BE Civil Engineering (IOE, TU) Engineering Mathematics I (IOE, SH 401) Question Paper 2080 Nepal

(a) Statements

Rolle's theorem: If $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then there exists at least one $c \in (a,b)$ such that $f'(c)=0$.

Lagrange MVT: If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists $c \in (a,b)$ such that

$$f'(c) = \frac{f(b)-f(a)}{b-a}.$$

Verification for $f(x)=x^3-6x^2+11x-6$ on $[1,3]$.

$f$ is a polynomial, hence continuous on $[1,3]$ and differentiable on $(1,3)$; the hypotheses hold.

Endpoint values:

• $f(1)=1-6+11-6=0$
• $f(3)=27-54+33-6=0$

So the average rate of change is $\dfrac{f(3)-f(1)}{3-1}=\dfrac{0-0}{2}=0$.

Derivative: $f'(x)=3x^2-12x+11$. Set $f'(c)=0$:

$$3c^2-12c+11=0 \implies c=\frac{12\pm\sqrt{144-132}}{6}=\frac{12\pm\sqrt{12}}{6}=2\pm\frac{\sqrt{3}}{3}.$$

Numerically $c \approx 2 \pm 0.5774$, i.e. $c_1 \approx 1.4226$ and $c_2 \approx 2.5774$. Both lie in $(1,3)$.

Both values $c = 2 \pm \dfrac{\sqrt{3}}{3}$ satisfy the MVT. (Since $f(1)=f(3)$, this is also a verification of Rolle's theorem.)

(b) Limit

As $x \to 0$ the form is $\frac{0}{0}$. Apply L'Hôpital repeatedly (Maclaurin expansion confirms each order).

Numerator $N=e^x-e^{-x}-2x$, denominator $D=x-\sin x$.

1st derivatives: $N'=e^x+e^{-x}-2$, $D'=1-\cos x$ — still $\frac{0}{0}$.

2nd derivatives: $N''=e^x-e^{-x}$, $D''=\sin x$ — still $\frac{0}{0}$.

3rd derivatives: $N'''=e^x+e^{-x}$, $D'''=\cos x$.

Now substitute $x=0$:

$$\lim_{x\to0}\frac{N'''}{D'''}=\frac{e^0+e^0}{\cos 0}=\frac{1+1}{1}=2.$$

The limit equals $2$.

Check via series: $N = 2\left(\frac{x^3}{3!}+\cdots\right)=\frac{x^3}{3}+\cdots$ and $D=\frac{x^3}{6}+\cdots$, so the ratio $\to \frac{1/3}{1/6}=2$. Consistent.

(a) Let $v = \tan u = \dfrac{x^3+y^3}{x-y}$.

Check homogeneity: numerator is degree 3, denominator degree 1, so $v$ is homogeneous of degree $n = 3 - 1 = 2$. Indeed $v(tx,ty)=\dfrac{t^3(x^3+y^3)}{t(x-y)}=t^2 v(x,y)$.

By Euler's theorem for a homogeneous function of degree $n$:

$$x v_x + y v_y = n v = 2v.$$

Since $v=\tan u$, $v_x = \sec^2 u \, u_x$ and $v_y = \sec^2 u \, u_y$. Substituting:

$$\sec^2 u\,(x u_x + y u_y) = 2\tan u.$$ $$x u_x + y u_y = \frac{2\tan u}{\sec^2 u} = 2\sin u\cos u = \sin 2u.$$

Hence $x\,u_x + y\,u_y = \sin 2u$. (proved)

(b) With $x=r\cos\theta,\ y=r\sin\theta$. By the chain rule:

$$z_r = z_x x_r + z_y y_r = z_x\cos\theta + z_y\sin\theta,$$ $$z_\theta = z_x x_\theta + z_y y_\theta = z_x(-r\sin\theta) + z_y(r\cos\theta).$$

Then

$$z_r^2 = z_x^2\cos^2\theta + 2z_xz_y\cos\theta\sin\theta + z_y^2\sin^2\theta,$$ $$\frac{1}{r^2}z_\theta^2 = z_x^2\sin^2\theta - 2z_xz_y\cos\theta\sin\theta + z_y^2\cos^2\theta.$$

Adding, the cross terms cancel:

$$z_r^2 + \frac{1}{r^2}z_\theta^2 = z_x^2(\cos^2\theta+\sin^2\theta) + z_y^2(\sin^2\theta+\cos^2\theta) = z_x^2 + z_y^2.$$

Hence $\left(z_x\right)^2 + \left(z_y\right)^2 = \left(z_r\right)^2 + \dfrac{1}{r^2}\left(z_\theta\right)^2$. (proved)

Curve $y=\sqrt{x}$, so $y^2=x$, on $0 \le x \le 4$.

(a) Volume (disk method about the $x$-axis):

$$V=\pi\int_0^4 y^2\,dx = \pi\int_0^4 x\,dx = \pi\left[\frac{x^2}{2}\right]_0^4 = \pi\cdot\frac{16}{2}=8\pi.$$

$V = 8\pi \approx 25.13$ cubic units.

(b) Curved surface area about the $x$-axis:

$$S = 2\pi\int_0^4 y\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx.$$

Here $\dfrac{dy}{dx}=\dfrac{1}{2\sqrt{x}}$, so $1+\left(\dfrac{dy}{dx}\right)^2 = 1+\dfrac{1}{4x}=\dfrac{4x+1}{4x}$.

Thus $y\sqrt{1+(y')^2}=\sqrt{x}\cdot\sqrt{\dfrac{4x+1}{4x}}=\sqrt{x}\cdot\dfrac{\sqrt{4x+1}}{2\sqrt{x}}=\dfrac{\sqrt{4x+1}}{2}.$

Therefore

$$S=2\pi\int_0^4 \frac{\sqrt{4x+1}}{2}\,dx=\pi\int_0^4\sqrt{4x+1}\,dx.$$

Let $u=4x+1$, $du=4\,dx$. When $x=0,\,u=1$; $x=4,\,u=17$.

$$S=\pi\int_1^{17}\sqrt{u}\,\frac{du}{4}=\frac{\pi}{4}\cdot\frac{2}{3}\left[u^{3/2}\right]_1^{17}=\frac{\pi}{6}\left(17^{3/2}-1\right).$$

Now $17^{3/2}=17\sqrt{17}=17(4.1231)=70.093$.

$$S=\frac{\pi}{6}(70.093-1)=\frac{\pi}{6}(69.093)=11.516\,\pi.$$

$S = \dfrac{\pi}{6}\left(17\sqrt{17}-1\right) \approx 36.18$ square units.

Step 1 — Rotation to remove the $xy$ term.

For $Ax^2+Bxy+Cy^2+\dots$ with $A=3,\ B=2,\ C=3$, rotate by angle $\theta$ where

$$\cot 2\theta = \frac{A-C}{B}=\frac{3-3}{2}=0 \implies 2\theta=90^\circ \implies \theta=45^\circ.$$

Use $x=X\cos45^\circ - Y\sin45^\circ = \dfrac{X-Y}{\sqrt2}$, $\;y=X\sin45^\circ + Y\cos45^\circ=\dfrac{X+Y}{\sqrt2}$.

The new quadratic coefficients are the eigenvalue-type values $A'=A\cos^2\theta+B\cos\theta\sin\theta+C\sin^2\theta$ and $C'$ similarly. With $\theta=45^\circ$:

$$A' = 3\cdot\tfrac12 + 2\cdot\tfrac12 + 3\cdot\tfrac12 = \tfrac32+1+\tfrac32 = 4,$$ $$C' = 3\cdot\tfrac12 - 2\cdot\tfrac12 + 3\cdot\tfrac12 = \tfrac32-1+\tfrac32 = 2.$$

The $xy$ coefficient becomes $0$. (Check: $A'+C'=4+2=6=A+C$.)

Step 2 — Transform the linear term $-16y$.

$-16y = -16\cdot\dfrac{X+Y}{\sqrt2} = -\dfrac{16}{\sqrt2}(X+Y)= -8\sqrt2\,X - 8\sqrt2\,Y.$

The equation becomes

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

Step 3 — Translation (complete the square).

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

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

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

Step 4 — Identification.

Both squared terms are positive with unequal denominators: this is an ellipse.

• Semi-axis along $X'$: $a = \sqrt{1/4} = \dfrac{1}{2} = 0.5$.
• Semi-axis along $Y'$: $b = \sqrt{1/2} = \dfrac{1}{\sqrt2} \approx 0.707$ (this is the semi-major axis).

Centre in rotated coordinates: $(X,Y)=(\sqrt2,\,2\sqrt2)$. Transforming back to original axes,

$$x=\frac{X-Y}{\sqrt2}=\frac{\sqrt2-2\sqrt2}{\sqrt2}=-1,\qquad y=\frac{X+Y}{\sqrt2}=\frac{\sqrt2+2\sqrt2}{\sqrt2}=3.$$

The conic is an ellipse with centre $(-1, 3)$, semi-minor axis $\tfrac12$ and semi-major axis $\tfrac{1}{\sqrt2}$, standard form $4X'^2 + 2Y'^2 = 1$.

(a) Linear ODE.

Standard form $y' + P(x)y = Q(x)$ with $P=\dfrac1x$, $Q=x^2$.

Integrating factor: $\mu = e^{\int \frac1x dx}=e^{\ln x}=x$.

Multiply through: $(xy)' = x\cdot x^2 = x^3$.

Integrate: $xy = \dfrac{x^4}{4} + C$.

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

(b) Mixing problem.

Volume is constant at $200$ L (inflow rate = outflow rate = $4$ L/min).

Rate in of salt $=$ (concentration in)(flow in) $= 0.5\ \text{kg/L} \times 4\ \text{L/min} = 2\ \text{kg/min}.$

Rate out of salt $=$ (concentration in tank)(flow out) $= \dfrac{Q}{200}\times 4 = \dfrac{Q}{50}\ \text{kg/min}.$

Governing ODE:

$$\frac{dQ}{dt} = 2 - \frac{Q}{50}, \qquad Q(0)=0.$$

Write as $\dfrac{dQ}{dt} + \dfrac{1}{50}Q = 2$. Integrating factor $\mu = e^{t/50}$:

$$\left(Q\,e^{t/50}\right)' = 2 e^{t/50} \implies Q e^{t/50} = 2\cdot 50\, e^{t/50} + C = 100 e^{t/50}+C.$$ $$Q(t) = 100 + C e^{-t/50}.$$

Apply $Q(0)=0$: $0 = 100 + C \implies C=-100$. So

$$Q(t) = 100\left(1 - e^{-t/50}\right)\ \text{kg}.$$

At $t=50$ min: $Q(50)=100\left(1-e^{-1}\right)=100(1-0.36788)=100(0.63212).$

$Q(50) \approx 63.21$ kg of salt. (The salt content approaches the steady-state value $100$ kg as $t\to\infty$, consistent with $0.5\ \text{kg/L}\times200\ \text{L}$.)

Radius of curvature for $y=f(x)$:

$$\rho = \frac{\left(1+(y')^2\right)^{3/2}}{|y''|}.$$

From $y^2=4x$ take the upper branch $y=2\sqrt{x}=2x^{1/2}$.

First derivative: $y' = 2\cdot\tfrac12 x^{-1/2}=x^{-1/2}=\dfrac{1}{\sqrt x}$. At $x=1$: $y'=1$.

Second derivative: $y'' = -\tfrac12 x^{-3/2}=-\dfrac{1}{2x^{3/2}}$. At $x=1$: $y''=-\dfrac12$, so $|y''|=\dfrac12$.

Substitute:

$$\rho = \frac{\left(1+1^2\right)^{3/2}}{1/2} = \frac{2^{3/2}}{1/2} = 2\cdot 2^{3/2}=2^{5/2}=4\sqrt2.$$

$\rho = 4\sqrt2 \approx 5.66$ units.

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

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

$$x^2 + 4xh = 300 \implies h = \frac{300 - x^2}{4x}.$$

Volume to maximise:

$$V = x^2 h = x^2\cdot\frac{300-x^2}{4x} = \frac{300x - x^3}{4} = 75x - \frac{x^3}{4}.$$

Differentiate and set to zero:

$$\frac{dV}{dx} = 75 - \frac{3x^2}{4} = 0 \implies x^2 = \frac{75\cdot4}{3}=100 \implies x = 10\ \text{cm}.$$

Second derivative: $\dfrac{d^2V}{dx^2} = -\dfrac{3x}{2}<0$ at $x=10$, confirming a maximum.

Height: $h = \dfrac{300-100}{4\cdot10}=\dfrac{200}{40}=5\ \text{cm}.$

Maximum volume: $V = x^2 h = 10^2\cdot 5 = 500\ \text{cm}^3.$

Dimensions: base $10\,\text{cm}\times10\,\text{cm}$, height $5\,\text{cm}$; maximum volume $= 500\ \text{cm}^3$.

Walli's formula. For an even positive integer $n$,

$$\int_0^{\pi/2}\sin^n x\,dx = \frac{(n-1)(n-3)\cdots 1}{n(n-2)\cdots 2}\cdot\frac{\pi}{2}.$$

Here $n=6$ (even):

$$\int_0^{\pi/2}\sin^6 x\,dx = \frac{5\cdot3\cdot1}{6\cdot4\cdot2}\cdot\frac{\pi}{2}.$$

Compute the fraction: numerator $5\cdot3\cdot1 = 15$; denominator $6\cdot4\cdot2 = 48$.

$$= \frac{15}{48}\cdot\frac{\pi}{2} = \frac{5}{16}\cdot\frac{\pi}{2} = \frac{5\pi}{32}.$$

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

The Jacobian is the determinant of the matrix of partial derivatives:

$$J = \frac{\partial(u,v,w)}{\partial(x,y,z)} = \begin{vmatrix} u_x & u_y & u_z \\ v_x & v_y & v_z \\ w_x & w_y & w_z \end{vmatrix}.$$

Partial derivatives:

• $u_x=1,\ u_y=1,\ u_z=1$
• $v_x=y+z,\ v_y=x+z,\ v_z=x+y$
• $w_x=2x,\ w_y=2y,\ w_z=2z$

So

$$J = \begin{vmatrix} 1 & 1 & 1 \\ y+z & x+z & x+y \\ 2x & 2y & 2z \end{vmatrix}.$$

Factor $2$ from row 3: $J = 2\begin{vmatrix} 1 & 1 & 1 \\ y+z & x+z & x+y \\ x & y & z \end{vmatrix}.$

Add row 3 to row 2 ($R_2 \to R_2 + R_3$): each entry becomes $x+y+z$, a common factor:

$$J = 2(x+y+z)\begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ x & y & z \end{vmatrix}.$$

Rows 1 and 2 are identical, so the determinant is $0$.

$J = 0$.

Since the Jacobian vanishes identically, $u, v, w$ are functionally dependent. Indeed the relation is $w = u^2 - 2v$ (because $x^2+y^2+z^2 = (x+y+z)^2 - 2(xy+yz+zx)$).

Let the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.

Substitute each point.

Point $(1,1)$: $1+1+2g+2f+c=0 \Rightarrow 2g+2f+c=-2.\quad(1)$

Point $(2,-1)$: $4+1+4g-2f+c=0 \Rightarrow 4g-2f+c=-5.\quad(2)$

Point $(3,2)$: $9+4+6g+4f+c=0 \Rightarrow 6g+4f+c=-13.\quad(3)$

$(2)-(1)$: $2g - 4f = -3.\quad(4)$

$(3)-(1)$: $4g + 2f = -11.\quad(5)$

From $(5)$: $2f = -11 - 4g \Rightarrow f = \dfrac{-11-4g}{2}.$ Substitute into $(4)$:

$$2g - 4\cdot\frac{-11-4g}{2} = -3 \Rightarrow 2g -2(-11-4g) = -3 \Rightarrow 2g + 22 + 8g = -3.$$ $$10g = -25 \Rightarrow g = -2.5.$$

Then $f = \dfrac{-11-4(-2.5)}{2} = \dfrac{-11+10}{2} = -0.5.$

From $(1)$: $c = -2 - 2g - 2f = -2 -2(-2.5) -2(-0.5) = -2 +5 +1 = 4.$

Equation: $x^2 + y^2 - 5x - y + 4 = 0.$

Centre $(-g,-f) = (2.5,\ 0.5).$

Radius $r = \sqrt{g^2+f^2-c} = \sqrt{6.25 + 0.25 - 4} = \sqrt{2.5} = \dfrac{\sqrt{10}}{2} \approx 1.581.$

Centre $(2.5,\,0.5)$, radius $\dfrac{\sqrt{10}}{2}\approx 1.58$ units.

Check with $(1,1)$: $(1-2.5)^2+(1-0.5)^2 = 2.25+0.25 = 2.5 = r^2.$ Correct.

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.

Solve. There exists $F(x,y)$ with $F_x = M$, $F_y = N$, and the solution is $F(x,y)=C$.

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

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

Differentiate w.r.t. $y$ and match to $N$:

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

Therefore the general solution is

$$\boxed{\,x^2 y + x y^2 = C\,} \quad\text{equivalently}\quad xy(x+y) = C.$$