BE Computer Engineering (Pokhara University) Calculus I (PU, MTH 110) Question Paper 2079 Nepal

(a) $\varepsilon$–$\delta$ definition and proof

Definition. We say $\lim_{x \to a} f(x) = L$ if for every $\varepsilon > 0$ there exists a $\delta > 0$ such that

$$0 < |x - a| < \delta \implies |f(x) - L| < \varepsilon.$$

To prove: $\lim_{x \to 3}(2x - 1) = 5$.

Here $f(x) = 2x - 1$, $a = 3$, $L = 5$. Let $\varepsilon > 0$ be given. Consider

$$|f(x) - L| = |(2x - 1) - 5| = |2x - 6| = 2|x - 3|.$$

We want this $< \varepsilon$, i.e. $2|x - 3| < \varepsilon \iff |x - 3| < \dfrac{\varepsilon}{2}$.

Choose $\delta = \dfrac{\varepsilon}{2}$. Then whenever $0 < |x - 3| < \delta$,

$$|f(x) - 5| = 2|x - 3| < 2\delta = 2\cdot\frac{\varepsilon}{2} = \varepsilon.$$

Since for every $\varepsilon>0$ such a $\delta$ exists, by definition $\lim_{x \to 3}(2x - 1) = 5$. $\blacksquare$

(b) Continuity at $x = 2$

For $x \ne 2$, $\dfrac{x^2 - 4}{x - 2} = \dfrac{(x-2)(x+2)}{x-2} = x + 2$. Hence

$$\lim_{x \to 2} f(x) = \lim_{x \to 2}(x + 2) = 4.$$

But $f(2) = 3$. The three-part continuity test requires $\lim_{x\to 2} f(x) = f(2)$. Here

$$\lim_{x \to 2} f(x) = 4 \ne 3 = f(2),$$

so $f$ is discontinuous at $x = 2$.

Type: Since the limit exists (and is finite) but does not equal the function value, this is a removable discontinuity.

Removal: Redefine the value at $x = 2$ as $f(2) = 4$ (instead of $3$). With this redefinition $\lim_{x\to2}f(x)=f(2)=4$ and $f$ becomes continuous at $x = 2$.

(a) Rolle's theorem, MVT and verification

Rolle's Theorem. If $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a) = f(b)$, then there exists $c \in (a,b)$ with $f'(c) = 0$.

Mean Value Theorem (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 $f(x) = x^3 - 3x$ on $[-2, 2]$.

$f$ is a polynomial, so it is continuous on $[-2,2]$ and differentiable on $(-2,2)$; the hypotheses hold.

$$f(-2) = -8 + 6 = -2, \qquad f(2) = 8 - 6 = 2.$$ $$\frac{f(2) - f(-2)}{2 - (-2)} = \frac{2 - (-2)}{4} = \frac{4}{4} = 1.$$ $$f'(x) = 3x^2 - 3.$$

Set $f'(c) = 1$: $\;3c^2 - 3 = 1 \Rightarrow 3c^2 = 4 \Rightarrow c^2 = \dfrac{4}{3} \Rightarrow c = \pm\dfrac{2}{\sqrt 3}.$

Both $c = \pm\dfrac{2}{\sqrt3} \approx \pm 1.155$ lie in $(-2, 2)$, so the MVT is verified with $c = \pm\dfrac{2}{\sqrt3}$.

(b) Open-top box of minimum material

Let the (square) base have side $x$ and height $h$. Volume constraint:

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

Material = area of base + 4 sides (open top):

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

Minimize:

$$\frac{dS}{dx} = 2x - \frac{128}{x^2} = 0 \implies 2x^3 = 128 \implies x^3 = 64 \implies x = 4.$$

Then $h = \dfrac{32}{4^2} = \dfrac{32}{16} = 2.$

Second-derivative test:

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

in particular $S''(4) = 2 + \dfrac{256}{64} = 2 + 4 = 6 > 0$, so $x = 4$ gives a minimum.

Dimensions: base $4\text{ cm} \times 4\text{ cm}$, height $2\text{ cm}$; minimum material $S = 16 + 32 = 48\,\text{cm}^2.$

(a) Convergence of a series and the Integral Test

Definition. The infinite series $\sum_{n=1}^{\infty} a_n$ converges to sum $S$ if the sequence of partial sums $S_N = \sum_{n=1}^{N} a_n$ has a finite limit $\lim_{N\to\infty} S_N = S$. Otherwise the series diverges.

Integral Test. Let $a_n = f(n)$ where $f$ is positive, continuous and decreasing on $[1,\infty)$. Then $\sum_{n=1}^{\infty} a_n$ and $\displaystyle\int_{1}^{\infty} f(x)\,dx$ both converge or both diverge.

Proof. Because $f$ decreases, for $n \le x \le n+1$ we have $f(n+1) \le f(x) \le f(n)$. Integrating over $[n, n+1]$:

$$f(n+1) \le \int_{n}^{n+1} f(x)\,dx \le f(n).$$

Summing from $n=1$ to $N-1$:

$$\sum_{n=2}^{N} a_n \le \int_{1}^{N} f(x)\,dx \le \sum_{n=1}^{N-1} a_n.$$

Denote $S_N = \sum_{n=1}^N a_n$ and $I_N = \int_1^N f$.

• If $\int_1^\infty f$ converges, the right inequality gives $S_N \le a_1 + I_N \le a_1 + \int_1^\infty f$, so $\{S_N\}$ is increasing and bounded above, hence converges.
• If $\int_1^\infty f$ diverges, $I_N \to \infty$, and the left inequality $S_N - a_1 \ge I_N$ forces $S_N \to \infty$, so the series diverges.

Thus the series and integral share the same convergence behaviour. $\blacksquare$

(b) Testing for convergence/divergence

(i) $\displaystyle\sum_{n=1}^{\infty} \frac{n}{2^n}$ — Ratio Test.

$$\frac{a_{n+1}}{a_n} = \frac{(n+1)/2^{n+1}}{n/2^{n}} = \frac{n+1}{2n} \to \frac{1}{2} < 1.$$

Since $L = \tfrac12 < 1$, the series converges.

(ii) $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n \ln n}$ — Integral Test. Take $f(x) = \dfrac{1}{x\ln x}$ (positive, decreasing on $[2,\infty)$):

$$\int_{2}^{\infty} \frac{dx}{x\ln x} = \Big[\ln(\ln x)\Big]_{2}^{\infty} = \infty.$$

The integral diverges, so the series diverges.

(iii) $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\sqrt{n}}$ — Alternating Series (Leibniz) Test. With $b_n = \dfrac{1}{\sqrt n}$: $b_n > 0$, $b_{n+1} < b_n$ (decreasing), and $b_n \to 0$. Hence the series converges (conditionally, since $\sum 1/\sqrt n$ diverges as a $p$-series with $p = \tfrac12 \le 1$).

(a) Polar coordinates and conversion

Representation. A point $P$ in the plane is located by an ordered pair $(r, \theta)$, where $r$ is the directed distance from the origin (pole) $O$ to $P$, and $\theta$ is the angle (measured anticlockwise) from the positive $x$-axis (polar axis) to the ray $OP$.

Relations with Cartesian coordinates. Dropping a perpendicular from $P=(x,y)$ to the $x$-axis forms a right triangle, giving

$$x = r\cos\theta, \qquad y = r\sin\theta,$$

and conversely

$$r = \sqrt{x^2 + y^2}, \qquad \theta = \tan^{-1}\!\frac{y}{x}.$$

Conversion of $x^2 + y^2 = 4x$. Since $x^2 + y^2 = r^2$ and $x = r\cos\theta$:

$$r^2 = 4r\cos\theta \implies r = 4\cos\theta.$$

The Cartesian form $x^2 + y^2 = 4x \Rightarrow (x-2)^2 + y^2 = 4$ is a circle of radius $2$ centred at $(2,0)$ (passing through the origin).

(b) Cardioid $r = 1 + \cos\theta$ and enclosed area

Sketch (described). At $\theta = 0$, $r = 2$ (rightmost point); at $\theta = \tfrac{\pi}{2}$, $r = 1$; at $\theta = \pi$, $r = 0$ (cusp at the pole); at $\theta = \tfrac{3\pi}{2}$, $r = 1$. The curve is a heart-shaped loop symmetric about the polar ($x$) axis, with its cusp pointing in the negative-$x$ direction.

Area. Using $A = \dfrac{1}{2}\displaystyle\int_{0}^{2\pi} r^2\,d\theta$:

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

Using $\cos^2\theta = \tfrac{1 + \cos 2\theta}{2}$, the integrand becomes $\tfrac{3}{2} + 2\cos\theta + \tfrac12\cos2\theta$. Over $[0, 2\pi]$ the cosine terms integrate to $0$, so

$$A = \frac{1}{2}\cdot \frac{3}{2}\cdot 2\pi = \frac{3\pi}{2}.$$

Area enclosed $= \dfrac{3\pi}{2}$ square units.

(a) $\displaystyle\lim_{x \to 0} \frac{\sin 3x}{\tan 5x}$. Using $\sin 3x \sim 3x$ and $\tan 5x \sim 5x$ as $x\to0$:

$$\lim_{x\to0}\frac{\sin 3x}{\tan 5x} = \lim_{x\to0}\frac{\frac{\sin 3x}{3x}\cdot 3x}{\frac{\tan 5x}{5x}\cdot 5x} = \frac{1\cdot 3}{1\cdot 5} = \boxed{\frac{3}{5}}.$$

(b) $\displaystyle\lim_{x \to \infty}\left(1 + \frac{2}{x}\right)^{x}$. Using the standard limit $\lim_{x\to\infty}\left(1+\tfrac{a}{x}\right)^x = e^{a}$ with $a = 2$:

$$\lim_{x\to\infty}\left(1+\frac{2}{x}\right)^x = \boxed{e^{2}}.$$

(c) $\displaystyle\lim_{x \to 0}\frac{e^x - 1 - x}{x^2}$ (form $\tfrac00$, L'Hôpital twice):

$$\lim_{x\to0}\frac{e^x - 1 - x}{x^2} \overset{\text{L'H}}{=} \lim_{x\to0}\frac{e^x - 1}{2x} \overset{\text{L'H}}{=} \lim_{x\to0}\frac{e^x}{2} = \boxed{\frac{1}{2}}.$$

(a) Differentiate $x^y = y^x$

Take natural logs of both sides: $y \ln x = x \ln y$. Differentiate implicitly w.r.t. $x$:

$$\frac{dy}{dx}\ln x + y\cdot\frac{1}{x} = \ln y + x\cdot\frac{1}{y}\frac{dy}{dx}.$$

Collect $\dfrac{dy}{dx}$ terms:

$$\frac{dy}{dx}\left(\ln x - \frac{x}{y}\right) = \ln y - \frac{y}{x}.$$ $$\boxed{\frac{dy}{dx} = \frac{\ln y - \dfrac{y}{x}}{\ln x - \dfrac{x}{y}} = \frac{y\,(x\ln y - y)}{x\,(y\ln x - x)}.}$$

(b) $y = e^{ax}\sin bx$

$$y_1 = \frac{dy}{dx} = a e^{ax}\sin bx + b e^{ax}\cos bx = e^{ax}(a\sin bx + b\cos bx).$$ $$y_2 = a\,y_1 + e^{ax}(ab\cos bx - b^2\sin bx).$$

Note $e^{ax}(ab\cos bx - b^2 \sin bx) = b\,e^{ax}(a\cos bx - b\sin bx)$. It is cleaner to compute directly:

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

Now form the combination:

$$y_2 - 2a y_1 + (a^2 + b^2)y.$$

Substitute ($S=\sin bx,\ C=\cos bx$), factoring out $e^{ax}$:

$$\big[(a^2 - b^2)S + 2abC\big] - 2a\big[aS + bC\big] + (a^2 + b^2)S.$$ $$= \big[(a^2 - b^2) - 2a^2 + (a^2 + b^2)\big]S + \big[2ab - 2ab\big]C = 0\cdot S + 0\cdot C = 0.$$

Hence $y_2 - 2a y_1 + (a^2 + b^2)y = 0.$ $\blacksquare$

(a) $\displaystyle\int x^2 e^{x}\,dx$ — integration by parts (reduce the power of $x$ twice).

$$\int x^2 e^x\,dx = x^2 e^x - \int 2x\,e^x\,dx = x^2 e^x - 2\Big(x e^x - \int e^x dx\Big)$$ $$= x^2 e^x - 2x e^x + 2e^x + C = \boxed{e^x\big(x^2 - 2x + 2\big) + C}.$$

(b) $\displaystyle\int \frac{2x + 3}{(x - 1)(x + 2)}\,dx$ — partial fractions. Write

$$\frac{2x + 3}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}.$$

$2x + 3 = A(x+2) + B(x-1)$. At $x = 1$: $5 = 3A \Rightarrow A = \tfrac{5}{3}$. At $x = -2$: $-1 = -3B \Rightarrow B = \tfrac13$. Then

$$\int\!\left(\frac{5/3}{x-1} + \frac{1/3}{x+2}\right)dx = \boxed{\frac{5}{3}\ln|x-1| + \frac{1}{3}\ln|x+2| + C}.$$

(c) $\displaystyle\int \frac{dx}{\sqrt{4 - x^2}}$ — standard form $\int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\frac{x}{a}$ with $a = 2$:

$$\int \frac{dx}{\sqrt{4 - x^2}} = \boxed{\sin^{-1}\!\left(\frac{x}{2}\right) + C}.$$

(a) $\displaystyle\int_{0}^{\pi/2} \sin^4 x \cos^2 x \, dx$

Use the Wallis / beta-function formula for $\int_0^{\pi/2}\sin^m x\cos^n x\,dx$ with $m = 4$ (even), $n = 2$ (even):

$$\int_0^{\pi/2}\sin^m x\cos^n x\,dx = \frac{(m-1)!!\,(n-1)!!}{(m+n)!!}\cdot\frac{\pi}{2}.$$ $$= \frac{(3\cdot1)(1)}{6\cdot4\cdot2}\cdot\frac{\pi}{2} = \frac{3}{48}\cdot\frac{\pi}{2} = \frac{1}{16}\cdot\frac{\pi}{2} = \boxed{\frac{\pi}{32}}.$$

(Check: $3!!\,1!! = 3,\ 6!! = 6\cdot4\cdot2 = 48,\ \tfrac{3}{48}\cdot\tfrac{\pi}{2}=\tfrac{\pi}{32}$.)

(b) Area between $y = x^2$ and $y = x + 2$

Intersections: $x^2 = x + 2 \Rightarrow x^2 - x - 2 = 0 \Rightarrow (x-2)(x+1) = 0$, so $x = -1$ and $x = 2$. On $[-1, 2]$ the line lies above the parabola.

$$A = \int_{-1}^{2}\big[(x + 2) - x^2\big]\,dx = \left[\frac{x^2}{2} + 2x - \frac{x^3}{3}\right]_{-1}^{2}.$$

At $x = 2$: $2 + 4 - \tfrac{8}{3} = 6 - \tfrac{8}{3} = \tfrac{10}{3}$. At $x = -1$: $\tfrac12 - 2 + \tfrac13 = -\tfrac{7}{6}$.

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

(a) Improper integral of the first kind

Definition. An improper integral of the first kind has an infinite limit of integration, e.g. $\int_a^\infty f(x)\,dx$, defined as

$$\int_a^\infty f(x)\,dx = \lim_{t\to\infty}\int_a^{t} f(x)\,dx,$$

and it converges if this limit is finite.

Evaluation:

$$\int_{1}^{\infty}\frac{dx}{x^2} = \lim_{t\to\infty}\int_1^t x^{-2}\,dx = \lim_{t\to\infty}\left[-\frac{1}{x}\right]_1^t = \lim_{t\to\infty}\left(1 - \frac{1}{t}\right) = 1.$$

The integral converges to $1$.

(b) $\displaystyle\int_{0}^{1}\frac{dx}{\sqrt{x}}$

This is improper of the second kind (integrand unbounded at $x = 0$):

$$\int_0^1 x^{-1/2}\,dx = \lim_{\epsilon\to 0^+}\int_\epsilon^1 x^{-1/2}\,dx = \lim_{\epsilon\to0^+}\Big[2\sqrt{x}\Big]_\epsilon^1 = \lim_{\epsilon\to0^+}\big(2 - 2\sqrt{\epsilon}\big) = 2.$$

The integral converges and its value is $2$.

(a) Radius and interval of convergence of $\displaystyle\sum_{n=1}^{\infty} \frac{(x-2)^n}{n\,3^n}$

Apply the Ratio Test with $a_n = \dfrac{(x-2)^n}{n\,3^n}$:

$$\left|\frac{a_{n+1}}{a_n}\right| = \frac{|x-2|^{n+1}}{(n+1)3^{n+1}}\cdot\frac{n\,3^n}{|x-2|^n} = \frac{|x-2|}{3}\cdot\frac{n}{n+1} \to \frac{|x-2|}{3}.$$

Converges when $\dfrac{|x-2|}{3} < 1 \iff |x-2| < 3$, so radius of convergence $R = 3$, centred at $2$: the interval $(-1, 5)$ before checking endpoints.

Endpoints:

• $x = 5$: $\sum \dfrac{3^n}{n\,3^n} = \sum \dfrac{1}{n}$ — harmonic series, diverges.
• $x = -1$: $\sum \dfrac{(-3)^n}{n\,3^n} = \sum \dfrac{(-1)^n}{n}$ — alternating harmonic, converges.

Interval of convergence: $\boxed{[-1, 5)}$, radius $R = 3$.

(b) Maclaurin series of $\cos x$ to $x^4$

Using $f(0)=1,\ f'(0)=0,\ f''(0)=-1,\ f'''(0)=0,\ f^{(4)}(0)=1$:

$$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots = \boxed{1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots}.$$

Related rates: spherical balloon

Given: $\dfrac{dV}{dt} = 100\,\text{cm}^3/\text{s}$, $r = 5\,\text{cm}$, $V = \tfrac{4}{3}\pi r^3$, $S = 4\pi r^2$.

(a) Rate of change of radius

Differentiate $V = \tfrac{4}{3}\pi r^3$ with respect to $t$:

$$\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt} \implies \frac{dr}{dt} = \frac{1}{4\pi r^2}\frac{dV}{dt}.$$

At $r = 5$:

$$\frac{dr}{dt} = \frac{100}{4\pi (5)^2} = \frac{100}{100\pi} = \frac{1}{\pi} \approx 0.318\,\text{cm/s}.$$

(b) Rate of change of surface area

Differentiate $S = 4\pi r^2$:

$$\frac{dS}{dt} = 8\pi r\frac{dr}{dt}.$$

At $r = 5$ with $\dfrac{dr}{dt} = \dfrac{1}{\pi}$:

$$\frac{dS}{dt} = 8\pi (5)\cdot\frac{1}{\pi} = 40\,\text{cm}^2/\text{s}.$$

Answers: $\dfrac{dr}{dt} = \dfrac{1}{\pi}\approx 0.318\,\text{cm/s}$ and $\dfrac{dS}{dt} = 40\,\text{cm}^2/\text{s}.$

(a) Slope of tangent to $r = 2 + 2\sin\theta$ at $\theta = \pi/6$

For a polar curve, with $x = r\cos\theta,\ y = r\sin\theta$, the slope is

$$\frac{dy}{dx} = \frac{\dfrac{dr}{d\theta}\sin\theta + r\cos\theta}{\dfrac{dr}{d\theta}\cos\theta - r\sin\theta}.$$

Here $\dfrac{dr}{d\theta} = 2\cos\theta$. At $\theta = \tfrac{\pi}{6}$: $\sin\tfrac\pi6 = \tfrac12,\ \cos\tfrac\pi6 = \tfrac{\sqrt3}{2}$, $r = 2 + 2\cdot\tfrac12 = 3$, $\dfrac{dr}{d\theta} = 2\cdot\tfrac{\sqrt3}{2} = \sqrt3.$

$$\frac{dy}{dx} = \frac{\sqrt3\cdot\tfrac12 + 3\cdot\tfrac{\sqrt3}{2}}{\sqrt3\cdot\tfrac{\sqrt3}{2} - 3\cdot\tfrac12} = \frac{\tfrac{\sqrt3}{2} + \tfrac{3\sqrt3}{2}}{\tfrac{3}{2} - \tfrac{3}{2}} = \frac{2\sqrt3}{0}.$$

The denominator is $0$ while the numerator $\ne 0$, so the tangent is vertical at $\theta = \tfrac{\pi}{6}$ (slope undefined / $\infty$).

(b) Arc length of $r = e^{\theta}$ from $\theta = 0$ to $\theta = \pi$

Polar arc length: $L = \displaystyle\int_{\alpha}^{\beta}\sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta.$ With $r = e^\theta,\ \dfrac{dr}{d\theta} = e^\theta$:

$$L = \int_0^\pi \sqrt{e^{2\theta} + e^{2\theta}}\,d\theta = \int_0^\pi \sqrt{2}\,e^{\theta}\,d\theta = \sqrt2\Big[e^{\theta}\Big]_0^\pi = \sqrt{2}\,(e^{\pi} - 1).$$ $$L = \sqrt2\,(e^\pi - 1) \approx 1.414\times(23.14 - 1) \approx 31.3 \text{ units}.$$