Rate of Change of Quantities Questions in English

(C) Marginal revenue is defined as the rate of change of total revenue with respect to the number of units sold.
$\therefore$ Marginal Revenue $(MR) = \frac{dR}{dx} = \frac{d}{dx}(3x^2 + 36x + 5) = 6x + 36$.
Now,we calculate the marginal revenue when $x = 15$:
$MR = 6(15) + 36 = 90 + 36 = 126$.
Hence,the required marginal revenue is $Rs. 126$.
The correct answer is $C$.

(A) Let $v$ be the velocity of the car at $t$ seconds.
The distance covered is given by $x = t^{2}(2 - \frac{t}{3}) = 2t^{2} - \frac{t^{3}}{3}$.
The velocity $v$ is the rate of change of distance with respect to time,so $v = \frac{dx}{dt}$.
$v = \frac{d}{dt}(2t^{2} - \frac{t^{3}}{3}) = 4t - t^{2} = t(4 - t)$.
The car stops at point $Q$ when the velocity $v = 0$.
Setting $v = 0$,we get $t(4 - t) = 0$,which implies $t = 0$ or $t = 4$.
Since the car starts at $t = 0$ at point $P$,it reaches point $Q$ at $t = 4$ seconds.
The distance between $P$ and $Q$ is the value of $x$ at $t = 4$.
$x(4) = 4^{2}(2 - \frac{4}{3}) = 16(\frac{6-4}{3}) = 16(\frac{2}{3}) = \frac{32}{3} \, m$.

(A) Let $r$ be the radius and $h$ be the depth of the water at any time $t$. The semi-vertical angle $\alpha$ is given by $\tan \alpha = 0.5 = \frac{1}{2}$.
From the geometry of the cone,$\tan \alpha = \frac{r}{h}$,so $\frac{r}{h} = \frac{1}{2}$,which implies $r = \frac{h}{2}$.
The volume $V$ of the water in the cone is given by $V = \frac{1}{3} \pi r^2 h$.
Substituting $r = \frac{h}{2}$,we get $V = \frac{1}{3} \pi \left(\frac{h}{2}\right)^2 h = \frac{\pi h^3}{12}$.
Differentiating with respect to time $t$,we get $\frac{dV}{dt} = \frac{\pi}{12} \cdot 3h^2 \cdot \frac{dh}{dt} = \frac{\pi h^2}{4} \cdot \frac{dh}{dt}$.
Given $\frac{dV}{dt} = 5 \ m^3/h$ and $h = 4 \ m$,we substitute these values:
$5 = \frac{\pi (4)^2}{4} \cdot \frac{dh}{dt} = 4\pi \cdot \frac{dh}{dt}$.
Therefore,$\frac{dh}{dt} = \frac{5}{4\pi} \ m/h$.

(A) Let $AB$ be the lamp post of height $6 \text{ m}$ and $MN$ be the man of height $2 \text{ m}$. Let the man be at a distance $l$ from the lamp post at time $t$,so $AM = l$. Let $MS = s$ be the length of the shadow.
Since $\triangle MSN \sim \triangle ASB$,we have:
$\frac{MS}{AS} = \frac{MN}{AB}$
Substituting the values,we get:
$\frac{s}{l + s} = \frac{2}{6}$
$\frac{s}{l + s} = \frac{1}{3}$
$3s = l + s$
$l = 2s$
Differentiating both sides with respect to time $t$:
$\frac{dl}{dt} = 2 \frac{ds}{dt}$
Given that the man walks at a speed of $\frac{dl}{dt} = 5 \text{ km/h}$,we have:
$5 = 2 \frac{ds}{dt}$
$\frac{ds}{dt} = \frac{5}{2} = 2.5 \text{ km/h}$
Thus,the length of the shadow increases at the rate of $2.5 \text{ km/h}$.

(A) Let $r$ be the radius of the circular disc and $A$ be its area.
The area of the disc is given by $A = \pi r^2$.
Differentiating both sides with respect to time $t$,we get:
$\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$.
Given that the rate of increase of the radius is $\frac{dr}{dt} = 0.05 \text{ cm/s}$.
We need to find the rate of increase of the area when $r = 3.2 \text{ cm}$.
Substituting the values into the derivative formula:
$\frac{dA}{dt} = 2 \times \pi \times (3.2 \text{ cm}) \times (0.05 \text{ cm/s})$
$\frac{dA}{dt} = 2 \times 3.2 \times 0.05 \times \pi \text{ cm}^2\text{/s}$
$\frac{dA}{dt} = 0.320\pi \text{ cm}^2\text{/s}$.
Thus,the rate at which the area is increasing is $0.320\pi \text{ cm}^2\text{/s}$.

(B) Let $\triangle ABC$ be an isosceles triangle where $BC$ is the base of fixed length $b$. Let the length of the two equal sides be $a$.
Draw $AD \perp BC$. In $\triangle ADC$,by the Pythagoras theorem,the height $h = AD = \sqrt{a^2 - (b/2)^2} = \sqrt{a^2 - b^2/4}$.
The area $A$ of the triangle is given by $A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times b \times \sqrt{a^2 - b^2/4} = \frac{b}{2} \sqrt{a^2 - b^2/4}$.
Differentiating with respect to time $t$,we get $\frac{dA}{dt} = \frac{b}{2} \times \frac{1}{2\sqrt{a^2 - b^2/4}} \times 2a \times \frac{da}{dt} = \frac{ab}{2\sqrt{a^2 - b^2/4}} \times \frac{da}{dt}$.
Given $\frac{da}{dt} = -3 \ cm/s$. When $a = b$,the height $h = \sqrt{b^2 - b^2/4} = \sqrt{3b^2/4} = \frac{\sqrt{3}b}{2}$.
Substituting $a = b$ and $\frac{da}{dt} = -3$ into the derivative:
$\frac{dA}{dt} = \frac{b \cdot b}{2(\sqrt{3}b/2)} \times (-3) = \frac{b^2}{\sqrt{3}b} \times (-3) = -\frac{3b}{\sqrt{3}} = -\sqrt{3}b$.
The area is decreasing at the rate of $\sqrt{3}b \ cm^2/s$.

(C) Let $r$ be the radius of the cylinder and $h$ be the depth (height) of the wheat.
The volume $V$ of the cylinder is given by $V = \pi r^2 h$.
Given $r = 10 \ m$,we have $V = \pi (10)^2 h = 100 \pi h$.
Differentiating with respect to time $t$,we get $\frac{dV}{dt} = 100 \pi \frac{dh}{dt}$.
We are given $\frac{dV}{dt} = 314 \ m^3/h$.
Substituting the values,$314 = 100 \pi \frac{dh}{dt}$.
Using $\pi \approx 3.14$,we get $314 = 100(3.14) \frac{dh}{dt} = 314 \frac{dh}{dt}$.
Therefore,$\frac{dh}{dt} = \frac{314}{314} = 1 \ m/h$.
The depth of the wheat is increasing at the rate of $1 \ m/h$.
The correct answer is $C$.

(A) The slope of the curve is given by $m = \frac{dy}{dx} = 5 - 6x^{2}$.
To find the rate of change of the slope with respect to time $t$,we differentiate $m$ with respect to $t$ using the chain rule:
$\frac{dm}{dt} = \frac{d}{dt}(5 - 6x^{2}) = -12x \cdot \frac{dx}{dt}$.
Given that $x = 3$ and $\frac{dx}{dt} = 2 \text{ units/sec}$,we substitute these values into the equation:
$\frac{dm}{dt} = -12(3)(2) = -72 \text{ units/sec}$.
Thus,the slope of the curve is changing at the rate of $-72 \text{ units/sec}$,which means it is decreasing at a rate of $72 \text{ units/sec}$.

(N/A) Let $v$ be the volume of water in the conical vessel at time $t$. We are given that $\frac{dv}{dt} = -1 \text{ cm}^3/\text{s}$ (negative because water is dripping out).
Let $l$ be the slant height,$h$ be the vertical height,and $r$ be the radius of the water surface.
The semi-vertical angle is $\alpha = \frac{\pi}{6}$.
From the geometry of the cone,we have $h = l \cos \alpha = l \cos \frac{\pi}{6} = l \frac{\sqrt{3}}{2}$ and $r = l \sin \alpha = l \sin \frac{\pi}{6} = \frac{l}{2}$.
The volume of the cone is $v = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \left(\frac{l}{2}\right)^2 \left(l \frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3} \pi}{24} l^3$.
Differentiating with respect to $t$,we get $\frac{dv}{dt} = \frac{\sqrt{3} \pi}{24} \cdot 3l^2 \frac{dl}{dt} = \frac{\sqrt{3} \pi}{8} l^2 \frac{dl}{dt}$.
Given $l = 4 \text{ cm}$ and $\frac{dv}{dt} = -1 \text{ cm}^3/\text{s}$,we substitute these values:
$-1 = \frac{\sqrt{3} \pi}{8} (4)^2 \frac{dl}{dt} = \frac{\sqrt{3} \pi}{8} \cdot 16 \frac{dl}{dt} = 2\sqrt{3} \pi \frac{dl}{dt}$.
Thus,$\frac{dl}{dt} = -\frac{1}{2\sqrt{3} \pi} \text{ cm/s}$.
The rate of decrease of the slant height is $\frac{1}{2\sqrt{3} \pi} \text{ cm/s}$.

(8 M/S) Let the height of the kite be $CD = 151.5 \ m$ and the height of the boy be $AB = 1.5 \ m$. Let the horizontal distance between the boy and the kite be $x$ and the length of the string be $y$.
From the geometry of the problem,the effective height of the kite above the boy's eye level is $h = 151.5 - 1.5 = 150 \ m$.
Using the Pythagorean theorem in the right-angled triangle formed by the string,the horizontal distance,and the effective height:
$x^2 + h^2 = y^2$
$x^2 + (150)^2 = y^2$ --- $(i)$
Given that the kite moves horizontally at $10 \ m/s$,we have $\frac{dx}{dt} = 10 \ m/s$.
When the string length $y = 250 \ m$,we find $x$ from $(i)$:
$x^2 + 150^2 = 250^2$
$x^2 = 62500 - 22500 = 40000$
$x = 200 \ m$.
Differentiating $(i)$ with respect to $t$:
$2x \frac{dx}{dt} + 0 = 2y \frac{dy}{dt}$
$x \frac{dx}{dt} = y \frac{dy}{dt}$
Substituting the values:
$200 \times 10 = 250 \times \frac{dy}{dt}$
$\frac{dy}{dt} = \frac{2000}{250} = 8 \ m/s$.
Thus,the string is being let out at a rate of $8 \ m/s$.

(A) Let two men start from the point $C$ with velocity $v$ each at the same time. Also,$\angle BCA = 45^{\circ}$.
Since $A$ and $B$ are moving with the same velocity $v$,they will cover the same distance in the same time.
Therefore,$\Delta ABC$ is an isosceles triangle with $AC = BC$. Let $AC = BC = x$ and the distance between them at any instant $t$ be $y = AB$.
Draw $CD \perp AB$. In $\Delta ACD$ and $\Delta DCB$:
$\angle CAD = \angle CBD$ (since $AC = BC$)
$\angle CDA = \angle CDB = 90^{\circ}$
Therefore,$\angle ACD = \angle DCB = \frac{1}{2} \times \angle ACB = \frac{1}{2} \times 45^{\circ} = 22.5^{\circ} = \frac{\pi}{8}$.
In $\Delta ACD$,$\sin(\frac{\pi}{8}) = \frac{AD}{AC} = \frac{y/2}{x}$.
Thus,$y = 2x \sin(\frac{\pi}{8})$.
Differentiating with respect to $t$:
$\frac{dy}{dt} = 2 \sin(\frac{\pi}{8}) \frac{dx}{dt} = 2v \sin(\frac{\pi}{8})$.
Using $\sin(\frac{\pi}{8}) = \frac{\sqrt{2-\sqrt{2}}}{2}$,we get:
$\frac{dy}{dt} = 2v \cdot \frac{\sqrt{2-\sqrt{2}}}{2} = v \sqrt{2-\sqrt{2}}$.

(A) Let $AB$ be the street light post and $CD$ be the height of the man,i.e.,$CD = 2 \ m$.
Let $BC = x \ m$,$CE = y \ m$,and $\frac{dx}{dt} = -\frac{5}{3} \ m/s$ (since the man is moving towards the light post).
From $\Delta ABE$ and $\Delta DCE$,we see that $\Delta ABE \sim \Delta DCE$ by $AAA$ similarity.
Therefore,$\frac{AB}{DC} = \frac{BE}{CE} \Rightarrow \frac{16/3}{2} = \frac{x+y}{y}$.
$\Rightarrow \frac{16}{6} = \frac{x+y}{y} \Rightarrow \frac{8}{3} = \frac{x+y}{y}$.
$\Rightarrow 8y = 3x + 3y \Rightarrow 5y = 3x \Rightarrow y = \frac{3}{5}x$.
On differentiating both sides with respect to $t$,we get $\frac{dy}{dt} = \frac{3}{5} \cdot \frac{dx}{dt} = \frac{3}{5} \cdot \left(-\frac{5}{3}\right) = -1 \ m/s$.
Thus,the length of the shadow is decreasing at the rate of $1 \ m/s$.
Let $z = x + y$ be the distance of the tip of the shadow from the light post.
Now,differentiating both sides with respect to $t$,we get $\frac{dz}{dt} = \frac{dx}{dt} + \frac{dy}{dt} = -\frac{5}{3} - 1 = -\frac{8}{3} = -2 \frac{2}{3} \ m/s$.
Hence,the tip of the shadow is moving at the rate of $2 \frac{2}{3} \ m/s$ towards the light source.

(A) Given $f(x) = x^{2}$.
We need to calculate $\frac{f(1.1) - f(1)}{1.1 - 1}$.
First,calculate $f(1.1) = (1.1)^{2} = 1.21$.
Next,calculate $f(1) = (1)^{2} = 1$.
Substitute these values into the expression:
$\frac{1.21 - 1}{1.1 - 1} = \frac{0.21}{0.1} = 2.1$.

(N/A) Let the radius of the spherical ball of salt at any time $t$ be $r$.
The volume of the ball is $V = \frac{4}{3} \pi r^{3}$ and the surface area is $S = 4 \pi r^{2}$.
According to the problem,the rate of decrease of volume is proportional to the surface area:
$-\frac{dV}{dt} \propto S$
This implies $-\frac{dV}{dt} = kS$,where $k$ is a positive constant of proportionality.
Substituting the expressions for $V$ and $S$:
$-\frac{d}{dt} \left( \frac{4}{3} \pi r^{3} \right) = k(4 \pi r^{2})$
Using the chain rule:
$-\frac{4}{3} \pi \cdot 3r^{2} \cdot \frac{dr}{dt} = k(4 \pi r^{2})$
$-4 \pi r^{2} \cdot \frac{dr}{dt} = k(4 \pi r^{2})$
Dividing both sides by $4 \pi r^{2}$ (assuming $r \neq 0$):
$-\frac{dr}{dt} = k$
$\frac{dr}{dt} = -k$
Since $k$ is a constant,the rate of change of the radius $\frac{dr}{dt}$ is a constant. Thus,the radius is decreasing at a constant rate.

(N/A) Let the radius of the circle be $r$ and the area of the circle be $A = \pi r^2$.
Differentiating both sides with respect to time $t$,we get:
$\frac{dA}{dt} = 2\pi r \cdot \frac{dr}{dt} \quad \dots(i)$
Given that the area of the circle increases at a uniform rate,let $\frac{dA}{dt} = k$,where $k$ is a constant.
From equation $(i)$,we have:
$k = 2\pi r \cdot \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{k}{2\pi r} \quad \dots(ii)$
Let the perimeter of the circle be $P = 2\pi r$.
Differentiating the perimeter with respect to time $t$:
$\frac{dP}{dt} = \frac{d}{dt}(2\pi r) = 2\pi \cdot \frac{dr}{dt}$
Substituting the value of $\frac{dr}{dt}$ from equation $(ii)$:
$\frac{dP}{dt} = 2\pi \cdot \left( \frac{k}{2\pi r} \right) = \frac{k}{r}$
Since $k$ is a constant,we have $\frac{dP}{dt} \propto \frac{1}{r}$.
Thus,the rate of change of the perimeter varies inversely as the radius.

(A) Given the volume of water $L$ at time $t$ is $L = 200(10-t)^2$.
The rate at which water flows out is given by $-\frac{dL}{dt}$.
First,find the derivative: $\frac{dL}{dt} = 200 \cdot 2(10-t) \cdot (-1) = -400(10-t)$.
Thus,the rate of flow is $-\frac{dL}{dt} = 400(10-t)$.
At $t = 5$ seconds,the rate is $400(10-5) = 400(5) = 2000 \text{ L/s}$.
To find the average rate during the first $5$ seconds,we calculate the total change in volume divided by the time interval: $\text{Average Rate} = \frac{L(0) - L(5)}{5 - 0}$.
$L(0) = 200(10-0)^2 = 200(100) = 20000 \text{ L}$.
$L(5) = 200(10-5)^2 = 200(25) = 5000 \text{ L}$.
$\text{Average Rate} = \frac{20000 - 5000}{5} = \frac{15000}{5} = 3000 \text{ L/s}$.

(N/A) Let the side of a cube be $x$ units.
Volume of cube $V = x^{3}$.
Differentiating both sides with respect to $t$,we get $\frac{dV}{dt} = 3x^{2} \frac{dx}{dt} = k$ (where $k$ is a constant).
$\Rightarrow \frac{dx}{dt} = \frac{k}{3x^{2}} \dots (i)$.
Surface area of the cube $S = 6x^{2}$.
Differentiating with respect to $t$,we get $\frac{dS}{dt} = 12x \cdot \frac{dx}{dt}$.
Substituting the value of $\frac{dx}{dt}$ from equation $(i)$:
$\frac{dS}{dt} = 12x \cdot \left( \frac{k}{3x^{2}} \right) = \frac{4k}{x}$.
Since $4k$ is a constant,$\frac{dS}{dt} \propto \frac{1}{x}$.
Thus,the rate of increase of the surface area varies inversely as the length of the side $x$.

(A) Let $A_{1}$ be the area of the first square and $A_{2}$ be the area of the second square.
Given that the side of the first square is $x$,so $A_{1} = x^{2}$.
The side of the second square is $y = x - x^{2}$,so $A_{2} = y^{2} = (x - x^{2})^{2}$.
We need to find the rate of change of $A_{2}$ with respect to $A_{1}$,which is $\frac{dA_{2}}{dA_{1}}$.
Using the chain rule,$\frac{dA_{2}}{dA_{1}} = \frac{dA_{2}/dx}{dA_{1}/dx}$.
First,calculate $\frac{dA_{1}}{dx} = \frac{d}{dx}(x^{2}) = 2x$.
Next,calculate $\frac{dA_{2}}{dx} = \frac{d}{dx}(x - x^{2})^{2} = 2(x - x^{2}) \cdot \frac{d}{dx}(x - x^{2}) = 2(x - x^{2})(1 - 2x)$.
Now,$\frac{dA_{2}}{dA_{1}} = \frac{2(x - x^{2})(1 - 2x)}{2x} = \frac{2x(1 - x)(1 - 2x)}{2x}$.
Simplifying this,we get $\frac{dA_{2}}{dA_{1}} = (1 - x)(1 - 2x) = 1 - 2x - x + 2x^{2} = 2x^{2} - 3x + 1$.

(A) Let the side length of the cube be $a$. The surface area $S$ of the cube is given by $S = 6a^2$.
Given that $\frac{dS}{dt} = 3.6 \text{ cm}^2/\text{sec}$.
Differentiating $S$ with respect to $t$,we get $\frac{dS}{dt} = 12a \frac{da}{dt}$.
Substituting the given values: $3.6 = 12(10) \frac{da}{dt} \Rightarrow 3.6 = 120 \frac{da}{dt} \Rightarrow \frac{da}{dt} = \frac{3.6}{120} = 0.03 \text{ cm}/\text{sec}$.
The volume $V$ of the cube is $V = a^3$.
Differentiating $V$ with respect to $t$,we get $\frac{dV}{dt} = 3a^2 \frac{da}{dt}$.
Substituting $a = 10$ and $\frac{da}{dt} = 0.03$: $\frac{dV}{dt} = 3(10)^2(0.03) = 3(100)(0.03) = 300 \times 0.03 = 9 \text{ cm}^3/\text{sec}$.

(A) Let $r$ be the radius of the spherical balloon.
The surface area $S$ is given by $S = 4 \pi r^2$.
Given that the surface area increases at a constant rate,we have $\frac{dS}{dt} = k$,where $k$ is a constant.
Integrating with respect to $t$,we get $S = kt + C$,where $C$ is the constant of integration.
Substituting $S = 4 \pi r^2$,we have $4 \pi r^2 = kt + C$.
At $t = 0$,$r = 3$,so $4 \pi (3)^2 = k(0) + C \Rightarrow C = 36 \pi$.
At $t = 5$,$r = 7$,so $4 \pi (7)^2 = k(5) + 36 \pi \Rightarrow 196 \pi = 5k + 36 \pi \Rightarrow 5k = 160 \pi \Rightarrow k = 32 \pi$.
Thus,the equation becomes $4 \pi r^2 = 32 \pi t + 36 \pi$.
Dividing by $4 \pi$,we get $r^2 = 8t + 9$.
For $t = 9$,$r^2 = 8(9) + 9 = 72 + 9 = 81$.
Therefore,$r = \sqrt{81} = 9$ units.

(C) Let the radius of the cone be $R = 7 \, cm$ and height be $H = 35 \, cm$. By similar triangles,$\frac{r}{h} = \frac{R}{H} = \frac{7}{35} = \frac{1}{5}$,so $h = 5r$.
The volume of water is $V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi r^2 (5r) = \frac{5}{3} \pi r^3$.
Given $\frac{dV}{dt} = 1 \, cm^3/sec$,we have $\frac{d}{dt} (\frac{5}{3} \pi r^3) = 5 \pi r^2 \frac{dr}{dt} = 1$,so $\frac{dr}{dt} = \frac{1}{5 \pi r^2}$.
The wet conical surface area is $S = \pi r l = \pi r \sqrt{r^2 + h^2} = \pi r \sqrt{r^2 + (5r)^2} = \pi r \sqrt{26r^2} = \sqrt{26} \pi r^2$.
Differentiating with respect to $t$,$\frac{dS}{dt} = 2 \sqrt{26} \pi r \frac{dr}{dt}$.
Substituting $\frac{dr}{dt} = \frac{1}{5 \pi r^2}$,we get $\frac{dS}{dt} = 2 \sqrt{26} \pi r (\frac{1}{5 \pi r^2}) = \frac{2 \sqrt{26}}{5r}$.
When $h = 10 \, cm$,$r = \frac{h}{5} = \frac{10}{5} = 2 \, cm$.
Thus,$\frac{dS}{dt} = \frac{2 \sqrt{26}}{5(2)} = \frac{\sqrt{26}}{5} \, cm^2/sec$.

(C) Let $h$ be the depth of water,$r$ be the radius of the water surface,and $\theta$ be the semi-vertical angle. Given $\tan \theta = \frac{r}{h} = \frac{3}{4}$,so $r = \frac{3}{4}h$.
The volume of water is $V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \left(\frac{3}{4}h\right)^2 h = \frac{3 \pi}{16} h^3$.
Differentiating with respect to time $t$,we get $\frac{dV}{dt} = \frac{3 \pi}{16} \cdot 3h^2 \frac{dh}{dt} = \frac{9 \pi}{16} h^2 \frac{dh}{dt}$.
Given $\frac{dV}{dt} = 6 \text{ m}^3/\text{hr}$. At $h = 4 \text{ m}$,$6 = \frac{9 \pi}{16} (4)^2 \frac{dh}{dt} = 9 \pi \frac{dh}{dt}$,so $\frac{dh}{dt} = \frac{6}{9 \pi} = \frac{2}{3 \pi} \text{ m/hr}$.
The slant height is $\ell = \sqrt{r^2 + h^2} = \sqrt{(\frac{3}{4}h)^2 + h^2} = \sqrt{\frac{9}{16}h^2 + h^2} = \sqrt{\frac{25}{16}h^2} = \frac{5}{4}h$.
The curved surface area is $S = \pi r \ell = \pi (\frac{3}{4}h) (\frac{5}{4}h) = \frac{15 \pi}{16} h^2$.
Differentiating with respect to $t$,$\frac{dS}{dt} = \frac{15 \pi}{16} \cdot 2h \frac{dh}{dt} = \frac{15 \pi}{8} h \frac{dh}{dt}$.
Substituting $h = 4$ and $\frac{dh}{dt} = \frac{2}{3 \pi}$,we get $\frac{dS}{dt} = \frac{15 \pi}{8} (4) (\frac{2}{3 \pi}) = \frac{15 \pi}{8} \cdot \frac{8}{3 \pi} = 5 \text{ m}^2/\text{hr}$.

(B) Let the radius of the first base be $R_1 = 50 \, mm$ and the radius of the second base be $R_2 = r \, mm$. The height is $h$. The volume of a truncated cone is given by $V = \frac{\pi h}{3} (R_1^2 + R_1 R_2 + R_2^2)$.
Given $R_1 = 50 \, mm$. The new radius $R_1' = R_1 + 0.21 R_1 = 1.21 R_1 = 1.21 \times 50 = 60.5 \, mm$.
The new volume $V' = 1.21 V$. Substituting the values:
$1.21 \times \frac{\pi h}{3} (50^2 + 50r + r^2) = \frac{\pi h}{3} (60.5^2 + 60.5r + r^2)$.
Dividing both sides by $\frac{\pi h}{3}$:
$1.21 (2500 + 50r + r^2) = 3660.25 + 60.5r + r^2$.
$3025 + 60.5r + 1.21r^2 = 3660.25 + 60.5r + r^2$.
Subtracting $60.5r$ from both sides:
$3025 + 1.21r^2 = 3660.25 + r^2$.
$0.21r^2 = 635.25$.
$r^2 = \frac{635.25}{0.21} = 3025$.
$r = \sqrt{3025} = 55 \, mm$.

(D) Let $R$ and $H$ be the radius and height of the full cone,respectively. The volume of the full cone is $V = \frac{1}{3} \pi R^2 H$.
Since the water is let out at a constant speed,the rate of change of volume is constant,i.e.,$\frac{dV}{dt} = -k$ for some constant $k > 0$.
At any height $h$,the radius $r$ of the water surface is given by $\frac{r}{h} = \frac{R}{H}$,so $r = \frac{R}{H} h$.
The volume of water at height $h$ is $V(h) = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \left(\frac{R}{H} h\right)^2 h = \frac{\pi R^2}{3 H^2} h^3$.
Given that at $t = 0$,$h = H$,so $V(H) = \frac{1}{3} \pi R^2 H$.
At $t = 21 \, min$,$h = \frac{H}{2}$,so the volume of water remaining is $V\left(\frac{H}{2}\right) = \frac{\pi R^2}{3 H^2} \left(\frac{H}{2}\right)^3 = \frac{1}{8} \left(\frac{1}{3} \pi R^2 H\right) = \frac{V}{8}$.
The volume of water removed in $21 \, min$ is $V - \frac{V}{8} = \frac{7V}{8}$.
Since the rate of outflow is constant,the time taken to remove the remaining volume $\frac{V}{8}$ is $t'$.
Using the proportion: $\frac{\text{Volume removed}}{\text{Time taken}} = \text{constant}$.
$\frac{7V/8}{21} = \frac{V/8}{t'}$.
$t' = 21 \times \frac{V/8}{7V/8} = 21 \times \frac{1}{7} = 3 \, min$.
Thus,it requires $3 \, min$ more to empty the vessel.

(A) The mass $m$ of the sphere is given by $m = \rho \cdot \frac{4}{3} \pi R^3$.
Since the total mass $m$ is constant,its derivative with respect to time $t$ is zero: $\frac{dm}{dt} = 0$.
Differentiating the mass equation with respect to $t$:
$0 = \frac{d\rho}{dt} \cdot \frac{4}{3} \pi R^3 + \rho \cdot 4 \pi R^2 \frac{dR}{dt}$.
Dividing by $\rho \cdot \frac{4}{3} \pi R^3$,we get:
$0 = \frac{1}{\rho} \frac{d\rho}{dt} + \frac{3}{R} \frac{dR}{dt}$.
Rearranging for the velocity of the surface $v = \frac{dR}{dt}$:
$\frac{dR}{dt} = -\frac{R}{3} \left(\frac{1}{\rho} \frac{d\rho}{dt}\right)$.
Given that the rate of fractional change in density $\left(\frac{1}{\rho} \frac{d\rho}{dt}\right)$ is a constant,let this constant be $k$.
Then $v = \frac{dR}{dt} = -\frac{k}{3} R$.
Thus,$v \propto R$.

(D) Let $r$ be the radius of the chocolate ball and $x$ be the thickness of the ice-cream layer. The total radius of the sphere (chocolate + ice-cream) is $R = r + x$.
Given $x = 1 \ cm$, the total volume $V$ of the sphere is $V = \frac{4}{3}\pi R^3$.
Differentiating with respect to time $t$, we get $\frac{dV}{dt} = 4\pi R^2 \frac{dR}{dt}$.
Since the chocolate ball is constant, $\frac{dR}{dt} = \frac{dx}{dt}$.
We are given $\frac{dV}{dt} = -81 \ cm^3/min$ (as it melts) and $\frac{dx}{dt} = -\frac{1}{4\pi} \ cm/min$.
Substituting these values: $-81 = 4\pi R^2 \left(-\frac{1}{4\pi}\right)$.
$-81 = -R^2 \implies R^2 = 81 \implies R = 9 \ cm$.
Since $R = r + x$ and $x = 1 \ cm$, we have $r = 9 - 1 = 8 \ cm$.
The surface area of the chocolate ball is $4\pi r^2 = 4\pi(8)^2 = 4\pi(64) = 256\pi \ cm^2$.

(D) Given the curve equation is $y = \frac{2x^3 - 1}{3}$.
We are given that the rate of change of the $y$-coordinate is $18$ times the rate of change of the $x$-coordinate,which implies $\frac{dy}{dt} = 18 \frac{dx}{dt}$.
Dividing both sides by $\frac{dx}{dt}$,we get $\frac{dy}{dx} = 18$.
Now,differentiate the curve equation with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} (\frac{2x^3 - 1}{3}) = \frac{1}{3} \cdot 6x^2 = 2x^2$.
Equating the derivative to $18$:
$2x^2 = 18 \implies x^2 = 9 \implies x = \pm 3$.
For $x = 3$,$y = \frac{2(3)^3 - 1}{3} = \frac{54 - 1}{3} = \frac{53}{3}$.
For $x = -3$,$y = \frac{2(-3)^3 - 1}{3} = \frac{-54 - 1}{3} = -\frac{55}{3}$.
Thus,the points are $(3, \frac{53}{3})$ and $(-3, -\frac{55}{3})$.
Comparing this with the given options,the correct option is $D$.

(B) Given the position function $x = 2 + 27t - t^3$.
To find the velocity,we differentiate $x$ with respect to $t$:
$v = \frac{dx}{dt} = 27 - 3t^2$.
The direction of motion reverses when the velocity becomes zero:
$27 - 3t^2 = 0$
$3t^2 = 27$
$t^2 = 9$
$t = 3$ (since $t > 0$).
Now,we calculate the distance $x$ at $t = 3$:
$x = 2 + 27(3) - (3)^3$
$x = 2 + 81 - 27$
$x = 56 \text{ units}$.

(C) Let $r$ be the radius of the circular wave and $A$ be the area of the circular region.
Given that the rate of change of the radius is $\frac{dr}{dt} = 2.1 \text{ cm/sec}$.
The area of the circle is given by $A = \pi r^2$.
Differentiating both sides with respect to $t$,we get $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$.
Given $r = 10 \text{ cm}$ and $\frac{dr}{dt} = 2.1 \text{ cm/sec}$.
Substituting these values,$\frac{dA}{dt} = 2 \times \frac{22}{7} \times 10 \times 2.1$.
$\frac{dA}{dt} = 2 \times \frac{22}{7} \times 10 \times \frac{21}{10}$.
$\frac{dA}{dt} = 2 \times 22 \times 3 = 132 \text{ cm}^2/\text{sec}$.

(D) Let $r$ be the radius and $h$ be the altitude of the cone. The slant height $l$ is given by $l = \sqrt{r^2 + h^2}$.
Given: $\frac{dr}{dt} = 3 \text{ cm/min}$ and $\frac{dh}{dt} = -4 \text{ cm/min}$.
The lateral surface area $S$ of a cone is $S = \pi rl = \pi r \sqrt{r^2 + h^2}$.
At $r = 7$ and $h = 24$,$l = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \text{ cm}$.
Differentiating $S$ with respect to $t$:
$\frac{dS}{dt} = \pi \left( \frac{dr}{dt} \sqrt{r^2 + h^2} + r \cdot \frac{1}{2\sqrt{r^2 + h^2}} \cdot (2r \frac{dr}{dt} + 2h \frac{dh}{dt}) \right)$.
$\frac{dS}{dt} = \pi \left( 3 \cdot 25 + 7 \cdot \frac{1}{25} \cdot (7 \cdot 3 + 24 \cdot (-4)) \right)$.
$\frac{dS}{dt} = \pi \left( 75 + \frac{7}{25} \cdot (21 - 96) \right) = \pi \left( 75 + \frac{7}{25} \cdot (-75) \right)$.
$\frac{dS}{dt} = \pi (75 - 7 \cdot 3) = \pi (75 - 21) = 54 \pi \text{ cm}^2/\text{min}$.

(B) The area $A$ of an isosceles triangle with equal sides $x$ and included angle $\theta$ is given by $A = \frac{1}{2} x^2 \sin \theta$.
To find the rate of change of the area,we differentiate $A$ with respect to time $t$ using the product rule:
$\frac{dA}{dt} = \frac{1}{2} \left( 2x \frac{dx}{dt} \sin \theta + x^2 \cos \theta \frac{d\theta}{dt} \right) = x \frac{dx}{dt} \sin \theta + \frac{1}{2} x^2 \cos \theta \frac{d\theta}{dt}$.
Given values: $x = 12 \text{ m}$,$\theta = \frac{\pi}{4}$,$\frac{dx}{dt} = \frac{1}{12} \text{ m/h}$,and $\frac{d\theta}{dt} = \frac{\pi}{180} \text{ rad/h}$.
Substituting these values into the derivative:
$\frac{dA}{dt} = (12) \left( \frac{1}{12} \right) \sin \left( \frac{\pi}{4} \right) + \frac{1}{2} (12)^2 \cos \left( \frac{\pi}{4} \right) \left( \frac{\pi}{180} \right)$.
$\frac{dA}{dt} = (1) \left( \frac{1}{\sqrt{2}} \right) + \frac{1}{2} (144) \left( \frac{1}{\sqrt{2}} \right) \left( \frac{\pi}{180} \right)$.
$\frac{dA}{dt} = \frac{1}{\sqrt{2}} + \frac{72}{\sqrt{2}} \left( \frac{\pi}{180} \right) = \frac{1}{\sqrt{2}} + \frac{2\pi}{5\sqrt{2}} = \frac{1}{\sqrt{2}} \left( 1 + \frac{2\pi}{5} \right)$.
Note: The provided options seem to contain a typo in the coefficient. Re-evaluating the expression: $\frac{1}{\sqrt{2}} (1 + \frac{2\pi}{5}) = \frac{\sqrt{2}}{2} (1 + \frac{2\pi}{5}) = \sqrt{2} (\frac{1}{2} + \frac{\pi}{5})$.
Thus,the correct option is $B$.

(B) Given the equation of motion: $s(t) = at^2 + bt + c$.
Velocity $v(t) = \frac{ds}{dt} = 2at + b$.
Acceleration $a_{acc}(t) = \frac{dv}{dt} = 2a$.
Given acceleration is $10 \ m/s^2$,so $2a = 10 \implies a = 5$.
Given velocity after $2 \ s$ is $30 \ m/s$: $v(2) = 2a(2) + b = 30 \implies 4a + b = 30$.
Substituting $a = 5$: $4(5) + b = 30 \implies 20 + b = 30 \implies b = 10$.
Given displacement after $1 \ s$ is $20 \ m$: $s(1) = a(1)^2 + b(1) + c = 20 \implies a + b + c = 20$.
Substituting $a = 5$ and $b = 10$: $5 + 10 + c = 20 \implies 15 + c = 20 \implies c = 5$.
Now check the options:
$a + c = 5 + 5 = 10$.
Since $b = 10$,we have $a + c = b$.

(A) Let $x$ be the length and $y$ be the breadth of the rectangle. Given $\frac{dx}{dt} = -5 \text{ cm/min}$ and $\frac{dy}{dt} = 3 \text{ cm/min}$.
Perimeter $P = 2(x + y)$.
The rate of change of perimeter is $\frac{dP}{dt} = 2(\frac{dx}{dt} + \frac{dy}{dt}) = 2(-5 + 3) = 2(-2) = -4 \text{ cm/min}$.
Area $A = xy$.
The rate of change of area is $\frac{dA}{dt} = x \frac{dy}{dt} + y \frac{dx}{dt}$.
Substituting the values $x = 5$,$y = 2$,$\frac{dx}{dt} = -5$,and $\frac{dy}{dt} = 3$:
$\frac{dA}{dt} = (5)(3) + (2)(-5) = 15 - 10 = 5 \text{ cm}^2\text{/min}$.
Thus,the rates of change are $-4 \text{ cm/min}$ and $5 \text{ cm}^2\text{/min}$.

(A) Let $r$ be the radius of the sphere.
The volume $V$ of the sphere is given by $V = \frac{4}{3} \pi r^3$.
The surface area $S$ of the sphere is given by $S = 4 \pi r^2$.
We need to find the rate of change of volume with respect to surface area,which is $\frac{dV}{dS}$.
Using the chain rule,$\frac{dV}{dS} = \frac{dV/dr}{dS/dr}$.
Differentiating $V$ with respect to $r$: $\frac{dV}{dr} = \frac{d}{dr}(\frac{4}{3} \pi r^3) = 4 \pi r^2$.
Differentiating $S$ with respect to $r$: $\frac{dS}{dr} = \frac{d}{dr}(4 \pi r^2) = 8 \pi r$.
Therefore,$\frac{dV}{dS} = \frac{4 \pi r^2}{8 \pi r} = \frac{r}{2}$.
When the radius $r = 5 \ m$,the rate of change is $\frac{5}{2} \ m$.

(C) Let $A_1$ be the area of the first square and $A_2$ be the area of the second square.
Given that the side of the first square is $x$,so $A_1 = x^2$.
Given that the side of the second square is $y = x - x^2$,so $A_2 = y^2 = (x - x^2)^2$.
We need to find the rate of change of $A_2$ with respect to $A_1$,which is $\frac{dA_2}{dA_1}$.
Using the chain rule,$\frac{dA_2}{dA_1} = \frac{dA_2/dx}{dA_1/dx}$.
First,calculate $\frac{dA_1}{dx} = \frac{d}{dx}(x^2) = 2x$.
Next,calculate $\frac{dA_2}{dx} = \frac{d}{dx}((x - x^2)^2) = 2(x - x^2) \cdot \frac{d}{dx}(x - x^2) = 2(x - x^2)(1 - 2x)$.
Now,substitute these into the chain rule formula:
$\frac{dA_2}{dA_1} = \frac{2(x - x^2)(1 - 2x)}{2x} = \frac{2x(1 - x)(1 - 2x)}{2x} = (1 - x)(1 - 2x) = 1 - 2x - x + 2x^2 = 2x^2 - 3x + 1$.

(B) Let $S$ be the surface area and $r$ be the radius of the spherical ball. The surface area is given by $S = 4 \pi r^2$.
Given that $\frac{dS}{dt} = 4 \pi \,cm^2/s$.
Differentiating $S$ with respect to $t$,we get $\frac{dS}{dt} = 8 \pi r \frac{dr}{dt}$.
When $S = 16 \pi \,cm^2$,we have $4 \pi r^2 = 16 \pi$,which implies $r^2 = 4$,so $r = 2 \,cm$.
Substituting the values into the derivative equation: $4 \pi = 8 \pi (2) \frac{dr}{dt}$.
$4 \pi = 16 \pi \frac{dr}{dt}$.
$\frac{dr}{dt} = \frac{4 \pi}{16 \pi} = 0.25 \,cm/s$.

(D) The volume of a spherical balloon is given by $V = \frac{4}{3} \pi r^3$.
Given the initial volume $V_0 = 4500 \pi \ m^3$.
The rate of change of volume is $\frac{dV}{dt} = -72 \pi \ m^3/\text{min}$.
After $t = 49$ minutes,the volume $V$ is $V = V_0 + (\frac{dV}{dt}) \times t = 4500 \pi - 72 \pi \times 49$.
$V = 4500 \pi - 3528 \pi = 972 \pi \ m^3$.
Using $V = \frac{4}{3} \pi r^3$,we find the radius $r$ at $t = 49$:
$972 \pi = \frac{4}{3} \pi r^3 \implies r^3 = 972 \times \frac{3}{4} = 729$.
$r = \sqrt[3]{729} = 9 \ m$.
Now,differentiate $V = \frac{4}{3} \pi r^3$ with respect to $t$:
$\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$.
Substitute the known values: $-72 \pi = 4 \pi (9)^2 \frac{dr}{dt}$.
$-72 \pi = 4 \pi (81) \frac{dr}{dt} \implies -72 \pi = 324 \pi \frac{dr}{dt}$.
$\frac{dr}{dt} = -\frac{72}{324} = -\frac{2}{9} \ m/\text{min}$.
The rate at which the radius decreases is $\frac{2}{9} \ m/\text{min}$.

(B) The distance equation is given by $S = 1200t - 15t^2$.
To find the velocity $v$, we differentiate $S$ with respect to $t$:
$v = \frac{dS}{dt} = \frac{d}{dt}(1200t - 15t^2) = 1200 - 30t$.
When the bullet comes to rest, its velocity $v$ becomes $0$:
$1200 - 30t = 0$
$30t = 1200$
$t = 40 \text{ seconds}$.
Now, substitute $t = 40$ into the distance equation to find the total distance covered:
$S = 1200(40) - 15(40)^2$
$S = 48000 - 15(1600)$
$S = 48000 - 24000$
$S = 24000 \text{ cm}$.

(C) Given, the rate of change of the radius is $\frac{dr}{dt} = 8 \,cm/sec$.
The area of the circular wave is $A = \pi r^2$.
Differentiating both sides with respect to time $t$, we get $\frac{dA}{dt} = 2 \pi r \frac{dr}{dt}$.
At the instant when $r = 12 \,cm$, substituting the values:
$\frac{dA}{dt} = 2 \pi (12) (8)$.
$\frac{dA}{dt} = 192 \pi \,cm^2/sec$.

(D) Let $x$ be the distance of the foot of the ladder from the wall and $y$ be the height of the top of the ladder from the ground.
According to the Pythagorean theorem,$x^2 + y^2 = 5^2 = 25$.
Given that the top slides downwards at a rate of $10 \ cm/sec$,so $\frac{dy}{dt} = -10 \ cm/sec = -0.1 \ m/sec$.
When $x = 4 \ m$,we have $4^2 + y^2 = 25$,which gives $y^2 = 25 - 16 = 9$,so $y = 3 \ m$.
Differentiating $x^2 + y^2 = 25$ with respect to $t$,we get:
$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$
$x \frac{dx}{dt} = -y \frac{dy}{dt}$
Substituting the values $x = 4$,$y = 3$,and $\frac{dy}{dt} = -0.1$:
$4 \frac{dx}{dt} = -3(-0.1)$
$4 \frac{dx}{dt} = 0.3$
$\frac{dx}{dt} = \frac{0.3}{4} = 0.075 \ m/sec$.
Thus,the foot of the ladder is sliding at the rate of $0.075 \ m/sec$.

(B) Let $A$, $P$, and $X$ be the area, perimeter, and length of the side of the square respectively at time $t$ seconds.
Given that the area is contracting at a rate of $\frac{dA}{dt} = -3 \,cm^2 / sec$.
The area of the square is $A = X^2$ and the perimeter is $P = 4X$.
From $A = X^2$, we have $X = \sqrt{A}$.
Substituting this into the perimeter formula: $P = 4\sqrt{A}$.
Differentiating with respect to $t$:
$\frac{dP}{dt} = 4 \cdot \frac{1}{2\sqrt{A}} \cdot \frac{dA}{dt} = \frac{2}{X} \cdot \frac{dA}{dt}$.
Given $X = 15 \,cm$ and $\frac{dA}{dt} = -3 \,cm^2 / sec$ (since it is contracting):
$\frac{dP}{dt} = \frac{2}{15} \cdot (-3) = -\frac{6}{15} = -\frac{2}{5} \,cm / sec$.
The negative sign indicates that the perimeter is decreasing at a rate of $\frac{2}{5} \,cm / sec$.

(B) Given,$\frac{dx}{dt} = 2 \text{ units/sec}$.
Given equation of the parabola is $y = 2x^2$.
Differentiating with respect to $t$,we get $\frac{dy}{dt} = 4x \cdot \frac{dx}{dt}$.
Substituting $\frac{dx}{dt} = 2$,we get $\frac{dy}{dt} = 4x(2) = 8x$ ... $(i)$.
Let $r$ be the distance of the point $(x, y)$ from the origin $(0, 0)$,so $r = \sqrt{x^2 + y^2}$.
The rate of change of distance is $\frac{dr}{dt} = \frac{1}{2\sqrt{x^2 + y^2}} \cdot \frac{d}{dt}(x^2 + y^2) = \frac{2x \frac{dx}{dt} + 2y \frac{dy}{dt}}{2\sqrt{x^2 + y^2}} = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{\sqrt{x^2 + y^2}}$.
Substituting $\frac{dx}{dt} = 2$ and $\frac{dy}{dt} = 8x$,we get $\frac{dr}{dt} = \frac{x(2) + y(8x)}{\sqrt{x^2 + y^2}} = \frac{2x + 8xy}{\sqrt{x^2 + y^2}}$.
At the point $(1, 2)$,$x = 1$ and $y = 2$.
$\frac{dr}{dt} = \frac{2(1) + 8(1)(2)}{\sqrt{1^2 + 2^2}} = \frac{2 + 16}{\sqrt{1 + 4}} = \frac{18}{\sqrt{5}} \text{ units/sec}$.

(C) Radius of the hemispherical bowl $R = 180 \text{ cm}$.
Rate of flow of water $\frac{dV}{dt} = 108 \text{ dm}^3/\text{min} = 108 \times 1000 \text{ cm}^3/\text{min} = 108000 \text{ cm}^3/\text{min}$.
Converting to seconds: $\frac{dV}{dt} = \frac{108000}{60} \text{ cm}^3/\text{s} = 1800 \text{ cm}^3/\text{s}$.
Let the depth of water be $x$. The volume of water in a hemispherical bowl is given by $V = \frac{\pi}{3} x^2(3R - x)$.
Substituting $R = 180$: $V = \frac{\pi}{3} x^2(540 - x) = 180 \pi x^2 - \frac{\pi}{3} x^3$.
Differentiating with respect to time $t$: $\frac{dV}{dt} = (360 \pi x - \pi x^2) \frac{dx}{dt}$.
At $x = 120 \text{ cm}$,we have $1800 = (360 \pi(120) - \pi(120)^2) \frac{dx}{dt}$.
$1800 = (43200 \pi - 14400 \pi) \frac{dx}{dt} = 28800 \pi \frac{dx}{dt}$.
$\frac{dx}{dt} = \frac{1800}{28800 \pi} = \frac{18}{288 \pi} = \frac{1}{16 \pi} \text{ cm/s}$.

(B) Given,the volume of a ball $V = \frac{4}{3} \pi r^3$.
When $V = 288 \pi$,we have:
$288 \pi = \frac{4}{3} \pi r^3$
$r^3 = 288 \times \frac{3}{4} = 216$
$r = 6 \text{ cm}$.
Now,differentiating $V$ with respect to $t$:
$\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$
Given $\frac{dV}{dt} = 4 \pi \text{ cc/sec}$,substituting the values:
$4 \pi = 4 \pi (6)^2 \frac{dr}{dt}$
$1 = 36 \frac{dr}{dt}$
$\frac{dr}{dt} = \frac{1}{36} \text{ cm/sec}$.

(C) The distance covered by the body is given by $s = 3t^2 - 8t + 5$.
To find the velocity $v$,we differentiate $s$ with respect to time $t$:
$v = \frac{ds}{dt} = \frac{d}{dt}(3t^2 - 8t + 5) = 6t - 8$.
The body stops when its velocity becomes zero,i.e.,$v = 0$.
Setting the velocity to zero: $6t - 8 = 0$.
Solving for $t$: $6t = 8 \Rightarrow t = \frac{8}{6} = \frac{4}{3} \text{ sec}$.
Therefore,the body will stop after $\frac{4}{3} \text{ sec}$.

(C) Let the radius of the sphere be $r$.
Volume of the sphere $V = \frac{4}{3} \pi r^3$.
Surface area of the sphere $A = 4 \pi r^2$.
We need to find the rate of change of volume with respect to surface area,which is $\frac{dV}{dA}$.
Using the chain rule,$\frac{dV}{dA} = \frac{dV/dr}{dA/dr}$.
First,differentiate $V$ with respect to $r$: $\frac{dV}{dr} = \frac{d}{dr}(\frac{4}{3} \pi r^3) = 4 \pi r^2$.
Next,differentiate $A$ with respect to $r$: $\frac{dA}{dr} = \frac{d}{dr}(4 \pi r^2) = 8 \pi r$.
Now,calculate $\frac{dV}{dA} = \frac{4 \pi r^2}{8 \pi r} = \frac{r}{2}$.
Given the radius $r = 2 \text{ cm}$,substitute this value into the expression:
$\frac{dV}{dA} = \frac{2}{2} = 1 \text{ cm}^3 / \text{ cm}^2$.

(D) Let $x$ be the distance of the foot of the ladder from the wall and $y$ be the height of the top of the ladder on the wall.
Given that the length of the ladder is $5 \ m$,by the Pythagorean theorem,we have $x^2 + y^2 = 5^2 = 25$.
Differentiating both sides with respect to time $t$,we get:
$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$
$\frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt}$
Given $\frac{dx}{dt} = 2 \ m/sec$ and at the instant when $x = 4 \ m$,we have $y = \sqrt{25 - 4^2} = \sqrt{9} = 3 \ m$.
Substituting these values into the derivative equation:
$\frac{dy}{dt} = -\frac{4}{3} \times 2 = -\frac{8}{3} \ m/sec$.
The negative sign indicates that the height is decreasing.
Therefore,the height on the wall is decreasing at the rate of $\frac{8}{3} \ m/sec$.

(B) Given that the rate of change of volume is $\frac{dV}{dt} = 36 \ m^3/min$ and the radius of the base is $r = 3 \ m$.
The volume of a cylinder is given by $V = \pi r^2 h$.
Differentiating both sides with respect to time $t$,we get $\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}$.
Substituting the given values,$36 = \pi \times (3)^2 \times \frac{dh}{dt}$.
$36 = 9 \pi \times \frac{dh}{dt}$.
Therefore,$\frac{dh}{dt} = \frac{36}{9 \pi} = \frac{4}{\pi} \ m/min$.

(C) Volume of a sphere $(V) = \frac{4}{3} \pi r^3$.
Surface area of a sphere $(A) = 4 \pi r^2$.
Differentiating both with respect to $r$:
$\frac{dV}{dr} = 4 \pi r^2$ and $\frac{dA}{dr} = 8 \pi r$.
We need to find the rate of change of volume with respect to surface area,which is $\frac{dV}{dA}$.
Using the chain rule: $\frac{dV}{dA} = \frac{dV/dr}{dA/dr} = \frac{4 \pi r^2}{8 \pi r} = \frac{r}{2}$.
Given the radius $r = 2 \text{ cm}$,we substitute this value:
$\frac{dV}{dA} = \frac{2}{2} = 1 \text{ cm}^3 / \text{cm}^2$.