Vardaan Learning Institute

Find the rate of change of the area of a circle with respect to its radius $r$ when $r = 5\text{ cm}$.Solution: $A = \pi r^2 \Rightarrow \frac{dA}{dr} = 2\pi r$. At $r = 5$, $\frac{dA}{dr} = 2\pi(5) = 10\pi\text{ cm}^2\text{/cm}$.

Find the rate of change of the area of a circle with respect to its radius $r$ when $r = 12\text{ cm}$.Solution: $\frac{dA}{dr} = 2\pi(12) = 24\pi\text{ cm}^2\text{/cm}$.

The radius of a circle is increasing uniformly at the rate of $3\text{ cm/s}$. Find the rate at which the area of the circle is increasing when the radius is $10\text{ cm}$.Solution: $\frac{dr}{dt} = 3$. $A = \pi r^2 \Rightarrow \frac{dA}{dt} = 2\pi r \cdot \frac{dr}{dt}$. At $r=10$, $\frac{dA}{dt} = 2\pi(10)(3) = 60\pi\text{ cm}^2\text{/s}$.

An edge of a variable cube is increasing at the rate of $3\text{ cm/s}$. How fast is the volume of the cube increasing when the edge is $10\text{ cm}$ long?Solution: $V = x^3 \Rightarrow \frac{dV}{dt} = 3x^2 \frac{dx}{dt}$. Here $\frac{dx}{dt} = 3, x=10$. $\frac{dV}{dt} = 3(10^2)(3) = 900\text{ cm}^3\text{/s}$.

A stone is dropped into a quiet lake and waves move in circles at a speed of $4\text{ cm/s}$. At the instant when the radius of the circular wave is $10\text{ cm}$, how fast is the enclosed area increasing?Solution: $\frac{dr}{dt} = 4$. $A = \pi r^2 \Rightarrow \frac{dA}{dt} = 2\pi r \frac{dr}{dt} = 2\pi(10)(4) = 80\pi\text{ cm}^2\text{/s}$.

The length $x$ of a rectangle is decreasing at the rate of $3\text{ cm/min}$ and the width $y$ is increasing at the rate of $2\text{ cm/min}$. When $x = 10\text{ cm}$ and $y = 6\text{ cm}$, find the rate of change of the perimeter.Solution: $\frac{dx}{dt} = -3$, $\frac{dy}{dt} = 2$. $P = 2(x+y) \Rightarrow \frac{dP}{dt} = 2(\frac{dx}{dt} + \frac{dy}{dt}) = 2(-3 + 2) = -2\text{ cm/min}$.

Using the data from the previous question (Q6), find the rate of change of the area of the rectangle.Solution: $A = xy \Rightarrow \frac{dA}{dt} = x\frac{dy}{dt} + y\frac{dx}{dt} = 10(2) + 6(-3) = 20 - 18 = 2\text{ cm}^2\text{/min}$.

A balloon, which always remains spherical, has a variable diameter $\frac{3}{2}(2x + 1)$. Find the rate of change of its volume with respect to $x$.Solution: Radius $r = \frac{3}{4}(2x+1)$. $V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi \left[ \frac{3}{4}(2x+1) \right]^3 = \frac{9}{16}\pi (2x+1)^3$. $\frac{dV}{dx} = \frac{9}{16}\pi \cdot 3(2x+1)^2 \cdot 2 = \frac{27}{8}\pi (2x+1)^2$.

A spherical bubble is expanding so that its radius is increasing at the rate of $0.02\text{ cm/s}$. Find the rate of increase of its surface area when the radius is $5\text{ cm}$.Solution: $\frac{dr}{dt} = 0.02$. $S = 4\pi r^2 \Rightarrow \frac{dS}{dt} = 8\pi r \frac{dr}{dt} = 8\pi(5)(0.02) = 0.8\pi\text{ cm}^2\text{/s}$.

The total cost $C(x)$ in Rupees associated with the production of $x$ units of an item is given by $C(x) = 0.005x^3 - 0.02x^2 + 30x + 5000$. Find the marginal cost when 3 units are produced.Solution: Marginal Cost (MC) $= C'(x) = 0.015x^2 - 0.04x + 30$. At $x=3$, MC $= 0.015(9) - 0.04(3) + 30 = 0.135 - 0.12 + 30 = 30.015\text{ Rupees}$.

The total revenue in Rupees received from the sale of $x$ units of a product is given by $R(x) = 13x^2 + 26x + 15$. Find the marginal revenue when $x = 7$.Solution: Marginal Revenue (MR) $= R'(x) = 26x + 26$. At $x=7$, MR $= 26(7) + 26 = 182 + 26 = 208\text{ Rupees}$.

A particle moves along the curve $y = \frac{2}{3}x^3 + 1$. Find the points on the curve at which the y-coordinate is changing twice as fast as the x-coordinate.Solution: $\frac{dy}{dt} = 2\frac{dx}{dt}$. Differentiating curve: $\frac{dy}{dt} = 2x^2 \frac{dx}{dt}$. Therefore, $2x^2 = 2 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1$. If $x=1, y = 5/3$. If $x=-1, y = 1/3$. Points are $(1, 5/3)$ and $(-1, 1/3)$.

A ladder $5\text{ m}$ long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of $2\text{ cm/s}$. How fast is its height on the wall decreasing when the foot of the ladder is $4\text{ m}$ away from the wall?Solution: $x^2 + y^2 = 25$. $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$. Here $\frac{dx}{dt} = 0.02\text{ m/s}$ ($2\text{ cm/s}$), $x=4\text{m}$. Then $y = \sqrt{25-16} = 3\text{m}$. $2(4)(0.02) + 2(3)\frac{dy}{dt} = 0 \Rightarrow 6\frac{dy}{dt} = -0.16 \Rightarrow \frac{dy}{dt} = -0.16 / 6 = -0.0267\text{ m/s} = -8/3\text{ cm/s}$. Decreasing at $8/3\text{ cm/s}$.

The surface area of a spherical balloon is increasing at the rate of $2\text{ cm}^2\text{/s}$. Find the rate of change of its volume when the radius is $6\text{ cm}$.Solution: $\frac{dS}{dt} = 2 \Rightarrow 8\pi r \frac{dr}{dt} = 2 \Rightarrow \frac{dr}{dt} = \frac{2}{8\pi(6)} = \frac{1}{24\pi}$. Now $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} = 4\pi (36) (\frac{1}{24\pi}) = \frac{144}{24} = 6\text{ cm}^3\text{/s}$.

Sand is pouring from a pipe at the rate of $12\text{ cm}^3\text{/s}$. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is $4\text{ cm}$?Solution: $h = \frac{r}{6} \Rightarrow r = 6h$. $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (36h^2) h = 12\pi h^3$. $\frac{dV}{dt} = 36\pi h^2 \frac{dh}{dt}$. Given $\frac{dV}{dt} = 12, h=4$. $12 = 36\pi (16) \frac{dh}{dt} \Rightarrow \frac{dh}{dt} = \frac{12}{576\pi} = \frac{1}{48\pi}\text{ cm/s}$.

Show that the function $f(x) = 3x + 17$ is strictly increasing on $\mathbb{R}$.Solution: $f'(x) = 3$. Since $3 > 0$ for all real $x$, $f(x)$ is strictly increasing on $\mathbb{R}$.

Show that the function $f(x) = e^{2x}$ is strictly increasing on $\mathbb{R}$.Solution: $f'(x) = 2e^{2x}$. Since $e^{2x} > 0$ for all $x$, $2e^{2x} > 0$, hence strictly increasing.

Find the intervals in which the function $f(x) = x^2 - 4x + 6$ is strictly increasing.Solution: $f'(x) = 2x - 4$. For strictly increasing, $2x - 4 > 0 \Rightarrow 2x > 4 \Rightarrow x > 2$. Interval is $(2, \infty)$.

Find the intervals in which the function $f(x) = x^2 - 4x + 6$ is strictly decreasing.Solution: From previous Q, $2x - 4 < 0 \Rightarrow x < 2$. Interval is $(-\infty, 2)$.

Find the intervals in which the function $f(x) = 2x^2 - 3x$ is strictly increasing or decreasing.Solution: $f'(x) = 4x - 3$. Strictly increasing for $x > 3/4$ i.e., $(3/4, \infty)$. Strictly decreasing for $x < 3/4$ i.e., $(-\infty, 3/4)$.

Determine the intervals where $f(x) = 2x^3 - 3x^2 - 36x + 7$ is strictly increasing.Solution: $f'(x) = 6x^2 - 6x - 36 = 6(x^2 - x - 6) = 6(x-3)(x+2)$. Critical points are $-2, 3$. Using wavy curve, $f'(x) > 0$ on $(-\infty, -2) \cup (3, \infty)$.

Determine the intervals where $f(x) = 2x^3 - 3x^2 - 36x + 7$ is strictly decreasing.Solution: From Q21, $f'(x) < 0$ in the middle interval. So, strictly decreasing on $(-2, 3)$.

Find the intervals in which $f(x) = 4x^3 - 6x^2 - 72x + 30$ is strictly increasing.Solution: $f'(x) = 12x^2 - 12x - 72 = 12(x^2 - x - 6) = 12(x-3)(x+2)$. Strictly increasing on $(-\infty, -2) \cup (3, \infty)$.

Find the intervals in which $f(x) = -2x^3 - 9x^2 - 12x + 1$ is strictly decreasing.Solution: $f'(x) = -6x^2 - 18x - 12 = -6(x^2 + 3x + 2) = -6(x+1)(x+2)$. Roots are $-1, -2$. By sign chart, it is negative outside the roots. Strictly decreasing on $(-\infty, -2) \cup (-1, \infty)$.

Prove that the function given by $f(x) = \cos x$ is strictly decreasing in $(0, \pi)$.Solution: $f'(x) = -\sin x$. In $(0, \pi)$, $\sin x > 0$, so $-\sin x < 0$. Hence strictly decreasing.

Prove that the function given by $f(x) = \cos x$ is strictly increasing in $(\pi, 2\pi)$.Solution: $f'(x) = -\sin x$. In $(\pi, 2\pi)$, $\sin x < 0$, so $-\sin x > 0$. Hence strictly increasing.

Prove that $f(x) = x - \sin x$ is strictly increasing for all real $x$.Solution: $f'(x) = 1 - \cos x$. Since $\cos x \le 1$ for all $x$, $1 - \cos x \ge 0$. It is $0$ only at discrete points $2n\pi$, not an interval. Hence strictly increasing on $\mathbb{R}$.

Find the intervals in which the function $f(x) = \sin x + \cos x$, $0 \le x \le 2\pi$ is strictly increasing.Solution: $f'(x) = \cos x - \sin x$. $f'(x) > 0 \Rightarrow \cos x > \sin x$. In $[0, 2\pi]$, this holds for $x \in [0, \pi/4) \cup (5\pi/4, 2\pi]$.

Find the intervals in which the function $f(x) = \sin 3x$, $x \in [0, \pi/2]$ is strictly increasing or decreasing.Solution: $f'(x) = 3\cos 3x$. In $[0, \pi/2]$, $3x \in [0, 3\pi/2]$. $\cos 3x > 0$ when $3x \in [0, \pi/2) \Rightarrow x \in [0, \pi/6)$ (Strictly increasing). $\cos 3x < 0$ when $3x \in (\pi/2, 3\pi/2) \Rightarrow x \in (\pi/6, \pi/2]$ (Strictly decreasing).

Show that $f(x) = \ln(\sin x)$ is strictly increasing on $(0, \pi/2)$.Solution: $f'(x) = \frac{1}{\sin x} \cdot \cos x = \cot x$. In $(0, \pi/2)$, $\cot x > 0$. Hence strictly increasing.

Show that $f(x) = \ln(\cos x)$ is strictly decreasing on $(0, \pi/2)$.Solution: $f'(x) = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x$. In $(0, \pi/2)$, $\tan x > 0 \Rightarrow -\tan x < 0$. Hence strictly decreasing.

Find the slope of the tangent to the curve $y = 3x^4 - 4x$ at $x = 4$.Solution: $\frac{dy}{dx} = 12x^3 - 4$. At $x=4$, slope $= 12(64) - 4 = 768 - 4 = 764$.

Find the slope of the tangent to the curve $y = \frac{x-1}{x-2}$ at $x = 10$.Solution: $\frac{dy}{dx} = \frac{(x-2)(1) - (x-1)(1)}{(x-2)^2} = \frac{-1}{(x-2)^2}$. At $x=10$, slope $= \frac{-1}{64}$.

Find the slope of the normal to the curve $y = x^3 - 3x + 2$ at the point whose x-coordinate is 3.Solution: $\frac{dy}{dx} = 3x^2 - 3$. At $x=3$, $m_t = 3(9) - 3 = 24$. Slope of normal $m_n = -1/24$.

Find the point at which the tangent to the curve $y = \sqrt{4x-3} - 1$ has its slope equal to $\frac{2}{3}$.Solution: $\frac{dy}{dx} = \frac{4}{2\sqrt{4x-3}} = \frac{2}{\sqrt{4x-3}}$. We set $\frac{2}{\sqrt{4x-3}} = \frac{2}{3} \Rightarrow \sqrt{4x-3} = 3 \Rightarrow 4x-3 = 9 \Rightarrow x=3$. When $x=3, y = \sqrt{9}-1 = 2$. Point is $(3, 2)$.

Find points on the curve $\frac{x^2}{9} + \frac{y^2}{16} = 1$ at which the tangents are parallel to the x-axis.Solution: Diff w.r.t $x$: $\frac{2x}{9} + \frac{2y}{16}\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \frac{-16x}{9y}$. Parallel to x-axis $\Rightarrow \frac{dy}{dx} = 0 \Rightarrow x = 0$. If $x=0$, $\frac{y^2}{16} = 1 \Rightarrow y = \pm 4$. Points are $(0, 4), (0, -4)$.

Find points on the curve $\frac{x^2}{9} + \frac{y^2}{16} = 1$ at which the tangents are parallel to the y-axis.Solution: Parallel to y-axis means denominator of $\frac{dy}{dx}$ is $0$, so $y = 0$. If $y=0$, $\frac{x^2}{9} = 1 \Rightarrow x = \pm 3$. Points are $(3, 0), (-3, 0)$.

Find the equation of all lines having slope $-1$ that are tangents to the curve $y = \frac{1}{x-1}$.Solution: $\frac{dy}{dx} = \frac{-1}{(x-1)^2}$. Set this to $-1$: $\frac{-1}{(x-1)^2} = -1 \Rightarrow (x-1)^2 = 1 \Rightarrow x-1 = \pm 1 \Rightarrow x=2, x=0$. If $x=2, y=1$. If $x=0, y=-1$. Lines: $y-1 = -1(x-2) \Rightarrow x+y=3$, and $y+1 = -1(x-0) \Rightarrow x+y=-1$.

Find the equation of the tangent to the curve $y = x^4 - 6x^3 + 13x^2 - 10x + 5$ at $(0, 5)$.Solution: $\frac{dy}{dx} = 4x^3 - 18x^2 + 26x - 10$. At $x=0$, $m = -10$. Eq: $y - 5 = -10(x - 0) \Rightarrow 10x + y - 5 = 0$.

Find the equation of the normal to the curve $y = x^4 - 6x^3 + 13x^2 - 10x + 5$ at $(0, 5)$.Solution: From Q39, $m_t = -10$, so $m_n = 1/10$. Eq: $y - 5 = \frac{1}{10}(x - 0) \Rightarrow x - 10y + 50 = 0$.

Find the equation of the tangent to the curve $y = \sin^2 x$ at $x = \pi/4$.Solution: At $x=\pi/4$, $y = (\frac{1}{\sqrt{2}})^2 = 1/2$. $\frac{dy}{dx} = 2\sin x \cos x = \sin 2x$. At $x=\pi/4$, slope $= \sin(\pi/2) = 1$. Eq: $y - \frac{1}{2} = 1(x - \pi/4)$.

Find the equation of the tangent to the curve $y = x^3 - 3x + 2$ which is parallel to the line $y = 9x - 5$.Solution: Slope must be 9. $\frac{dy}{dx} = 3x^2 - 3 = 9 \Rightarrow 3x^2 = 12 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$. If $x=2, y=4$. If $x=-2, y=0$. Tangents: $y - 4 = 9(x - 2)$ and $y - 0 = 9(x + 2)$.

Find the angle of intersection of the curves $y = x^2$ and $x^2 = y$ at $(1, 1)$.Solution: First curve $y=x^2 \Rightarrow \frac{dy}{dx} = 2x$. At $(1,1), m_1 = 2$. (Wait, $x^2=y$ is the same curve. The question likely meant $y^2 = x$). Let's assume $y^2=x$. Then $2y\frac{dy}{dx} = 1 \Rightarrow \frac{dy}{dx} = \frac{1}{2y}$. At $(1,1), m_2 = 1/2$. Angle $\tan \theta = |\frac{m_1 - m_2}{1+m_1 m_2}| = |\frac{2 - 1/2}{1+1}| = \frac{3/2}{2} = 3/4$. So $\theta = \tan^{-1}(3/4)$.

Show that the curves $x = y^2$ and $xy = k$ cut at right angles if $8k^2 = 1$.Solution: Intersect: $y^3 = k \Rightarrow y = k^{1/3}, x = k^{2/3}$. Slopes: $x=y^2 \Rightarrow m_1 = \frac{1}{2y}$. $xy=k \Rightarrow m_2 = \frac{-y}{x}$. For orthogonality, $m_1 m_2 = -1 \Rightarrow (\frac{1}{2y})(\frac{-y}{x}) = -1 \Rightarrow \frac{-1}{2x} = -1 \Rightarrow x = 1/2$. Since $x = k^{2/3}$, $k^{2/3} = 1/2 \Rightarrow (k^{2/3})^3 = (1/2)^3 \Rightarrow k^2 = 1/8 \Rightarrow 8k^2 = 1$.

Find the equation of the normal to the curve $x^2 = 4y$ which passes through the point $(1, 2)$.Solution: Let point of contact be $(h, k)$. $h^2 = 4k$. Slope of tangent $= 2x/4 = x/2 \Rightarrow h/2$. Slope of normal $= -2/h$. Normal Eq: $y - k = \frac{-2}{h}(x - h)$. It passes through $(1,2)$, so $2 - k = \frac{-2}{h}(1 - h)$. Substitute $k=h^2/4$: $2 - h^2/4 = -2/h + 2 \Rightarrow -h^2/4 = -2/h \Rightarrow h^3 = 8 \Rightarrow h=2$. Then $k=1$. Point is $(2, 1)$. Normal slope $= -2/2 = -1$. Eq: $y - 1 = -1(x - 2) \Rightarrow x+y=3$.

Prove that the curve $\left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n = 2$ touches the straight line $\frac{x}{a} + \frac{y}{b} = 2$ at the point $(a, b)$, whatever be the value of $n$.Solution: First verify point $(a, b)$ lies on curve: $(1)^n + (1)^n = 2$. True. Now slope: diff curve gives $n\left(\frac{x}{a}\right)^{n-1} \frac{1}{a} + n\left(\frac{y}{b}\right)^{n-1} \frac{1}{b} \frac{dy}{dx} = 0$. At $(a,b)$, this gives $\frac{n}{a} + \frac{n}{b} \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \frac{-b}{a}$. Eq of tangent: $y - b = \frac{-b}{a}(x - a) \Rightarrow \frac{y}{b} - 1 = -\frac{x}{a} + 1 \Rightarrow \frac{x}{a} + \frac{y}{b} = 2$. Since tangent matches given line, they touch.

Using differentials, find the approximate value of $\sqrt{36.6}$.Solution: Let $y = \sqrt{x}$. Take $x = 36, \Delta x = 0.6$. $dy = \frac{1}{2\sqrt{x}} \Delta x = \frac{1}{12}(0.6) = 0.05$. Approx value $= \sqrt{36} + 0.05 = 6.05$.

Using differentials, find the approximate value of $\sqrt{25.3}$.Solution: $x = 25, \Delta x = 0.3$. $dy = \frac{1}{10}(0.3) = 0.03$. Value $= 5 + 0.03 = 5.03$.

Using differentials, find the approximate value of $(26)^{1/3}$.Solution: Let $y = x^{1/3}$. Take $x=27, \Delta x = -1$. $dy = \frac{1}{3x^{2/3}}\Delta x = \frac{1}{3(9)}(-1) = -1/27 \approx -0.037$. Value $= 3 - 0.037 = 2.963$.

Using differentials, find the approximate value of $(0.999)^{1/10}$.Solution: Let $y = x^{1/10}$. Take $x=1, \Delta x = -0.001$. $dy = \frac{1}{10}x^{-9/10}\Delta x = \frac{1}{10}(-0.001) = -0.0001$. Value $= 1 - 0.0001 = 0.9999$.

Using differentials, find the approximate value of $\sqrt{0.48}$.Solution: Take $x = 0.49, \Delta x = -0.01$. $\sqrt{x} = 0.7$. $dy = \frac{1}{2(0.7)}(-0.01) = \frac{-0.01}{1.4} = -0.0071$. Value $= 0.7 - 0.0071 = 0.6929$.

Find the approximate change in the volume $V$ of a cube of side $x$ meters caused by increasing the side by $1\%$.Solution: $V = x^3 \Rightarrow dV = 3x^2 dx$. Given $dx = 0.01x$. $dV = 3x^2(0.01x) = 0.03x^3$. Change is $0.03x^3\text{ m}^3$.

Find the approximate change in the surface area of a cube of side $x$ meters caused by decreasing the side by $1\%$.Solution: $S = 6x^2 \Rightarrow dS = 12x dx$. Given $dx = -0.01x$. $dS = 12x(-0.01x) = -0.12x^2$. Decrease of $0.12x^2\text{ m}^2$.

If the radius of a sphere is measured as $9\text{ cm}$ with an error of $0.03\text{ cm}$, find the approximate error in calculating its volume.Solution: $V = \frac{4}{3}\pi r^3 \Rightarrow dV = 4\pi r^2 dr$. $dV = 4\pi (81)(0.03) = 9.72\pi\text{ cm}^3$.

If the radius of a sphere is measured as $7\text{ m}$ with an error of $0.02\text{ m}$, find the approximate error in calculating its surface area.Solution: $S = 4\pi r^2 \Rightarrow dS = 8\pi r dr$. $dS = 8\pi (7)(0.02) = 1.12\pi\text{ m}^2$.

Find the approximate value of $f(3.02)$, where $f(x) = 3x^2 + 5x + 3$.Solution: Let $x=3, \Delta x=0.02$. $f(3) = 3(9)+15+3 = 45$. $f'(x) = 6x+5 \Rightarrow f'(3) = 23$. $dy = f'(3)\Delta x = 23(0.02) = 0.46$. Value $= 45 + 0.46 = 45.46$.

Find the approximate value of $f(2.01)$, where $f(x) = 4x^2 + 5x + 2$.Solution: Let $x=2, \Delta x=0.01$. $f(2) = 16+10+2 = 28$. $f'(x) = 8x+5 \Rightarrow f'(2) = 21$. $dy = 21(0.01) = 0.21$. Value $= 28.21$.

Find the percentage error in calculating the volume of a cubical box if an error of $1\%$ is made in measuring the length of its edge.Solution: $V = x^3 \Rightarrow \frac{dV}{V} \times 100 = \frac{3x^2 dx}{x^3} \times 100 = 3 (\frac{dx}{x} \times 100) = 3(1\%) = 3\%$.

The pressure $P$ and volume $V$ of a gas are connected by the relation $PV^{1.4} = \text{const}$. Find the approximate percentage change in $V$ when $P$ is increased by $0.5\%$.Solution: Take ln: $\ln P + 1.4 \ln V = \ln C$. Diff: $\frac{dP}{P} + 1.4\frac{dV}{V} = 0$. $\frac{dV}{V} = -\frac{1}{1.4}\frac{dP}{P}$. Given $\frac{dP}{P} = 0.5\%$. $\frac{dV}{V} = -\frac{0.5\%}{1.4} \approx -0.357\%$. Decreases by $0.357\%$.

If the error in measuring the radius of a circle is $1.5\%$, find the percentage error in computing its area.Solution: $A = \pi r^2 \Rightarrow \frac{dA}{A} = 2\frac{dr}{r}$. $2(1.5\%) = 3\%$.

Show that the relative error in computing the volume of a sphere is three times the relative error in measuring its radius.Solution: $V = \frac{4}{3}\pi r^3 \Rightarrow dV = 4\pi r^2 dr$. Relative error $\frac{dV}{V} = \frac{4\pi r^2 dr}{\frac{4}{3}\pi r^3} = 3\frac{dr}{r}$. Proved.

Find the maximum and minimum values, if any, of the function $f(x) = (2x - 1)^2 + 3$.Solution: $(2x-1)^2 \ge 0 \Rightarrow (2x-1)^2 + 3 \ge 3$. Minimum value is 3 at $x=1/2$. Maximum does not exist.

Find the maximum and minimum values, if any, of the function $f(x) = -(x - 1)^2 + 10$.Solution: $-(x-1)^2 \le 0 \Rightarrow -(x-1)^2 + 10 \le 10$. Maximum value is 10 at $x=1$. Minimum does not exist.

Find the maximum and minimum values of $f(x) = |x + 2| - 1$.Solution: $|x+2| \ge 0 \Rightarrow |x+2| - 1 \ge -1$. Minimum is $-1$ at $x=-2$. Maximum does not exist.

Find the local maxima and local minima, if any, of the function $f(x) = x^2$.Solution: $f'(x) = 2x = 0 \Rightarrow x=0$. $f''(x) = 2 > 0$. So $x=0$ is local minimum point. Min value $= 0$.

Find the local maxima and local minima for $f(x) = x^3 - 3x$.Solution: $f'(x) = 3x^2 - 3 = 0 \Rightarrow x = \pm 1$. $f''(x) = 6x$. $f''(1) = 6>0$ (Local Min). $f''(-1) = -6<0$ (Local Max). Max value $= 2$. Min value $= -2$.

Find the local maxima and local minima for $f(x) = 3x^4 + 4x^3 - 12x^2 + 12$.Solution: $f'(x) = 12x^3 + 12x^2 - 24x = 12x(x^2+x-2) = 12x(x+2)(x-1) = 0$. Critical pts: $0, 1, -2$. $f''(x) = 36x^2 + 24x - 24$. $f''(0) = -24 < 0$ (Max). $f''(1) = 36 > 0$ (Min). $f''(-2) = 144-48-24 > 0$ (Min).

Find all the points of local maxima and local minima of the function $f(x) = \sin x - \cos x$, where $0 < x < 2\pi$.Solution: $f'(x) = \cos x + \sin x = 0 \Rightarrow \tan x = -1$. In $(0, 2\pi)$, $x = 3\pi/4, 7\pi/4$. $f''(x) = -\sin x + \cos x$. $f''(3\pi/4) = -1/\sqrt{2} - 1/\sqrt{2} < 0$ (Local Max). $f''(7\pi/4) = 1/\sqrt{2} + 1/\sqrt{2} > 0$ (Local Min).

Find the absolute maximum and absolute minimum values of $f(x) = x^3$ on the interval $[-2, 2]$.Solution: $f'(x) = 3x^2 = 0 \Rightarrow x=0$. Evaluate at points: $f(-2)=-8, f(0)=0, f(2)=8$. Abs max $= 8$, Abs min $= -8$.

Find the absolute maximum and absolute minimum values of $f(x) = 4x - \frac{1}{2}x^2$ on the interval $[-2, 4.5]$.Solution: $f'(x) = 4 - x = 0 \Rightarrow x=4$. Evaluate: $f(-2) = -8 - 2 = -10$. $f(4) = 16 - 8 = 8$. $f(4.5) = 18 - \frac{1}{2}(20.25) = 7.875$. Abs max $= 8$, Abs min $= -10$.

Find the absolute maximum and minimum values of $f(x) = \sin x + \cos x$ in the interval $[0, \pi]$.Solution: $f'(x) = \cos x - \sin x = 0 \Rightarrow x = \pi/4$. Evaluate: $f(0) = 1$. $f(\pi/4) = \sqrt{2}$. $f(\pi) = -1$. Abs max $= \sqrt{2}$, Abs min $= -1$.

Find the maximum profit that a company can make, if the profit function is given by $p(x) = 41 - 24x - 18x^2$.Solution: $p'(x) = -24 - 36x = 0 \Rightarrow x = -2/3$. $p''(-2/3) = -36 < 0$ (Max). Max profit $= 41 - 24(-2/3) - 18(4/9) = 41 + 16 - 8 = 49$.

Prove that the function $f(x) = e^x$ does not have any maxima or minima.Solution: $f'(x) = e^x$. Since $e^x$ is never zero for any real $x$, there are no critical points. Hence, no maxima or minima.

Show that the function $f(x) = x^3 - 6x^2 + 12x - 8$ has a point of inflection at $x = 2$.Solution: $f'(x) = 3x^2 - 12x + 12 = 3(x-2)^2$. $f'(x) \ge 0$ everywhere, and $f'(2)=0$. Since sign of $f'$ doesn't change, $x=2$ is a point of inflection. Alternatively $f''(2)=0$ and $f'''(2)=6 \neq 0$.

Find the points of local maxima, local minima, and the point of inflection for $f(x) = x^5 - 5x^4 + 5x^3 - 1$.Solution: $f'(x) = 5x^4 - 20x^3 + 15x^2 = 5x^2(x^2 - 4x + 3) = 5x^2(x-1)(x-3) = 0$. Roots: $0, 1, 3$. Number line for $f'$: + $(-\infty, 0)$, + $(0, 1)$, - $(1, 3)$, + $(3, \infty)$. At $x=0$, no sign change $\Rightarrow$ Inflection. At $x=1$, + to - $\Rightarrow$ Local Max. At $x=3$, - to + $\Rightarrow$ Local Min.

At what points in the interval $[0, 2\pi]$, does the function $\sin 2x$ attain its maximum value?Solution: $f(x) = \sin 2x$. Max value of sine is 1. $\sin 2x = 1 \Rightarrow 2x = \pi/2, 5\pi/2 \Rightarrow x = \pi/4, 5\pi/4$. These lie in $[0, 2\pi]$.

Find two positive numbers whose sum is 15 and the sum of whose squares is minimum.Solution: $x + y = 15$. Minimize $S = x^2 + y^2 = x^2 + (15-x)^2$. $S' = 2x + 2(15-x)(-1) = 4x - 30 = 0 \Rightarrow x = 7.5$. $S'' = 4 > 0$ (Min). Numbers are $7.5, 7.5$.

Find two positive numbers $x$ and $y$ such that their sum is 35 and the product $x^2 y^5$ is a maximum.Solution: $x = 35 - y$. Maximize $P = (35-y)^2 y^5$. $P' = 2(35-y)(-1)y^5 + 5y^4(35-y)^2 = y^4(35-y)[-2y + 175 - 5y] = y^4(35-y)(175 - 7y) = 0$. $y=25$ (since positive). $x = 10$.

Find two positive numbers whose product is 100 and whose sum is minimum.Solution: $xy = 100 \Rightarrow y = 100/x$. Minimize $S = x + 100/x$. $S' = 1 - 100/x^2 = 0 \Rightarrow x^2 = 100 \Rightarrow x = 10$ (positive). $y = 10$.

A wire of length $28\text{ m}$ is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the lengths of the two pieces so that the combined area of the square and the circle is minimum?Solution: Let square piece be $x$ and circle piece be $28-x$. Square side $a = x/4$, Circle radius $r = (28-x)/(2\pi)$. $A = x^2/16 + \pi((28-x)/2\pi)^2 = x^2/16 + (28-x)^2/(4\pi)$. $A' = x/8 - 2(28-x)/(4\pi) = 0 \Rightarrow \pi x = 4(56 - 2x) \Rightarrow x(\pi+8) = 224 \Rightarrow x = 224/(\pi+8)$. Lengths are $112/(\pi+4)$ and remaining. (Wait: $4(56-2x) \Rightarrow \pi x = 224 - 8x \Rightarrow x = 224/(\pi+8)$).

Prove that of all rectangles with a given perimeter, the square has the maximum area.Solution: $P = 2(x+y) \Rightarrow y = P/2 - x$. Maximize $A = x(P/2 - x)$. $A' = P/2 - 2x = 0 \Rightarrow x = P/4$. Then $y = P/4$. Since $x=y$, it is a square.

Prove that of all rectangles with a given fixed area, the square has the minimum perimeter.Solution: $xy = A \Rightarrow y = A/x$. Minimize $P = 2(x + A/x)$. $P' = 2(1 - A/x^2) = 0 \Rightarrow x^2 = A \Rightarrow x = \sqrt{A}$. Then $y = \sqrt{A}$. Hence a square.

A square piece of tin of side $18\text{ cm}$ is to be made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible?Solution: Cut square of side $x$. Box dimensions: $(18-2x), (18-2x), x$. $V = x(18-2x)^2$. $V' = (18-2x)^2 + x(2)(18-2x)(-2) = (18-2x)[18-2x-4x] = (18-2x)(18-6x) = 0$. $x=9$ (rejected) or $x=3$. Cut side $= 3\text{ cm}$.

A rectangular sheet of tin $45\text{ cm}$ by $24\text{ cm}$ is to be made into a box without top, by cutting off squares from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum?Solution: $V = x(45-2x)(24-2x) = 4x^3 - 138x^2 + 1080x$. $V' = 12x^2 - 276x + 1080 = 12(x^2 - 23x + 90) = 12(x-5)(x-18) = 0$. $x=18$ (rejected). $x=5$. Square side is $5\text{ cm}$.

Show that the volume of the largest cone that can be inscribed in a sphere of radius $R$ is $\frac{8}{27}$ of the volume of the sphere.Solution: Let cone height be $h = R+x$, radius $r = \sqrt{R^2 - x^2}$. $V_c = \frac{1}{3}\pi (R^2-x^2)(R+x)$. $V_c' = \frac{\pi}{3}[-2x(R+x) + (R^2-x^2)] = \frac{\pi}{3}(R+x)[-2x + R - x] = \frac{\pi}{3}(R+x)(R-3x) = 0 \Rightarrow x = R/3$. Max Vol $= \frac{\pi}{3}(R^2 - R^2/9)(4R/3) = \frac{32\pi R^3}{81} = \frac{8}{27} (\frac{4}{3}\pi R^3)$. Proved.

Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius $R$ is $\frac{2R}{\sqrt{3}}$.Solution: Cyl height $= 2x$, radius $r = \sqrt{R^2 - x^2}$. $V = \pi(R^2-x^2)(2x) = 2\pi(R^2 x - x^3)$. $V' = 2\pi(R^2 - 3x^2) = 0 \Rightarrow x = R/\sqrt{3}$. Height $= 2x = \frac{2R}{\sqrt{3}}$.

Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is $\tan^{-1}\sqrt{2}$.Solution: Let slant height $l$, angle $\theta$. $r = l \sin\theta, h = l \cos\theta$. $V = \frac{1}{3}\pi r^2 h = \frac{\pi l^3}{3} \sin^2\theta \cos\theta$. $V' = \frac{\pi l^3}{3}[2\sin\theta\cos^2\theta - \sin^3\theta] = 0 \Rightarrow 2\cos^2\theta = \sin^2\theta \Rightarrow \tan^2\theta = 2 \Rightarrow \theta = \tan^{-1}\sqrt{2}$.

An open cylindrical tank whose base is a circle is to be constructed of metal sheet so as to contain a volume of $\pi a^3$ cubic units. Find the dimensions (radius and height) so that the sheet required is minimum.Solution: $V = \pi r^2 h = \pi a^3 \Rightarrow h = a^3/r^2$. Surface $S = \pi r^2 + 2\pi r h = \pi r^2 + 2\pi r(a^3/r^2) = \pi r^2 + 2\pi a^3 / r$. $S' = 2\pi r - 2\pi a^3 / r^2 = 0 \Rightarrow r^3 = a^3 \Rightarrow r = a$. Then $h = a^3/a^2 = a$. Radius $= a$, height $= a$.

Show that a right circular cylinder which is open at the top, and has a given surface area and maximum volume, is such that its height is equal to the radius of the base.Solution: $S = \pi r^2 + 2\pi r h \Rightarrow h = \frac{S-\pi r^2}{2\pi r}$. $V = \pi r^2 \frac{S-\pi r^2}{2\pi r} = \frac{1}{2}(Sr - \pi r^3)$. $V' = \frac{1}{2}(S - 3\pi r^2) = 0 \Rightarrow S = 3\pi r^2$. Sub back into $h$: $h = \frac{3\pi r^2 - \pi r^2}{2\pi r} = \frac{2\pi r^2}{2\pi r} = r$. Proved.

Find the maximum area of an isosceles triangle inscribed in a given ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with its vertex at one end of the major axis.Solution: Let vertex be $(a,0)$. Base points $(a\cos\theta, \pm b\sin\theta)$. Base $= 2b\sin\theta$, Height $= a - a\cos\theta$. $A = \frac{1}{2}(2b\sin\theta)(a(1-\cos\theta)) = ab\sin\theta(1-\cos\theta)$. $A' = ab[\cos\theta(1-\cos\theta) + \sin^2\theta] = ab[\cos\theta - \cos^2\theta + 1 - \cos^2\theta] = ab[1+\cos\theta-2\cos^2\theta] = 0$. $(2\cos\theta+1)(\cos\theta-1)=0 \Rightarrow \cos\theta = -1/2 \Rightarrow \theta = 2\pi/3$. Area $= ab(\frac{\sqrt{3}}{2})(1 - (-1/2)) = \frac{3\sqrt{3}}{4}ab$.