BE Civil Engineering (IOE, TU) Engineering Mathematics I (IOE, SH 401) Question Paper 2077 Nepal

Leibnitz's theorem. If $u$ and $v$ are functions of $x$ possessing derivatives up to order $n$, then

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

Applying to $y = x^2 e^{3x}$. Take $u = e^{3x}$ and $v = x^2$, because the successive derivatives of $v=x^2$ terminate after the second.

Derivatives of $u = e^{3x}$:

$$u_{n} = 3^{n} e^{3x}, \quad u_{n-1} = 3^{n-1} e^{3x}, \quad u_{n-2} = 3^{n-2} e^{3x}.$$

Derivatives of $v = x^2$:

$$v = x^2, \quad v_1 = 2x, \quad v_2 = 2, \quad v_3 = v_4 = \cdots = 0.$$

By Leibnitz's theorem only the first three terms survive:

$$y_n = u_n v + \binom{n}{1} u_{n-1} v_1 + \binom{n}{2} u_{n-2} v_2.$$

Substituting:

$$y_n = 3^{n} e^{3x}\,x^2 + n\,(3^{n-1} e^{3x})(2x) + \frac{n(n-1)}{2}(3^{n-2} e^{3x})(2).$$

Simplify:

$$\boxed{\,y_n = 3^{n-2} e^{3x}\left[\,9x^2 + 6nx + n(n-1)\,\right].}$$

Evaluating $y_4$ at $x=0$. Put $n=4$ and $x=0$. The bracket becomes $9(0) + 6(4)(0) + 4(3) = 12$, and $3^{n-2}=3^{2}=9$, $e^{0}=1$:

$$y_4(0) = 9 \cdot 1 \cdot 12 = 108.$$

Final answer: $y_n = 3^{n-2}e^{3x}\big[9x^2 + 6nx + n(n-1)\big]$ and $\mathbf{y_4(0) = 108}$.

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

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

Setting up. Here $u$ itself is not homogeneous, but $\sin u$ is. Let

$$z = \sin u = \frac{x^3 + y^3}{x + y}.$$

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

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

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

Apply Euler's theorem to $z$:

$$x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y} = 2z.$$

Convert back to $u$. Since $z = \sin u$,

$$\frac{\partial z}{\partial x} = \cos u\,\frac{\partial u}{\partial x}, \qquad \frac{\partial z}{\partial y} = \cos u\,\frac{\partial u}{\partial y}.$$

Substitute:

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

Divide both sides by $\cos u$ (assuming $\cos u \neq 0$):

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

Hence $x\,u_x + y\,u_y = 2\tan u$, as required. $\blacksquare$

Setting up the reduction. Write $\sin^n x = \sin^{n-1}x \cdot \sin x$ and integrate by parts on $\int_0^{\pi/2}\sin^n x\,dx$ taking $u=\sin^{n-1}x$, $dv=\sin x\,dx$, so $v=-\cos x$:

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

The boundary term vanishes: at $x=\pi/2$, $\cos x=0$; at $x=0$, $\sin x=0$ (since $n-1\ge1$). So

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

Collect $I_n$ terms:

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

Base value. $I_0 = \displaystyle\int_0^{\pi/2} dx = \frac{\pi}{2}.$

Evaluate $I_6$. Apply the formula repeatedly:

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

Now $I_0 = \dfrac{\pi}{2}$, so

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

Final answer: $I_n = \dfrac{n-1}{n}I_{n-2}$ and $\displaystyle\int_0^{\pi/2}\sin^6 x\,dx = \dfrac{5\pi}{32} \approx 0.4909.$

Group and complete the square.

$$9x^2 - 36x + 4y^2 + 8y + 4 = 0.$$

Factor coefficients:

$$9(x^2 - 4x) + 4(y^2 + 2y) + 4 = 0.$$

Complete the square inside each bracket:

$$x^2 - 4x = (x-2)^2 - 4, \qquad y^2 + 2y = (y+1)^2 - 1.$$

Substitute:

$$9\big[(x-2)^2 - 4\big] + 4\big[(y+1)^2 - 1\big] + 4 = 0,$$ $$9(x-2)^2 + 4(y+1)^2 - 36 - 4 + 4 = 0,$$ $$9(x-2)^2 + 4(y+1)^2 = 36.$$

Divide by 36:

$$\boxed{\dfrac{(x-2)^2}{4} + \dfrac{(y+1)^2}{9} = 1.}$$

Identify the conic. Both squared terms are positive with different denominators, so this is an ellipse. The larger denominator is under the $y$-term, so the major axis is vertical.

Parameters. With centre form $\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$:

• $a^2 = 9 \Rightarrow a = 3$ (semi-major axis, along $y$)
• $b^2 = 4 \Rightarrow b = 2$ (semi-minor axis, along $x$)
• Centre: $(h,k) = (2,\,-1)$.

Eccentricity. For a vertical ellipse $b^2 = a^2(1-e^2)$:

$$e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \approx 0.745.$$

Foci. The distance from centre to each focus is $c = ae = a\,e = 3\cdot\dfrac{\sqrt5}{3} = \sqrt5$. (Check: $c^2=a^2-b^2 = 9-4 = 5$.) Foci lie on the vertical major axis through the centre:

$$(2,\,-1\pm\sqrt5) \;\Rightarrow\; (2,\,-1+\sqrt5)\approx(2,\,1.236) \text{ and } (2,\,-1-\sqrt5)\approx(2,\,-3.236).$$

Summary table.

Final answer: standard form $\dfrac{(x-2)^2}{4}+\dfrac{(y+1)^2}{9}=1$; an ellipse with centre $(2,-1)$, $a=3$, $b=2$, $\mathbf{e=\sqrt5/3}$, foci $\mathbf{(2,-1\pm\sqrt5)}$.

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

$$P(x) = \frac{1}{x}, \qquad Q(x) = x^2.$$

Integrating factor (I.F.).

$$\mu = e^{\int P\,dx} = e^{\int \frac{1}{x}\,dx} = e^{\ln x} = x \quad (x>0).$$

Multiply through by $\mu = x$:

$$x\frac{dy}{dx} + y = x^3.$$

The left side is the derivative of $(\mu y)$:

$$\frac{d}{dx}(x\,y) = x^3.$$

Integrate both sides:

$$x y = \int x^3\,dx = \frac{x^4}{4} + C.$$

General solution:

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

Apply initial condition $y(1)=2$:

$$2 = \frac{1^3}{4} + \frac{C}{1} = \frac{1}{4} + C \;\Rightarrow\; C = 2 - \frac{1}{4} = \frac{7}{4}.$$

Particular solution:

$$y = \frac{x^3}{4} + \frac{7}{4x}, \qquad x>0.$$

Check: $y(1) = \tfrac14 + \tfrac74 = \tfrac{8}{4} = 2.$ Correct.

Final answer: $y = \dfrac{x^3}{4} + \dfrac{C}{x}$; with $y(1)=2$, $\mathbf{y = \dfrac{x^3}{4} + \dfrac{7}{4x}}$.

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

Volume constraint:

$$V = x^2 h = 32 \;\Rightarrow\; h = \frac{32}{x^2}.$$

Surface area (open top = base + 4 sides):

$$S = x^2 + 4xh.$$

Substitute $h$:

$$S(x) = x^2 + 4x\cdot\frac{32}{x^2} = x^2 + \frac{128}{x}.$$

Minimise. Differentiate and set to zero:

$$\frac{dS}{dx} = 2x - \frac{128}{x^2} = 0 \;\Rightarrow\; 2x^3 = 128 \;\Rightarrow\; x^3 = 64 \;\Rightarrow\; x = 4\ \text{cm}.$$

Second-derivative test:

$$\frac{d^2S}{dx^2} = 2 + \frac{256}{x^3} > 0 \text{ for } x>0,$$

so $x=4$ gives a minimum.

Height: $h = \dfrac{32}{4^2} = \dfrac{32}{16} = 2\ \text{cm}.$

Minimum surface area:

$$S = 4^2 + \frac{128}{4} = 16 + 32 = 48\ \text{cm}^2.$$

Final answer: base $4\,\text{cm}\times 4\,\text{cm}$, height $2\,\text{cm}$, minimum surface area $\mathbf{48\ \text{cm}^2}$.

Maclaurin's theorem.

$$f(x) = f(0) + x f'(0) + \frac{x^2}{2!}f''(0) + \frac{x^3}{3!}f'''(0) + \frac{x^4}{4!}f^{(4)}(0) + \cdots$$

Compute derivatives of $f(x)=\ln(1+x)$ and evaluate at $0$.

Substitute into the series:

$$\ln(1+x) = 0 + x(1) + \frac{x^2}{2!}(-1) + \frac{x^3}{3!}(2) + \frac{x^4}{4!}(-6) + \cdots$$

Simplify the factorials: $\frac{2}{3!}=\frac{2}{6}=\frac13$, $\frac{-6}{4!}=\frac{-6}{24}=-\frac14$:

$$\boxed{\,\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots\,}$$

General term.

$$\ln(1+x) = \sum_{n=1}^{\infty} (-1)^{n-1}\,\frac{x^n}{n}.$$

Interval of validity. The series converges for

$$-1 < x \le 1,$$

i.e. it is valid for $|x|<1$ and also at the endpoint $x=1$ (where it gives $\ln 2$), but diverges at $x=-1$.

Final answer: $\ln(1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\cdots = \displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}\dfrac{x^n}{n}$, valid for $\mathbf{-1 < x \le 1}$.

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

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

Corresponding $x$: for $y=4$, $x = (4+4)/2 = 4$; for $y=-2$, $x=(-2+4)/2 = 1$. Intersection points: $(4,4)$ and $(1,-2)$.

Integrate with respect to $y$ (cleaner, since the parabola opens rightward). For a horizontal strip the right boundary is the line and the left boundary is the parabola:

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

Between $y=-2$ and $y=4$, the line lies to the right of the parabola, so

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

Evaluate.

$$A = \int_{-2}^{4}\left(\frac{y}{2} + 2 - \frac{y^2}{4}\right)dy = \left[\frac{y^2}{4} + 2y - \frac{y^3}{12}\right]_{-2}^{4}.$$

At $y=4$: $\dfrac{16}{4} + 8 - \dfrac{64}{12} = 4 + 8 - \dfrac{16}{3} = 12 - \dfrac{16}{3} = \dfrac{36-16}{3} = \dfrac{20}{3}.$

At $y=-2$: $\dfrac{4}{4} + 2(-2) - \dfrac{-8}{12} = 1 - 4 + \dfrac{2}{3} = -3 + \dfrac{2}{3} = -\dfrac{7}{3}.$

Subtract:

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

Final answer: the enclosed area is $\mathbf{9}$ square units.

Definition. The Jacobian of $(u,v)$ with respect to $(x,y)$ is the determinant

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

Partial derivatives.

$$\frac{\partial u}{\partial x} = 1, \quad \frac{\partial u}{\partial y} = 1, \quad \frac{\partial v}{\partial x} = y, \quad \frac{\partial v}{\partial y} = x.$$

Evaluate the determinant.

$$J = \begin{vmatrix} 1 & 1 \\ y & x \end{vmatrix} = (1)(x) - (1)(y) = x - y.$$ $$\boxed{\,\dfrac{\partial(u,v)}{\partial(x,y)} = x - y.\,}$$

Condition for local invertibility. By the inverse function theorem the transformation is locally invertible wherever the Jacobian is non-zero:

$$J = x - y \neq 0 \;\Longleftrightarrow\; x \neq y.$$

Thus the map $(x,y)\mapsto(u,v)$ fails to be locally invertible along the line $x=y$ (where $u^2 = 4v$, the boundary of the attainable region).

Final answer: $\dfrac{\partial(u,v)}{\partial(x,y)} = x - y$; locally invertible iff $\mathbf{x \neq y}$.

Test for exactness. Write $M\,dx + N\,dy = 0$ with

$$M = 2xy + 3y^2, \qquad N = x^2 + 6xy.$$

Compute the cross partials:

$$\frac{\partial M}{\partial y} = 2x + 6y, \qquad \frac{\partial N}{\partial x} = 2x + 6y.$$

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

Find the potential function $F(x,y)$ with $F_x = M$ and $F_y = N$.

Integrate $M$ with respect to $x$ (treating $y$ constant):

$$F = \int (2xy + 3y^2)\,dx = x^2 y + 3y^2 x + g(y),$$

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

Determine $g(y)$. Differentiate $F$ with respect to $y$ and set equal to $N$:

$$F_y = x^2 + 6xy + g'(y) = x^2 + 6xy.$$

Hence $g'(y) = 0 \Rightarrow g(y) =$ constant.

General solution. Setting $F(x,y) = C$:

$$\boxed{\,x^2 y + 3xy^2 = C.\,}$$

Check by implicit differentiation: $d(x^2y+3xy^2) = (2xy+3y^2)dx + (x^2+6xy)dy = 0$, which matches the original equation.

Final answer: the equation is exact and its general solution is $\mathbf{x^2 y + 3xy^2 = C}$.

Polar area formula. The area swept by a polar curve is

$$A = \frac{1}{2}\int_{0}^{2\pi} r^2\,d\theta.$$

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

$$A = \frac{1}{2}\int_{0}^{2\pi} a^2(1+\cos\theta)^2\,d\theta = \frac{a^2}{2}\int_{0}^{2\pi}\big(1 + 2\cos\theta + \cos^2\theta\big)\,d\theta.$$

Use $\cos^2\theta = \dfrac{1+\cos 2\theta}{2}$:

$$1 + 2\cos\theta + \cos^2\theta = 1 + 2\cos\theta + \frac{1}{2} + \frac{\cos 2\theta}{2} = \frac{3}{2} + 2\cos\theta + \frac{\cos 2\theta}{2}.$$

Integrate term by term over $[0,2\pi]$.

• $\displaystyle\int_0^{2\pi}\frac{3}{2}\,d\theta = \frac{3}{2}(2\pi) = 3\pi.$
• $\displaystyle\int_0^{2\pi}2\cos\theta\,d\theta = 2[\sin\theta]_0^{2\pi} = 0.$
• $\displaystyle\int_0^{2\pi}\frac{\cos 2\theta}{2}\,d\theta = \frac{1}{2}\left[\frac{\sin 2\theta}{2}\right]_0^{2\pi} = 0.$

So the integral equals $3\pi$, and

$$A = \frac{a^2}{2}\cdot 3\pi = \frac{3\pi a^2}{2}.$$

Final answer: the area enclosed by the cardioid is $\mathbf{\dfrac{3\pi a^2}{2}}$ square units.