BE Civil Engineering (IOE, TU) Engineering Mathematics III (IOE, SH 501) Question Paper 2077 Nepal

Statement of Green's theorem. If $C$ is a positively oriented, piecewise-smooth, simple closed curve in the plane enclosing a region $R$, and $M(x,y)$, $N(x,y)$ together with their first partial derivatives are continuous on $R$, then

$$\oint_C (M\,dx + N\,dy) = \iint_R \left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right)\,dA.$$

Here $M = 3x^2 - 8y^2$ and $N = 4y - 6xy$.

Right-hand side (double integral).

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

The region $R$ is the triangle with vertices $(0,0)$, $(1,0)$, $(0,1)$. For fixed $x \in [0,1]$, $y$ runs from $0$ to $1-x$:

$$\iint_R 10y\,dA = \int_0^1 \int_0^{1-x} 10y\,dy\,dx = \int_0^1 10\cdot\frac{(1-x)^2}{2}\,dx = 5\int_0^1 (1-x)^2\,dx.$$

With $\int_0^1 (1-x)^2\,dx = \tfrac13$, RHS $= 5 \cdot \tfrac13 = \dfrac{5}{3}$.

Left-hand side (line integral). $C$ has three pieces, counter-clockwise.

Segment $C_1$: $(0,0)\to(1,0)$ along $y=0$, $dy=0$, $x: 0\to1$.

$$\int_{C_1} (3x^2 - 0)\,dx = \int_0^1 3x^2\,dx = 1.$$

Segment $C_2$: $(1,0)\to(0,1)$ along $x+y=1$. Parametrize $x = 1-t$, $y = t$, $t:0\to1$, so $dx=-dt$, $dy=dt$.

• $M = 3(1-t)^2 - 8t^2$, $N = 4t - 6(1-t)t$.

$$\int_{C_2} = \int_0^1 \Big[(3(1-t)^2 - 8t^2)(-1) + (4t - 6(1-t)t)(1)\Big]dt.$$

Expand: $3(1-t)^2 = 3 - 6t + 3t^2$, so $M = 3 - 6t + 3t^2 - 8t^2 = 3 - 6t - 5t^2$; $-M = -3 + 6t + 5t^2$. $N = 4t - 6t + 6t^2 = -2t + 6t^2$. Integrand $= (-3 + 6t + 5t^2) + (-2t + 6t^2) = -3 + 4t + 11t^2$.

$$\int_0^1 (-3 + 4t + 11t^2)\,dt = -3 + 2 + \tfrac{11}{3} = -1 + \tfrac{11}{3} = \tfrac{8}{3}.$$

Segment $C_3$: $(0,1)\to(0,0)$ along $x=0$, $dx=0$, $y:1\to0$.

$$\int_{C_3} (4y - 0)\,dy = \int_1^0 4y\,dy = \big[2y^2\big]_1^0 = -2.$$

Total line integral $= 1 + \tfrac{8}{3} - 2 = -1 + \tfrac{8}{3} = \dfrac{5}{3}.$

Since LHS $= \dfrac{5}{3} =$ RHS, Green's theorem is verified.

Statement of the divergence theorem. If $V$ is a solid region bounded by a closed, piecewise-smooth, outward-oriented surface $S$, and $\mathbf{F}$ is a vector field with continuous first partial derivatives on $V$, then

$$\iint_S \mathbf{F}\cdot\mathbf{n}\,dS = \iiint_V (\nabla\cdot\mathbf{F})\,dV.$$

Compute the divergence.

$$\nabla\cdot\mathbf{F} = \frac{\partial}{\partial x}(2x) + \frac{\partial}{\partial y}(3y) + \frac{\partial}{\partial z}(z^2) = 2 + 3 + 2z = 5 + 2z.$$

Evaluate the volume integral over the cube $0\le x,y,z\le 2$:

$$\iiint_V (5 + 2z)\,dV = \int_0^2\!\int_0^2\!\int_0^2 (5 + 2z)\,dz\,dy\,dx.$$

The integrand does not depend on $x$ or $y$, so the $x$ and $y$ integrations each give a factor of $2$ (area $= 2\times2 = 4$):

$$= 4\int_0^2 (5 + 2z)\,dz = 4\big[5z + z^2\big]_0^2 = 4\,(10 + 4) = 4\cdot 14 = 56.$$

Final answer: $\displaystyle \iint_S \mathbf{F}\cdot\mathbf{n}\,dS = \mathbf{56}.$

Check by direct face integration (optional). Outflux contributions:

• $x=2$ face: $\int F_x\,dA = 2(2)\cdot 4 = 16$; $x=0$ face: $F_x=0 \Rightarrow 0$.
• $y=2$ face: $3(2)\cdot 4 = 24$; $y=0$: $0$.
• $z=2$ face: $z^2 = 4 \Rightarrow 4\cdot 4 = 16$; $z=0$: $0$. Sum $= 16 + 24 + 16 = 56.$ Confirms the result.

Apply the Laplace transform. Let $Y(s) = \mathcal{L}\{y(t)\}$. Using

$$\mathcal{L}\{y'\} = sY - y(0), \qquad \mathcal{L}\{y''\} = s^2 Y - s\,y(0) - y'(0),$$

and $\mathcal{L}\{e^{-2t}\} = \dfrac{1}{s+2}$:

$$\big[s^2 Y - s(1) - (-1)\big] + 4\big[sY - 1\big] + 3Y = \frac{6}{s+2}.$$

Collect terms:

$$(s^2 + 4s + 3)Y - s + 1 - 4 = \frac{6}{s+2},$$ $$(s^2 + 4s + 3)Y = \frac{6}{s+2} + s + 3.$$

Note $s^2 + 4s + 3 = (s+1)(s+3)$. Thus

$$Y = \frac{6}{(s+2)(s+1)(s+3)} + \frac{s+3}{(s+1)(s+3)} = \frac{6}{(s+1)(s+2)(s+3)} + \frac{1}{s+1}.$$

Partial fractions for the first term. Write

$$\frac{6}{(s+1)(s+2)(s+3)} = \frac{A}{s+1} + \frac{B}{s+2} + \frac{C}{s+3}.$$

• $A$: set $s=-1$: $A = \dfrac{6}{(1)(2)} = 3.$
• $B$: set $s=-2$: $B = \dfrac{6}{(-1)(1)} = -6.$
• $C$: set $s=-3$: $C = \dfrac{6}{(-2)(-1)} = 3.$ So

$$Y = \frac{3}{s+1} - \frac{6}{s+2} + \frac{3}{s+3} + \frac{1}{s+1} = \frac{4}{s+1} - \frac{6}{s+2} + \frac{3}{s+3}.$$

Inverse transform, using $\mathcal{L}^{-1}\{1/(s+a)\} = e^{-at}$:

$$\boxed{\,y(t) = 4e^{-t} - 6e^{-2t} + 3e^{-3t}\,}.$$

Check initial conditions. $y(0) = 4 - 6 + 3 = 1.$ ✓ $y'(t) = -4e^{-t} + 12e^{-2t} - 9e^{-3t}$, so $y'(0) = -4 + 12 - 9 = -1.$ ✓

Cauchy's residue theorem. If $f(z)$ is analytic inside and on a simple closed contour $C$ except at a finite number of isolated singular points $z_1, z_2, \dots, z_n$ inside $C$, then

$$\oint_C f(z)\,dz = 2\pi i \sum_{k=1}^{n} \operatorname{Res}_{z=z_k} f(z).$$

Identify singularities. $f(z) = \dfrac{z^2 + 1}{z^2 - 2z} = \dfrac{z^2 + 1}{z(z-2)}$. Simple poles at $z = 0$ and $z = 2$. Both satisfy $|z| < 3$, so both lie inside $C$.

Residue at $z = 0$ (simple pole):

$$\operatorname{Res}_{z=0} f = \lim_{z\to 0} z\cdot \frac{z^2+1}{z(z-2)} = \frac{0+1}{0-2} = -\frac{1}{2}.$$

Residue at $z = 2$ (simple pole):

$$\operatorname{Res}_{z=2} f = \lim_{z\to 2} (z-2)\cdot \frac{z^2+1}{z(z-2)} = \frac{4+1}{2} = \frac{5}{2}.$$

Apply the theorem.

$$\oint_C f(z)\,dz = 2\pi i \left(-\frac{1}{2} + \frac{5}{2}\right) = 2\pi i \cdot 2 = \mathbf{4\pi i}.$$

Remark. Since $\deg(\text{num}) = \deg(\text{den})$, one could also write $f(z) = 1 + \frac{2z+1}{z(z-2)}$; the constant $1$ integrates to $0$ around the closed contour, and the remaining proper fraction gives the same residue sum.

Separation of variables. Seek $u(x,t) = X(x)T(t)$. Substituting into $u_{tt} = c^2 u_{xx}$ and dividing by $c^2 XT$:

$$\frac{T''}{c^2 T} = \frac{X''}{X} = -\lambda \;(\text{constant}).$$

This gives $X'' + \lambda X = 0$ and $T'' + \lambda c^2 T = 0$.

Boundary conditions. $u(0,t)=u(\pi,t)=0 \Rightarrow X(0)=X(\pi)=0$. Non-trivial solutions require $\lambda = n^2$ ($n = 1,2,3,\dots$) with

$$X_n(x) = \sin(nx).$$

Time part. With $\lambda = n^2$, $T'' + n^2 c^2 T = 0 \Rightarrow T_n(t) = A_n\cos(nct) + B_n\sin(nct)$.

Apply the initial velocity condition. $u_t(x,0) = 0$ for all $x$ forces $B_n = 0$ for every $n$. Hence the general solution is

$$u(x,t) = \sum_{n=1}^{\infty} A_n \cos(nct)\sin(nx).$$

Apply the initial displacement. At $t=0$:

$$u(x,0) = \sum_{n=1}^{\infty} A_n \sin(nx) = 3\sin x - \sin 3x.$$

Matching coefficients term by term (the right side is already a finite sine series):

$$A_1 = 3, \qquad A_3 = -1, \qquad A_n = 0 \text{ otherwise}.$$

Final solution.

$$\boxed{\,u(x,t) = 3\cos(ct)\sin x - \cos(3ct)\sin 3x\,}.$$

Verification. Each term satisfies the wave equation: for $u_n = \cos(nct)\sin(nx)$, $u_{tt} = -n^2c^2 u_n$ and $c^2 u_{xx} = -n^2 c^2 u_n$, so $u_{tt} = c^2 u_{xx}$. At $t=0$: $u = 3\sin x - \sin 3x$ ✓ and $u_t = -3c\sin(ct)\sin x + 3c\sin(3ct)\sin 3x \big|_{t=0} = 0$ ✓; boundaries vanish since $\sin(0)=\sin(n\pi)=0$ ✓.

Identify the region. The inner limits give $x \le y \le 1$ and the outer give $0 \le x \le 1$. So the region is

$$R = \{(x,y): 0\le x\le 1,\; x\le y\le 1\},$$

the triangle with vertices $(0,0)$, $(0,1)$, $(1,1)$.

Reason for changing order. $\int e^{y^2}\,dy$ has no elementary antiderivative, so integrate in $x$ first.

Re-describe $R$ with $y$ outer. For fixed $y\in[0,1]$, $x$ runs from $0$ to $y$ (since $x\le y$):

$$R = \{(x,y): 0\le y\le 1,\; 0\le x\le y\}.$$

Thus

$$I = \int_0^1 \int_0^{y} e^{y^2}\,dx\,dy.$$

Inner integral (over $x$, $e^{y^2}$ constant in $x$):

$$\int_0^{y} e^{y^2}\,dx = e^{y^2}\cdot y.$$

Outer integral. Substitute $w = y^2$, $dw = 2y\,dy$:

$$I = \int_0^1 y\,e^{y^2}\,dy = \frac{1}{2}\int_0^1 e^{w}\,dw = \frac{1}{2}\big[e^{w}\big]_0^1 = \frac{1}{2}(e - 1).$$

Final answer: $\displaystyle I = \frac{e-1}{2} \approx \mathbf{0.8591}.$

(a) Gradient.

$$\nabla\phi = \left(\frac{\partial\phi}{\partial x},\frac{\partial\phi}{\partial y},\frac{\partial\phi}{\partial z}\right) = (2xy + z^2,\; x^2 + 2yz,\; y^2 + 2zx).$$

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

• $\phi_x = 2(1)(-1) + 2^2 = -2 + 4 = 2$
• $\phi_y = 1^2 + 2(-1)(2) = 1 - 4 = -3$
• $\phi_z = (-1)^2 + 2(2)(1) = 1 + 4 = 5$

$$\nabla\phi\big|_P = 2\mathbf{i} - 3\mathbf{j} + 5\mathbf{k}.$$

(b) Directional derivative. Unit vector along $\mathbf{a} = (2,-1,2)$: $|\mathbf{a}| = \sqrt{4+1+4} = 3$, so $\hat{\mathbf{a}} = \tfrac13(2,-1,2)$.

$$D_{\hat{\mathbf{a}}}\phi = \nabla\phi\cdot\hat{\mathbf{a}} = \frac{(2)(2) + (-3)(-1) + (5)(2)}{3} = \frac{4 + 3 + 10}{3} = \frac{17}{3}.$$ $$D_{\hat{\mathbf{a}}}\phi = \frac{17}{3} \approx \mathbf{5.667}.$$

(c) Maximum rate of change. This equals $|\nabla\phi|$ at $P$:

$$|\nabla\phi| = \sqrt{2^2 + (-3)^2 + 5^2} = \sqrt{4 + 9 + 25} = \sqrt{38} \approx \mathbf{6.164},$$

attained in the direction of $\nabla\phi = 2\mathbf{i} - 3\mathbf{j} + 5\mathbf{k}$.

Partial fractions. Since the numerator $s$ is odd, write

$$\frac{s}{(s^2+1)(s^2+4)} = \frac{As + B}{s^2+1} + \frac{Cs + D}{s^2+4}.$$

Multiply through:

$$s = (As+B)(s^2+4) + (Cs+D)(s^2+1).$$

Expand:

$$s = (A+C)s^3 + (B+D)s^2 + (4A + C)s + (4B + D).$$

Match coefficients:

• $s^3$: $A + C = 0$
• $s^2$: $B + D = 0$
• $s^1$: $4A + C = 1$
• $s^0$: $4B + D = 0$

From $A + C = 0 \Rightarrow C = -A$. Then $4A - A = 1 \Rightarrow 3A = 1 \Rightarrow A = \tfrac13$, $C = -\tfrac13$. From $B + D = 0$ and $4B + D = 0$: subtract to get $3B = 0 \Rightarrow B = 0$, $D = 0$.

So

$$\frac{s}{(s^2+1)(s^2+4)} = \frac{1}{3}\cdot\frac{s}{s^2+1} - \frac{1}{3}\cdot\frac{s}{s^2+4}.$$

Inverse transform, using $\mathcal{L}^{-1}\{s/(s^2+a^2)\} = \cos(at)$:

$$\mathcal{L}^{-1}\left\{\frac{s}{(s^2+1)(s^2+4)}\right\} = \frac{1}{3}\cos t - \frac{1}{3}\cos 2t = \frac{1}{3}(\cos t - \cos 2t).$$

Final answer: $\displaystyle f(t) = \frac{1}{3}\big(\cos t - \cos 2t\big).$

Consistency at $t=0$. $f(0) = \tfrac13(1 - 1) = 0$. By the initial value theorem, $\lim_{s\to\infty} sF(s) = \lim_{s\to\infty}\frac{s^2}{(s^2+1)(s^2+4)} = 0$, which matches $f(0)=0$. ✓

Definition. The Fourier cosine transform is

$$F_c(\omega) = \sqrt{\frac{2}{\pi}}\int_0^{\infty} f(x)\cos(\omega x)\,dx.$$

Compute the transform. Since $f(x) = 1$ on $[0,a)$ and $0$ beyond:

$$F_c(\omega) = \sqrt{\frac{2}{\pi}}\int_0^{a} \cos(\omega x)\,dx = \sqrt{\frac{2}{\pi}}\left[\frac{\sin(\omega x)}{\omega}\right]_0^{a} = \sqrt{\frac{2}{\pi}}\,\frac{\sin(\omega a)}{\omega}.$$

So

$$F_c(\omega) = \sqrt{\frac{2}{\pi}}\,\frac{\sin(a\omega)}{\omega}, \qquad \omega > 0.$$

Inversion (cosine integral representation). The inverse cosine transform is

$$f(x) = \sqrt{\frac{2}{\pi}}\int_0^{\infty} F_c(\omega)\cos(\omega x)\,d\omega = \frac{2}{\pi}\int_0^{\infty}\frac{\sin(a\omega)}{\omega}\cos(\omega x)\,d\omega.$$

Thus

$$\frac{2}{\pi}\int_0^{\infty}\frac{\sin(a\omega)\cos(\omega x)}{\omega}\,d\omega = \begin{cases} 1, & 0 \le x < a \\ \tfrac12, & x = a \\ 0, & x > a, \end{cases}$$

where the value $\tfrac12$ at $x=a$ is the average of the left and right limits at the jump discontinuity (Dirichlet condition).

Check harmonicity. Compute second partials of $u = x^3 - 3xy^2 + 2y$:

$$u_x = 3x^2 - 3y^2, \quad u_{xx} = 6x, \qquad u_y = -6xy + 2, \quad u_{yy} = -6x.$$

Then $u_{xx} + u_{yy} = 6x - 6x = 0$, so $u$ satisfies Laplace's equation and is harmonic.

Find the conjugate via Cauchy–Riemann. For $f = u + iv$ analytic: $v_y = u_x$ and $v_x = -u_y$.

From $v_y = u_x = 3x^2 - 3y^2$, integrate with respect to $y$:

$$v = 3x^2 y - y^3 + g(x),$$

where $g(x)$ is an arbitrary function of $x$.

Differentiate w.r.t. $x$: $v_x = 6xy + g'(x)$. The second CR equation requires

$$v_x = -u_y = -(-6xy + 2) = 6xy - 2.$$

Matching: $6xy + g'(x) = 6xy - 2 \Rightarrow g'(x) = -2 \Rightarrow g(x) = -2x + C.$

Thus (taking $C = 0$)

$$v(x,y) = 3x^2 y - y^3 - 2x.$$

Form $f(z)$.

$$f = u + iv = (x^3 - 3xy^2 + 2y) + i(3x^2 y - y^3 - 2x).$$

Group the cubic terms: $x^3 - 3xy^2 + i(3x^2 y - y^3) = (x+iy)^3 = z^3$. Group the linear terms: $2y - 2ix = -2i(x + iy) = -2iz$ (since $-2i\cdot iy = 2y$ and $-2i\cdot x = -2ix$). Therefore

$$\boxed{\,f(z) = z^3 - 2iz + C\,}, \quad C \text{ a real constant (take } C=0).$$

Part 1 — Eliminate $a$ and $b$. Given $z = (x^2 + a)(y^2 + b)$.

Differentiate partially:

$$p = \frac{\partial z}{\partial x} = 2x(y^2 + b), \qquad q = \frac{\partial z}{\partial y} = 2y(x^2 + a).$$

From these,

$$y^2 + b = \frac{p}{2x}, \qquad x^2 + a = \frac{q}{2y}.$$

Substitute into $z = (x^2+a)(y^2+b)$:

$$z = \frac{q}{2y}\cdot\frac{p}{2x} = \frac{pq}{4xy}.$$

Therefore the required PDE is

$$\boxed{\,4xyz = pq\,}, \qquad \text{i.e. } 4xy\,z = \frac{\partial z}{\partial x}\frac{\partial z}{\partial y}.$$

Part 2 — Classification. A second-order linear PDE $A u_{xx} + B u_{xy} + C u_{yy} + \dots = 0$ is classified by the discriminant $B^2 - 4AC$:

• $B^2 - 4AC < 0$ → elliptic,
• $B^2 - 4AC = 0$ → parabolic,
• $B^2 - 4AC > 0$ → hyperbolic.

For $u_{xx} + 4u_{xy} + 4u_{yy} = 0$: $A = 1$, $B = 4$, $C = 4$.

$$B^2 - 4AC = 16 - 4(1)(4) = 16 - 16 = 0.$$

Since the discriminant is $0$, the PDE is parabolic (everywhere in the plane).