi}{4}]$.

$$A = \frac12 \int_{-\pi/4}^{\pi/4} r^2\,d\theta = \frac{a^2}{2}\int_{-\pi/4}^{\pi/4}\cos^2 2\theta\,d\theta = a^2\int_{0}^{\pi/4}\cos^2 2\theta\,d\theta.$$

Using $\cos^2 2\theta = \tfrac12(1 + \cos 4\theta)$:

$$A = a^2\cdot\frac12\left[\theta + \frac{\sin 4\theta}{4}\right]_0^{\pi/4} = \frac{a^2}{2}\left(\frac{\pi}{4} + 0\right) = \frac{\pi a^2}{8}.$$ $$\boxed{A = \dfrac{\pi a^2}{8}}$$

(b) Volume generated by revolving $r = a(1+\cos\theta)$ about the initial line

Volume of revolution of a polar curve about the initial line:

$$V = \int \frac{2}{3}\pi r^3 \sin\theta\,d\theta,\qquad \theta: 0 \to \pi.$$

With $r = a(1+\cos\theta)$:

$$V = \frac{2\pi a^3}{3}\int_0^\pi (1+\cos\theta)^3 \sin\theta\,d\theta.$$

Let $u = 1 + \cos\theta$, $du = -\sin\theta\,d\theta$. When $\theta=0$, $u=2$; when $\theta=\pi$, $u=0$.

$$V = \frac{2\pi a^3}{3}\int_{2}^{0} u^3(-du) = \frac{2\pi a^3}{3}\int_0^2 u^3\,du = \frac{2\pi a^3}{3}\cdot\frac{u^4}{4}\Big|_0^2 = \frac{2\pi a^3}{3}\cdot\frac{16}{4}.$$ $$\boxed{V = \frac{2\pi a^3}{3}\cdot 4 = \frac{8\pi a^3}{3}}$$

(a) Conic $3x^2 + 2xy + 3y^2 - 16y + 23 = 0$

Removing the $xy$-term. For $Ax^2 + Bxy + Cy^2 + \dots$, rotate by angle $\theta$ with $\cot 2\theta = \dfrac{A-C}{B} = \dfrac{3-3}{2} = 0 \Rightarrow 2\theta = \tfrac{\pi}{2} \Rightarrow \theta = 45^\circ.$

The new quadratic coefficients are the eigenvalues of $\begin{pmatrix}3 & 1\\ 1 & 3\end{pmatrix}$: $\lambda = 3\pm 1 = 4,\ 2.$ Substituting $x = \tfrac{X-Y}{\sqrt2},\ y = \tfrac{X+Y}{\sqrt2}$ gives

$$4X^2 + 2Y^2 - 16\cdot\frac{X+Y}{\sqrt2} + 23 = 0,$$ $$4X^2 + 2Y^2 - 8\sqrt2\,X - 8\sqrt2\,Y + 23 = 0.$$

Complete the square:

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

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

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

This is an ellipse with $a^2 = \tfrac12$ (major, along $Y'$) and $b^2 = \tfrac14$. Eccentricity:

$$e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1/4}{1/2}} = \sqrt{1 - \tfrac12} = \frac{1}{\sqrt2}.$$ $$\boxed{\text{Ellipse},\quad e = \tfrac{1}{\sqrt2}}$$

(b) Conic with focus at pole, $e=\tfrac12$, directrix $r\cos\theta = -4$

The directrix $r\cos\theta = -4$ means $x = -4$, i.e. it lies to the left of the pole at distance $d = 4$. The polar equation of a conic with focus at the pole is

$$\frac{l}{r} = 1 - e\cos\theta\quad\text{(directrix at } x=-d),\qquad l = ed = \tfrac12\cdot 4 = 2.$$

Hence

$$\boxed{\ \frac{2}{r} = 1 - \frac12\cos\theta\quad\Longleftrightarrow\quad r = \frac{2}{1 - \tfrac12\cos\theta} = \frac{4}{2 - \cos\theta}\ }$$

Since $e = \tfrac12 < 1$, the conic is an ellipse.

$y = \sin(m\sin^{-1}x)$

Setting up. $y_1 = \cos(m\sin^{-1}x)\cdot\dfrac{m}{\sqrt{1-x^2}}$, so $\sqrt{1-x^2}\,y_1 = m\cos(m\sin^{-1}x)$. Squaring:

$$(1-x^2)y_1^2 = m^2\cos^2(m\sin^{-1}x) = m^2(1 - y^2).$$

Differentiate w.r.t. $x$:

$$(1-x^2)\,2y_1 y_2 - 2x\,y_1^2 = -2m^2 y\,y_1.$$

Dividing by $2y_1$:

$$(1-x^2)y_2 - x y_1 + m^2 y = 0. \qquad (\ast)$$

Apply Leibnitz's theorem to differentiate $(\ast)$ $n$ times. For the product $(1-x^2)y_2$:

$$(1-x^2)y_{n+2} + n(-2x)y_{n+1} + \tfrac{n(n-1)}{2}(-2)y_n;$$

for $-xy_1$: $-x y_{n+1} - n\,y_n$; for $m^2 y$: $m^2 y_n$. Adding:

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

Values at $x=0$. Put $x=0$ in the recurrence: $(y_{n+2})_0 = (n^2 - m^2)(y_n)_0.$ From $y(0)=0$ and $y_1(0)=m$:

• $(y_0)_0 = 0$, and from $(\ast)$ at $0$: $(y_2)_0 = -m^2 (y_0)_0 = 0$. All even orders $= 0$.
• $(y_1)_0 = m$, $(y_3)_0 = (1^2 - m^2)m = m(1-m^2)$, $(y_5)_0 = (3^2 - m^2)(y_3)_0 = (9-m^2)(1-m^2)m,\dots$

Thus for $n$ even, $(y_n)_0 = 0$; for $n$ odd,

$$\boxed{(y_n)_0 = m\big[(n-2)^2 - m^2\big]\big[(n-4)^2 - m^2\big]\cdots(1^2 - m^2)\ \text{(odd }n),\quad 0\ \text{(even }n).}$$

$\displaystyle \lim_{x\to0}\left(\frac{1}{x^2} - \frac{\cot x}{x}\right)$

Write with common denominator:

$$\frac{1}{x^2} - \frac{\cot x}{x} = \frac{1}{x^2} - \frac{\cos x}{x\sin x} = \frac{\sin x - x\cos x}{x^2\sin x}.$$

Series expansion near $0$:

$$\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots,\qquad x\cos x = x - \frac{x^3}{2} + \frac{x^5}{24}-\cdots.$$

Numerator:

$$\sin x - x\cos x = \left(-\frac{x^3}{6} + \frac{x^3}{2}\right) + O(x^5) = \frac{x^3}{3} + O(x^5).$$

Denominator: $x^2\sin x = x^3 + O(x^5).$ Hence

$$\lim_{x\to0}\frac{\tfrac{x^3}{3} + O(x^5)}{x^3 + O(x^5)} = \frac{1}{3}.$$ $$\boxed{\displaystyle \lim_{x\to0}\left(\frac{1}{x^2} - \frac{\cot x}{x}\right) = \frac{1}{3}}$$

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

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

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

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

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

Evaluate $\displaystyle\int_0^{\pi/2}\sin^6 x\,dx$

Apply repeatedly with $I_0 = \int_0^{\pi/2}dx = \tfrac{\pi}{2}$:

$$I_6 = \frac{5}{6}I_4 = \frac{5}{6}\cdot\frac{3}{4}I_2 = \frac{5}{6}\cdot\frac{3}{4}\cdot\frac{1}{2}I_0 = \frac{5\cdot3\cdot1}{6\cdot4\cdot2}\cdot\frac{\pi}{2}.$$ $$I_6 = \frac{15}{48}\cdot\frac{\pi}{2} = \frac{5}{16}\cdot\frac{\pi}{2} = \frac{5\pi}{32}.$$ $$\boxed{\int_0^{\pi/2}\sin^6 x\,dx = \frac{5\pi}{32}}$$

$\displaystyle I = \int_0^1 \frac{dx}{\sqrt{1-x^4}}$

Convergence. The integrand is singular only at $x=1$. Near $x=1$, $1 - x^4 = (1-x)(1+x)(1+x^2) \approx 4(1-x)$, so the integrand behaves like $\dfrac{1}{2\sqrt{1-x}}$, i.e. like $(1-x)^{-1/2}$. Since the exponent $\tfrac12 < 1$, by the comparison ($p$-test for improper integrals) the integral converges.

Reduction to Beta/Gamma. Substitute $x^4 = t$, so $x = t^{1/4}$, $dx = \tfrac14 t^{-3/4}\,dt$; limits $t:0\to1$:

$$I = \int_0^1 \frac{\tfrac14 t^{-3/4}}{\sqrt{1-t}}\,dt = \frac14\int_0^1 t^{-3/4}(1-t)^{-1/2}\,dt.$$

This is a Beta integral $B(p,q) = \int_0^1 t^{p-1}(1-t)^{q-1}dt$ with $p-1 = -\tfrac34 \Rightarrow p = \tfrac14$, and $q-1 = -\tfrac12 \Rightarrow q = \tfrac12$:

$$I = \frac14\,B\!\left(\tfrac14,\tfrac12\right) = \frac14\cdot\frac{\Gamma(\tfrac14)\,\Gamma(\tfrac12)}{\Gamma(\tfrac34)}.$$

Since $\Gamma(\tfrac12) = \sqrt\pi$:

$$\boxed{\,I = \frac{\sqrt\pi}{4}\cdot\frac{\Gamma(\tfrac14)}{\Gamma(\tfrac34)}\,}$$

which is finite, confirming convergence.

$(x^2 - ay)\,dx + (y^2 - ax)\,dy = 0$

Test for exactness. $M = x^2 - ay$, $N = y^2 - ax$.

$$\frac{\partial M}{\partial y} = -a,\qquad \frac{\partial N}{\partial x} = -a.$$

Since $\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x}$, the equation is exact.

Solution. The solution is $F(x,y) = c$ where

$$F = \int M\,dx + \int(\text{terms of }N\text{ free of }x)\,dy.$$ $$\int M\,dx = \int(x^2 - ay)\,dx = \frac{x^3}{3} - axy.$$

The term of $N = y^2 - ax$ not containing $x$ is $y^2$, giving $\int y^2\,dy = \dfrac{y^3}{3}$.

Therefore the general solution is

$$\boxed{\,\frac{x^3}{3} + \frac{y^3}{3} - axy = c\quad\text{i.e.}\quad x^3 + y^3 - 3axy = C.\,}$$

Bernoulli equation $\dfrac{dy}{dx} + \dfrac{y}{x} = y^2 x$

This is Bernoulli with $n=2$. Divide by $y^2$:

$$y^{-2}\frac{dy}{dx} + \frac{1}{x}y^{-1} = x.$$

Substitute $v = y^{-1}$, so $\dfrac{dv}{dx} = -y^{-2}\dfrac{dy}{dx}$, i.e. $y^{-2}\dfrac{dy}{dx} = -\dfrac{dv}{dx}$:

$$-\frac{dv}{dx} + \frac{v}{x} = x \quad\Longrightarrow\quad \frac{dv}{dx} - \frac{v}{x} = -x.$$

This is linear in $v$. Integrating factor: $\mu = e^{\int -\frac{1}{x}dx} = e^{-\ln x} = \dfrac{1}{x}.$

$$\frac{d}{dx}\!\left(\frac{v}{x}\right) = -x\cdot\frac{1}{x} = -1.$$

Integrating: $\dfrac{v}{x} = -x + c$, so $v = -x^2 + cx.$

Returning to $y$ via $v = 1/y$:

$$\boxed{\,\frac{1}{y} = cx - x^2\quad\text{i.e.}\quad y = \frac{1}{cx - x^2}.\,}$$

Surface $F(x,y,z) = x^2 + 2y^2 + 3z^2 - 12 = 0$ at $P(1,2,-1)$

Gradient (normal direction).

$$F_x = 2x,\quad F_y = 4y,\quad F_z = 6z.$$

At $P(1,2,-1)$: $F_x = 2,\ F_y = 8,\ F_z = -6.$ Normal vector $\vec n = (2, 8, -6)\parallel (1,4,-3).$

Tangent plane: $F_x(x-1) + F_y(y-2) + F_z(z+1)=0$:

$$2(x-1) + 8(y-2) - 6(z+1) = 0 \Rightarrow 2x + 8y - 6z - 24 = 0,$$ $$\boxed{\,x + 4y - 3z = 12.\,}$$

(Check: $1 + 8 + 3 = 12.$ ✓)

Normal line:

$$\boxed{\,\frac{x-1}{1} = \frac{y-2}{4} = \frac{z+1}{-3}.\,}$$

Radius of curvature of $x^3 + y^3 = 3axy$ at $\left(\tfrac{3a}{2}, \tfrac{3a}{2}\right)$

First derivative (implicit). Differentiate $x^3 + y^3 = 3axy$:

$$3x^2 + 3y^2 y_1 = 3a(y + x y_1) \Rightarrow y_1 = \frac{ay - x^2}{y^2 - ax}.$$

At $\left(\tfrac{3a}{2},\tfrac{3a}{2}\right)$: numerator $= a\cdot\tfrac{3a}{2} - \tfrac{9a^2}{4} = \tfrac{6a^2 - 9a^2}{4} = -\tfrac{3a^2}{4}$; denominator $= \tfrac{9a^2}{4} - a\cdot\tfrac{3a}{2} = \tfrac{9a^2 - 6a^2}{4} = \tfrac{3a^2}{4}.$ So $y_1 = -1.$

Second derivative. Differentiate $3x^2 + 3y^2 y_1 = 3a(y + xy_1)$ again and divide by 3:

$$2x + 2y y_1^2 + y^2 y_2 = a(y_1 + y_1 + x y_2) = a(2y_1 + x y_2).$$

Solve for $y_2$:

$$y_2(y^2 - ax) = 2a y_1 - 2x - 2y y_1^2.$$

At the point, with $x=y=\tfrac{3a}{2},\ y_1=-1$: $y^2 - ax = \tfrac{3a^2}{4}.$

$$\text{RHS} = 2a(-1) - 2\cdot\tfrac{3a}{2} - 2\cdot\tfrac{3a}{2}\cdot1 = -2a - 3a - 3a = -8a.$$ $$y_2 = \frac{-8a}{\tfrac{3a^2}{4}} = \frac{-32}{3a}.$$

Radius of curvature.

$$\rho = \frac{(1+y_1^2)^{3/2}}{|y_2|} = \frac{(1+1)^{3/2}}{32/(3a)} = \frac{2\sqrt2\cdot 3a}{32} = \frac{6\sqrt2\,a}{32} = \frac{3\sqrt2\,a}{16}.$$ $$\boxed{\rho = \frac{3\sqrt2}{16}\,a = \frac{3a}{8\sqrt2}.}$$