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

Leibnitz's Theorem

If $u$ and $v$ are functions of $x$ that are differentiable $n$ times, then the $n$-th derivative of the product $uv$ is

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

where a subscript $k$ denotes the $k$-th derivative with respect to $x$.

Setting up the recurrence

Given $y = e^{m\,\sin^{-1}x}$.

Step 1 — first derivative.

$$y_1 = e^{m\sin^{-1}x}\cdot \frac{m}{\sqrt{1-x^2}} = \frac{m\,y}{\sqrt{1-x^2}}.$$

Thus $\sqrt{1-x^2}\,y_1 = m y$. Squaring,

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

Step 2 — differentiate again. Differentiating $(1-x^2)y_1^2 = m^2 y^2$ with respect to $x$:

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

Divide through by $2y_1$ (assuming $y_1\neq 0$):

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

Step 3 — apply Leibnitz's theorem. Differentiate $(\star)$ $n$ times. Treat each term as a product and use Leibnitz's theorem.

For $(1-x^2)y_2$, take $u = y_2,\; v = (1-x^2)$. Only $v, v_1=-2x, v_2=-2$ are non-zero:

$$\big[(1-x^2)y_2\big]_n = (1-x^2)y_{n+2} + n(-2x)y_{n+1} + \binom{n}{2}(-2)y_n.$$

For $x y_1$, take $u = y_1,\; v = x$ (so $v_1=1$):

$$(x y_1)_n = x\,y_{n+1} + n\,y_n.$$

The right side $m^2 y$ differentiates to $m^2 y_n$. Hence

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

Step 4 — collect terms.

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

Since $n(n-1)+n = n^2$,

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

This is the required relation.

Euler's Theorem

If $f(x,y)$ is a homogeneous function of degree $n$ (i.e. $f(tx,ty)=t^n f(x,y)$), then

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

Applying to the given function

Step 1 — identify the homogeneous part. Here $u$ itself is not homogeneous, but $\tan u$ is:

$$v := \tan u = \frac{x^3 + y^3}{x - y}.$$

Replace $x\to tx,\,y\to ty$:

$$v(tx,ty) = \frac{t^3(x^3+y^3)}{t(x-y)} = t^2\,\frac{x^3+y^3}{x-y} = t^2 v.$$

So $v$ is homogeneous of degree $n = 2$.

Step 2 — apply Euler's theorem to $v$.

$$x\frac{\partial v}{\partial x} + y\frac{\partial v}{\partial y} = 2v. \qquad (1)$$

Step 3 — convert back to $u$. Since $v = \tan u$,

$$\frac{\partial v}{\partial x} = \sec^2 u\,\frac{\partial u}{\partial x}, \qquad \frac{\partial v}{\partial y} = \sec^2 u\,\frac{\partial u}{\partial y}.$$

Substitute into (1):

$$\sec^2 u\left(x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y}\right) = 2\tan u.$$

Step 4 — solve.

$$x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = \frac{2\tan u}{\sec^2 u} = 2\,\frac{\sin u}{\cos u}\cdot\cos^2 u = 2\sin u\cos u = \sin 2u.$$

Hence

$$\boxed{x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = \sin 2u.}$$

Deriving the reduction formula

Let $I_n = \displaystyle\int_0^{\pi/2}\sin^n x\,dx$. Write $\sin^n x = \sin^{n-1}x\cdot\sin x$ and integrate by parts with

$$u = \sin^{n-1}x,\quad dv = \sin x\,dx \;\Rightarrow\; du = (n-1)\sin^{n-2}x\cos x\,dx,\quad v = -\cos x.$$

Then

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

The boundary term vanishes (at $x=\tfrac\pi2$, $\cos x = 0$; at $x=0$, $\sin x = 0$, valid for $n\ge 2$). Using $\cos^2 x = 1-\sin^2 x$:

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

Solve for $I_n$:

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

Evaluating $I_6$

Apply repeatedly down to $I_0$:

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

Now $I_0 = \displaystyle\int_0^{\pi/2}dx = \frac{\pi}{2}$. Therefore

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

Compute the product of fractions: $\dfrac{5}{6}\cdot\dfrac{3}{4}\cdot\dfrac{1}{2} = \dfrac{15}{48} = \dfrac{5}{16}$. Multiply by $I_0 = \dfrac{\pi}{2}$:

$$I_6 = \frac{5}{16}\cdot\frac{\pi}{2} = \frac{5\pi}{32}.$$ $$\boxed{\int_0^{\pi/2}\sin^6 x\,dx = \frac{5\pi}{32} \approx 0.4909.}$$

Reducing to standard form

Step 1 — group and complete the square.

$$3x^2 - 12x + 2y^2 + 8y + 14 = 0$$ $$3(x^2 - 4x) + 2(y^2 + 4y) + 14 = 0.$$

Complete the squares:

$$x^2 - 4x = (x-2)^2 - 4, \qquad y^2 + 4y = (y+2)^2 - 4.$$

Substitute:

$$3\big[(x-2)^2 - 4\big] + 2\big[(y+2)^2 - 4\big] + 14 = 0$$ $$3(x-2)^2 + 2(y+2)^2 - 12 - 8 + 14 = 0$$ $$3(x-2)^2 + 2(y+2)^2 - 6 = 0$$ $$3(x-2)^2 + 2(y+2)^2 = 6.$$

Step 2 — divide by 6 to get standard form.

$$\frac{(x-2)^2}{2} + \frac{(y+2)^2}{3} = 1.$$

Both squared terms are positive with unequal denominators, so the conic is an ellipse.

Reading off the parameters

The centre is at $(x,y) = (2,-2)$.

Since $3 > 2$, the larger denominator is under $(y+2)^2$, so the major axis is vertical. Thus

$$a^2 = 3,\quad b^2 = 2 \;\Rightarrow\; a = \sqrt{3}\approx 1.732,\quad b = \sqrt{2}\approx 1.414.$$

• Semi-major axis $a = \sqrt 3$, semi-minor axis $b = \sqrt 2$.

Eccentricity (for a vertical major axis, $b^2 = a^2(1-e^2)$):

$$e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{2}{3} = \frac{1}{3} \;\Rightarrow\; e = \frac{1}{\sqrt 3} \approx 0.577.$$

Trace (description)

            y
            |       major axis vertical
        (2, -2+√3)
            *
          .   .
        .       .
  (2-√2,-2)*  C  *(2+√2,-2)     C = centre (2,-2)
        .       .
          .   .
            *
        (2, -2-√3)
            |

The ellipse is centred at $(2,-2)$, stretches $\pm\sqrt3$ vertically and $\pm\sqrt2$ horizontally.

Summary: Ellipse, centre $(2,-2)$, semi-axes $a=\sqrt3,\,b=\sqrt2$, eccentricity $e = 1/\sqrt3$.

Recognising the form

The equation $\dfrac{dy}{dx} + P(x)\,y = Q(x)$ is linear with

$$P(x) = \cot x, \qquad Q(x) = 2\cos x.$$

Step 1 — integrating factor.

$$\text{IF} = e^{\int P\,dx} = e^{\int \cot x\,dx} = e^{\ln(\sin x)} = \sin x.$$

Step 2 — multiply through by the IF.

$$\sin x\,\frac{dy}{dx} + y\cos x = 2\sin x\cos x.$$

The left side is $\dfrac{d}{dx}(y\sin x)$, and $2\sin x\cos x = \sin 2x$:

$$\frac{d}{dx}(y\sin x) = \sin 2x.$$

Step 3 — integrate.

$$y\sin x = \int \sin 2x\,dx = -\frac{1}{2}\cos 2x + C.$$

Step 4 — apply the initial condition $y=0$ at $x=\pi/2$. Then $\sin(\pi/2)=1$ and $\cos(\pi)= -1$:

$$0\cdot 1 = -\frac{1}{2}(-1) + C \;\Rightarrow\; 0 = \frac{1}{2} + C \;\Rightarrow\; C = -\frac{1}{2}.$$

Step 5 — final solution.

$$y\sin x = -\frac{1}{2}\cos 2x - \frac{1}{2} = -\frac{1}{2}(\cos 2x + 1).$$

Using $\cos 2x + 1 = 2\cos^2 x$:

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

Therefore

$$\boxed{y = -\frac{\cos^2 x}{\sin x}.}$$

Check: at $x=\pi/2$, $y = -\dfrac{\cos^2(\pi/2)}{\sin(\pi/2)} = -\dfrac{0}{1} = 0.$ ✓

Point of contact

At $x = 2$:

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

So the point is $(2, -2)$.

Slope of the tangent

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

At $x = 2$:

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

The tangent is horizontal.

Tangent line

With slope $0$ through $(2,-2)$:

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

Normal line

The normal is perpendicular to the tangent. Since the tangent is horizontal ($m_t = 0$), the normal is vertical:

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

Summary: Tangent $y = -2$; Normal $x = 2$, meeting at $(2,-2)$.

Variables and constraint

Let the square base have side $x$ cm and the height be $h$ cm.

Surface area (base + 4 sides, no top):

$$A = x^2 + 4xh = 108. \qquad (1)$$

Volume to maximise:

$$V = x^2 h. \qquad (2)$$

Express V in one variable

From (1): $h = \dfrac{108 - x^2}{4x}$. Substitute into (2):

$$V(x) = x^2\cdot\frac{108 - x^2}{4x} = \frac{x(108 - x^2)}{4} = \frac{108x - x^3}{4} = 27x - \frac{x^3}{4}.$$

Maximise

$$\frac{dV}{dx} = 27 - \frac{3x^2}{4}.$$

Set to zero:

$$27 = \frac{3x^2}{4} \;\Rightarrow\; x^2 = 36 \;\Rightarrow\; x = 6\ \text{cm} \ (x>0).$$

Second-derivative test:

$$\frac{d^2V}{dx^2} = -\frac{6x}{4} = -\frac{3x}{2}.$$

At $x=6$: $\dfrac{d^2V}{dx^2} = -9 < 0$, confirming a maximum.

Dimensions and volume

Height:

$$h = \frac{108 - 6^2}{4(6)} = \frac{108 - 36}{24} = \frac{72}{24} = 3\ \text{cm}.$$

Maximum volume:

$$V = x^2 h = 6^2 \times 3 = 36 \times 3 = 108\ \text{cm}^3.$$ $$\boxed{\text{Base } 6\,\text{cm}\times 6\,\text{cm},\ \text{height } 3\,\text{cm},\ V_{\max} = 108\ \text{cm}^3.}$$

Points of intersection

From the line, $x = \dfrac{y+4}{2}$. Substitute into $y^2 = 4x$:

$$y^2 = 4\cdot\frac{y+4}{2} = 2(y+4) = 2y + 8.$$ $$y^2 - 2y - 8 = 0 \;\Rightarrow\; (y-4)(y+2) = 0 \;\Rightarrow\; y = 4 \text{ or } y = -2.$$

The curves meet at $y = -2$ and $y = 4$.

Set up the integral (integrate in $y$)

For a horizontal strip at height $y$, the right boundary is the line and the left boundary is the parabola:

$$x_{\text{line}} = \frac{y+4}{2}, \qquad x_{\text{parabola}} = \frac{y^2}{4}.$$

On $-2 < y < 4$ the line lies to the right of the parabola, so the width is

$$x_{\text{line}} - x_{\text{parabola}} = \frac{y+4}{2} - \frac{y^2}{4}.$$

Evaluate

$$A = \int_{-2}^{4}\left(\frac{y+4}{2} - \frac{y^2}{4}\right)dy = \int_{-2}^{4}\left(\frac{y}{2} + 2 - \frac{y^2}{4}\right)dy.$$

Antiderivative:

$$F(y) = \frac{y^2}{4} + 2y - \frac{y^3}{12}.$$

At $y = 4$:

$$F(4) = \frac{16}{4} + 8 - \frac{64}{12} = 4 + 8 - \frac{16}{3} = 12 - \frac{16}{3} = \frac{36-16}{3} = \frac{20}{3}.$$

At $y = -2$:

$$F(-2) = \frac{4}{4} + (-4) - \frac{-8}{12} = 1 - 4 + \frac{2}{3} = -3 + \frac{2}{3} = -\frac{7}{3}.$$

Therefore

$$A = F(4) - F(-2) = \frac{20}{3} - \left(-\frac{7}{3}\right) = \frac{27}{3} = 9.$$ $$\boxed{\text{Area} = 9 \text{ square units}.}$$

Definition

The Jacobian is the determinant

$$\frac{\partial(u,v)}{\partial(x,y)} = \begin{vmatrix} \dfrac{\partial u}{\partial x} & \dfrac{\partial u}{\partial y} \\[2mm] \dfrac{\partial v}{\partial x} & \dfrac{\partial v}{\partial y} \end{vmatrix}.$$

Partial derivatives

For $u = x^2 - y^2$:

$$\frac{\partial u}{\partial x} = 2x, \qquad \frac{\partial u}{\partial y} = -2y.$$

For $v = 2xy$:

$$\frac{\partial v}{\partial x} = 2y, \qquad \frac{\partial v}{\partial y} = 2x.$$

Evaluate the determinant

$$\frac{\partial(u,v)}{\partial(x,y)} = \begin{vmatrix} 2x & -2y \\ 2y & 2x \end{vmatrix} = (2x)(2x) - (-2y)(2y) = 4x^2 + 4y^2.$$ $$\boxed{\frac{\partial(u,v)}{\partial(x,y)} = 4(x^2 + y^2).}$$

Remark: This equals $4(x^2+y^2)\neq 0$ except at the origin, so the transformation is locally invertible everywhere except $(0,0)$.

Formula

The perpendicular distance from a point $(x_0,y_0)$ to the line $ax + by + c = 0$ is

$$d = \frac{|a x_0 + b y_0 + c|}{\sqrt{a^2 + b^2}}.$$

Substitute

Here $a = 5,\ b = -12,\ c = 6$ and $(x_0,y_0) = (3,-4)$.

Numerator:

$$|5(3) + (-12)(-4) + 6| = |15 + 48 + 6| = |69| = 69.$$

Denominator:

$$\sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13.$$

Therefore

$$d = \frac{69}{13} \approx 5.31.$$ $$\boxed{d = \frac{69}{13} \text{ units} \approx 5.31 \text{ units}.}$$

Part (a) — Exact equation

Write $M\,dx + N\,dy = 0$ with

$$M = 2xy + y^2, \qquad N = x^2 + 2xy.$$

Test for exactness:

$$\frac{\partial M}{\partial y} = 2x + 2y, \qquad \frac{\partial N}{\partial x} = 2x + 2y.$$

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

Find the potential $F(x,y)$ with $F_x = M$:

$$F = \int (2xy + y^2)\,dx = x^2 y + x y^2 + g(y).$$

Differentiate with respect to $y$ and match $N$:

$$F_y = x^2 + 2xy + g'(y) = x^2 + 2xy \;\Rightarrow\; g'(y) = 0 \;\Rightarrow\; g(y) = \text{const}.$$

Hence the general solution is

$$\boxed{x^2 y + x y^2 = C,}$$

or equivalently $xy(x+y) = C$.

Part (b) — Separable equation

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

Separate the variables:

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

Integrate both sides. Each side has the form $\int \dfrac{t}{1+t^2}\,dt = \tfrac12\ln(1+t^2)$:

$$\frac{1}{2}\ln(1+y^2) = \frac{1}{2}\ln(1+x^2) + \frac{1}{2}\ln C_1.$$

Multiply by 2 and combine logs:

$$\ln(1+y^2) = \ln\big[C_1(1+x^2)\big].$$

Exponentiating,

$$1 + y^2 = C_1(1 + x^2).$$ $$\boxed{1 + y^2 = C\,(1 + x^2),\quad C \text{ an arbitrary positive constant.}}$$