BE Civil Engineering (IOE, TU) Engineering Mathematics I (IOE, SH 401) Question Paper 2076 Nepal

Leibnitz's theorem. If $u$ and $v$ are functions of $x$ possessing derivatives up to order $n$, then the $n$-th derivative of their product is

$$(uv)_n = \sum_{k=0}^{n}\binom{n}{k}\,u_{n-k}\,v_{k},$$

that is,

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

Proof of the differential equation.

Given $y = e^{ax}\cos(bx)$.

First derivative (product rule):

$$y_1 = a e^{ax}\cos(bx) - b e^{ax}\sin(bx) = e^{ax}\big(a\cos bx - b\sin bx\big).$$

Note that $y_1 = a y - b\,e^{ax}\sin(bx)$, so

$$e^{ax}\sin(bx) = \frac{a y - y_1}{b}. \qquad (\ast)$$

Second derivative — differentiate $y_1 = e^{ax}(a\cos bx - b\sin bx)$:

$$y_2 = a e^{ax}(a\cos bx - b\sin bx) + e^{ax}(-ab\sin bx - b^2\cos bx).$$

Group terms:

$$y_2 = e^{ax}\big[(a^2 - b^2)\cos bx - 2ab\sin bx\big].$$

Now form $y_2 - 2a y_1 + (a^2+b^2)y$. Using $y_1 = e^{ax}(a\cos bx - b\sin bx)$ and $y = e^{ax}\cos bx$:

$$-2a y_1 = e^{ax}(-2a^2\cos bx + 2ab\sin bx),$$ $$(a^2+b^2)y = e^{ax}(a^2+b^2)\cos bx.$$

Add all three (drop the common factor $e^{ax}$):

• Coefficient of $\cos bx$: $(a^2 - b^2) - 2a^2 + (a^2 + b^2) = a^2 - b^2 - 2a^2 + a^2 + b^2 = 0.$
• Coefficient of $\sin bx$: $-2ab + 2ab = 0.$

Hence

$$\boxed{\,y_2 - 2a y_1 + (a^2+b^2)y = 0\,}$$

as required. $\blacksquare$

Euler's theorem. If $f(x,y)$ is a homogeneous function of degree $n$, then

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

Application. Here $u$ is not homogeneous, but $\sin u$ is. Let

$$z = \sin u = \frac{x^2 + y^2}{x + y}.$$

Check homogeneity: replace $x\to tx,\ y\to ty$:

$$z(tx,ty) = \frac{t^2(x^2+y^2)}{t(x+y)} = t\,\frac{x^2+y^2}{x+y} = t^1 z.$$

So $z$ is homogeneous of degree $n = 1$. By Euler's theorem,

$$x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y} = 1\cdot z = \sin u. \qquad (1)$$

Now $z = \sin u$, so $\dfrac{\partial z}{\partial x} = \cos u\,\dfrac{\partial u}{\partial x}$ and $\dfrac{\partial z}{\partial y} = \cos u\,\dfrac{\partial u}{\partial y}$. Substituting in (1):

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

Divide by $\cos u$:

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

Hence

$$\boxed{\,x\,u_x + y\,u_y = \tan u\,}$$

as required. $\blacksquare$

Reduction formula. Write $I_n = \displaystyle\int_0^{\pi/2}\sin^n x\,dx = \int_0^{\pi/2}\sin^{n-1}x\,\sin x\,dx.$

Integrate by parts with $u = \sin^{n-1}x$ and $dv = \sin x\,dx$, so $du = (n-1)\sin^{n-2}x\cos x\,dx$ and $v = -\cos x$:

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

The boundary term is $0$ at both limits (at $\pi/2$, $\cos x = 0$; at $0$, $\sin x = 0$ for $n\ge 2$). Put $\cos^2 x = 1 - \sin^2 x$:

$$I_n = (n-1)\int_0^{\pi/2}\sin^{n-2}x\,dx - (n-1)\int_0^{\pi/2}\sin^{n}x\,dx = (n-1)I_{n-2} - (n-1)I_n.$$

Thus $I_n + (n-1)I_n = (n-1)I_{n-2}$, i.e. $n I_n = (n-1)I_{n-2}$, giving the reduction formula

$$\boxed{\,I_n = \frac{n-1}{n}\,I_{n-2}\,}\qquad (n\ge 2).$$

Evaluation of $I_6$. With $I_0 = \displaystyle\int_0^{\pi/2}dx = \frac{\pi}{2}$:

$$I_6 = \frac{5}{6}I_4 = \frac{5}{6}\cdot\frac{3}{4}I_2 = \frac{5}{6}\cdot\frac{3}{4}\cdot\frac{1}{2}I_0.$$

So

$$I_6 = \frac{5\cdot 3\cdot 1}{6\cdot 4\cdot 2}\cdot\frac{\pi}{2} = \frac{15}{48}\cdot\frac{\pi}{2} = \frac{5}{16}\cdot\frac{\pi}{2} = \frac{5\pi}{32}.$$

Final answer: $\displaystyle\int_0^{\pi/2}\sin^6 x\,dx = \mathbf{\dfrac{5\pi}{32}}\approx 0.4909.$

Step 1 — Remove the $xy$-term by rotation. For $Ax^2 + Bxy + Cy^2 + \cdots = 0$ with $A=3,\ B=2,\ C=3$, the rotation angle $\theta$ satisfies

$$\tan 2\theta = \frac{B}{A - C} = \frac{2}{3-3} = \frac{2}{0} = \infty \;\Rightarrow\; 2\theta = 90^\circ,\ \theta = 45^\circ.$$

With $x = X\cos45^\circ - Y\sin45^\circ = \tfrac{1}{\sqrt2}(X - Y)$ and $y = \tfrac{1}{\sqrt2}(X + Y)$.

Step 2 — New quadratic coefficients. The rotated quadratic part is $A'X^2 + C'Y^2$ where

$$A' = \frac{A+C}{2} + \frac{A-C}{2}\cos2\theta + \frac{B}{2}\sin2\theta = 3 + 0 + \frac{2}{2}(1) = 4,$$ $$C' = \frac{A+C}{2} - \frac{A-C}{2}\cos2\theta - \frac{B}{2}\sin2\theta = 3 - 0 - 1 = 2.$$

(Check: $A' + C' = A + C = 6$. $\checkmark$)

Step 3 — Transform the linear term. The term $-16y = -16\cdot\tfrac{1}{\sqrt2}(X+Y) = -\tfrac{16}{\sqrt2}(X+Y) = -8\sqrt2\,X - 8\sqrt2\,Y.$

So the equation becomes

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

Step 4 — Complete the square.

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

Step 5 — Standard form. Put $X' = X-\sqrt2,\ Y' = Y - 2\sqrt2$:

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

Since both coefficients are positive and the denominators differ, this is an ellipse. The semi-axes are

$$a = \sqrt{\tfrac12} = \frac{1}{\sqrt2}\approx 0.707 \ (\text{along } Y'),\qquad b = \sqrt{\tfrac14} = \frac12 = 0.5\ (\text{along } X').$$

Final answer: the conic is an ellipse with semi-major axis $\mathbf{a = 1/\sqrt2 \approx 0.707}$ and semi-minor axis $\mathbf{b = 1/2 = 0.5}$, after a $45^\circ$ rotation.

Step 1 — Identify the form. This is linear of the form $\dfrac{dy}{dx} + P(x)y = Q(x)$ with

$$P(x) = \frac{2}{x}, \qquad Q(x) = 6x^2.$$

Step 2 — Integrating factor.

$$\mu(x) = e^{\int P\,dx} = e^{\int \frac{2}{x}dx} = e^{2\ln x} = e^{\ln x^2} = x^2.$$

Step 3 — Multiply through by $\mu$.

$$x^2\frac{dy}{dx} + 2x\,y = 6x^4.$$

The left side is the exact derivative $\dfrac{d}{dx}\big(x^2 y\big)$:

$$\frac{d}{dx}\big(x^2 y\big) = 6x^4.$$

Step 4 — Integrate both sides.

$$x^2 y = \int 6x^4\,dx = \frac{6x^5}{5} + C.$$

Hence the general solution

$$\boxed{\,y = \frac{6x^3}{5} + \frac{C}{x^2}\,}.$$

Step 5 — Apply the initial condition $y(1) = 5$:

$$5 = \frac{6(1)^3}{5} + \frac{C}{1^2} = \frac{6}{5} + C \;\Rightarrow\; C = 5 - \frac{6}{5} = \frac{25-6}{5} = \frac{19}{5}.$$

Particular solution:

$$\boxed{\,y = \frac{6x^3}{5} + \frac{19}{5x^2}\,}.$$

Check at $x=1$: $y = \frac{6}{5} + \frac{19}{5} = \frac{25}{5} = 5.$ $\checkmark$

Setup. Volume constraint: $V = \pi r^2 h = 500 \Rightarrow h = \dfrac{500}{\pi r^2}.$

Total surface area (two circular ends + curved surface):

$$S = 2\pi r^2 + 2\pi r h.$$

Substitute $h$:

$$S(r) = 2\pi r^2 + 2\pi r\cdot\frac{500}{\pi r^2} = 2\pi r^2 + \frac{1000}{r}.$$

Minimise. Differentiate and set to zero:

$$\frac{dS}{dr} = 4\pi r - \frac{1000}{r^2} = 0 \;\Rightarrow\; 4\pi r = \frac{1000}{r^2} \;\Rightarrow\; r^3 = \frac{1000}{4\pi} = \frac{250}{\pi}.$$ $$r = \left(\frac{250}{\pi}\right)^{1/3} = (79.577)^{1/3} \approx 4.301\ \text{cm}.$$

Second-derivative test: $\dfrac{d^2S}{dr^2} = 4\pi + \dfrac{2000}{r^3} > 0$, so $S$ is minimised. $\checkmark$

Height.

$$h = \frac{500}{\pi r^2} = \frac{500}{\pi (4.301)^2} = \frac{500}{\pi\cdot 18.499} = \frac{500}{58.116} \approx 8.602\ \text{cm}.$$

Note $h = 2r$, the classic optimal-can result. $\checkmark$

Minimum surface area.

$$S = 2\pi r^2 + \frac{1000}{r} = 2\pi(18.499) + \frac{1000}{4.301} = 116.23 + 232.50 = 348.73\ \text{cm}^2.$$

Final answers: $r \approx \mathbf{4.30\ cm}$, $h \approx \mathbf{8.60\ cm}$, minimum surface area $\approx \mathbf{348.7\ cm^2}$.

Step 1 — Point on the curve. At $x = 2$:

$$y = (2)^3 - 3(2)^2 + 4 = 8 - 12 + 4 = 0.$$

So the point is $(2,\,0)$.

Step 2 — Slope of the tangent.

$$\frac{dy}{dx} = 3x^2 - 6x.$$

At $x = 2$:

$$m_t = 3(4) - 6(2) = 12 - 12 = 0.$$

The tangent has slope $0$ (horizontal line).

Step 3 — Tangent equation. Through $(2,0)$ with slope $0$:

$$y - 0 = 0\,(x - 2) \;\Rightarrow\; \boxed{\,y = 0\,}.$$

Step 4 — Normal equation. The normal is perpendicular to the tangent. Since the tangent is horizontal ($m_t = 0$), the normal is vertical:

$$\boxed{\,x = 2\,}.$$

Final answers: tangent $y = 0$ (the $x$-axis); normal $x = 2$.

Step 1 — Points of intersection. Set $x^2 = x + 2$:

$$x^2 - x - 2 = 0 \;\Rightarrow\; (x-2)(x+1) = 0 \;\Rightarrow\; x = -1,\ x = 2.$$

Step 2 — Which curve is on top? On $(-1,2)$, test $x = 0$: line gives $y = 2$, parabola gives $y = 0$. The line lies above the parabola.

Step 3 — Set up the area integral.

$$A = \int_{-1}^{2}\big[(x+2) - x^2\big]\,dx.$$

Step 4 — Integrate.

$$A = \left[\frac{x^2}{2} + 2x - \frac{x^3}{3}\right]_{-1}^{2}.$$

At $x = 2$: $\dfrac{4}{2} + 4 - \dfrac{8}{3} = 2 + 4 - 2.6667 = 3.3333 = \dfrac{10}{3}.$

At $x = -1$: $\dfrac{1}{2} - 2 - \dfrac{-1}{3} = 0.5 - 2 + 0.3333 = -1.1667 = -\dfrac{7}{6}.$

Step 5 — Subtract.

$$A = \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2}.$$

Final answer: $A = \mathbf{\dfrac{9}{2} = 4.5}$ square units.

Step 1 — Value of $z$ at the point.

$$f(1,2) = (1)^2 + 3(1)(2) - (2)^2 = 1 + 6 - 4 = 3.$$

So the point is $(1,\,2,\,3)$.

Step 2 — Partial derivatives.

$$f_x = \frac{\partial}{\partial x}(x^2 + 3xy - y^2) = 2x + 3y,$$ $$f_y = \frac{\partial}{\partial y}(x^2 + 3xy - y^2) = 3x - 2y.$$

At $(1,2)$:

$$f_x(1,2) = 2(1) + 3(2) = 2 + 6 = 8,$$ $$f_y(1,2) = 3(1) - 2(2) = 3 - 4 = -1.$$

Step 3 — Tangent-plane formula.

$$z - z_0 = f_x(x_0,y_0)(x - x_0) + f_y(x_0,y_0)(y - y_0).$$

Substitute $(x_0,y_0,z_0) = (1,2,3)$:

$$z - 3 = 8(x - 1) + (-1)(y - 2).$$ $$z - 3 = 8x - 8 - y + 2 = 8x - y - 6.$$ $$z = 8x - y - 3.$$

Final answer (tangent plane):

$$\boxed{\,8x - y - z - 3 = 0\,}\quad\text{equivalently}\quad z = 8x - y - 3.$$

Check at $(1,2)$: $z = 8 - 2 - 3 = 3.$ $\checkmark$

Step 1 — Recognise homogeneity. The right side is homogeneous of degree $0$ (numerator and denominator both degree 2). Substitute $y = vx$, so $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}.$

Step 2 — Rewrite the equation.

$$\frac{x^2 + y^2}{xy} = \frac{x^2 + v^2 x^2}{x\cdot vx} = \frac{x^2(1+v^2)}{vx^2} = \frac{1+v^2}{v}.$$

So

$$v + x\frac{dv}{dx} = \frac{1 + v^2}{v}.$$

Step 3 — Separate variables.

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

Step 4 — Integrate both sides.

$$\int v\,dv = \int \frac{dx}{x} \;\Rightarrow\; \frac{v^2}{2} = \ln|x| + C_1.$$

Step 5 — Back-substitute $v = \dfrac{y}{x}$:

$$\frac{1}{2}\cdot\frac{y^2}{x^2} = \ln|x| + C_1.$$

Multiply by $2$ and write $C = 2C_1$:

$$\boxed{\,\frac{y^2}{x^2} = 2\ln|x| + C\,}\quad\text{equivalently}\quad y^2 = x^2\big(2\ln|x| + C\big).$$

This is the required general solution.

Step 1 — Locate the centre. Let the centre be $(h, k)$. The centre is equidistant from the two given points, so it lies on the perpendicular bisector of the chord joining $(1,1)$ and $(3,1)$.

The chord is horizontal with midpoint $\left(\frac{1+3}{2}, \frac{1+1}{2}\right) = (2, 1)$, so its perpendicular bisector is the vertical line

$$h = 2. \qquad (1)$$

(Equivalently, set $(h-1)^2 + (k-1)^2 = (h-3)^2 + (k-1)^2 \Rightarrow -2h+1 = -6h+9 \Rightarrow 4h = 8 \Rightarrow h = 2.$)

Step 2 — Use the centre-line condition. The centre lies on $x + y = 2$:

$$h + k = 2. \qquad (2)$$

With $h = 2$, this gives $k = 0$. So the centre is $(2, 0)$.

Step 3 — Radius. Distance from centre $(2,0)$ to $(1,1)$:

$$r^2 = (2-1)^2 + (0-1)^2 = 1 + 1 = 2.$$

Check with the other point $(3,1)$: $r^2 = (2-3)^2 + (0-1)^2 = 1 + 1 = 2.$ $\checkmark$

Step 4 — Equation of the circle.

$$(x - 2)^2 + (y - 0)^2 = 2.$$

Expanding:

$$x^2 - 4x + 4 + y^2 = 2 \;\Rightarrow\; x^2 + y^2 - 4x + 2 = 0.$$

Final answer:

$$\boxed{\,(x-2)^2 + y^2 = 2\,}\quad\text{equivalently}\quad x^2 + y^2 - 4x + 2 = 0,$$

with centre $(2, 0)$ and radius $\sqrt{2}$.