Rate of Change of Quantities Questions in English

(D) Let $r$ be the radius and $C$ be the circumference of the circle.
Given that the rate of change of the radius is $\frac{dr}{dt} = 0.7 \text{ cm/s}$.
The formula for the circumference of a circle is $C = 2 \pi r$.
Differentiating both sides with respect to time $t$,we get:
$\frac{dC}{dt} = 2 \pi \frac{dr}{dt}$.
Substituting the given value $\frac{dr}{dt} = 0.7 \text{ cm/s}$:
$\frac{dC}{dt} = 2 \pi (0.7) = 1.4 \pi \text{ cm/s}$.
Thus,the circumference of the circle is increasing at a rate of $1.4 \pi \text{ cm/s}$.

(C) The marginal revenue is defined as the rate of change of total revenue with respect to the number of units sold,which is given by the derivative $R'(x)$.
Given $R(x) = x^2 + 6x + 5$.
Differentiating with respect to $x$,we get:
$R'(x) = \frac{d}{dx}(x^2 + 6x + 5) = 2x + 6$.
To find the marginal revenue when $x = 20$,we substitute $x = 20$ into the derivative:
$R'(20) = 2(20) + 6 = 40 + 6 = 46$.
Thus,the marginal revenue is $46$ Rupees.

(A) The area of a circle $A$ is given by the formula $A = \pi r^2$,where $r$ is the radius.
To find the rate of change of the area with respect to the radius,we differentiate $A$ with respect to $r$:
$\frac{dA}{dr} = \frac{d}{dr}(\pi r^2) = 2 \pi r$.
Now,we evaluate this derivative at $r = 3 \text{ cm}$:
$\left. \frac{dA}{dr} \right|_{r=3} = 2 \pi (3) = 6 \pi$.
Thus,the rate of change of the area with respect to its radius at $r = 3 \text{ cm}$ is $6 \pi \text{ cm}^2/\text{cm}$.

(A) Given the curve equation is $y^2 = 18x$.
Differentiating both sides with respect to $t$,we get:
$2y \frac{dy}{dt} = 18 \frac{dx}{dt}$
$y \frac{dy}{dt} = 9 \frac{dx}{dt}$
According to the problem,the rate of change of $Y$-coordinate is twice the rate of change of $X$-coordinate,i.e.,$\frac{dy}{dt} = 2 \frac{dx}{dt}$.
Substituting this into the differentiated equation:
$y(2 \frac{dx}{dt}) = 9 \frac{dx}{dt}$
Since $\frac{dx}{dt} \neq 0$,we can divide by $\frac{dx}{dt}$:
$2y = 9 \implies y = \frac{9}{2}$.
Now,substitute $y = \frac{9}{2}$ into the original curve equation $y^2 = 18x$:
$\left(\frac{9}{2}\right)^2 = 18x$
$\frac{81}{4} = 18x$
$x = \frac{81}{4 \times 18} = \frac{81}{72} = \frac{9}{8}$.
Thus,the required point is $\left(\frac{9}{8}, \frac{9}{2}\right)$.

(B) The velocity $v$ of a particle is the rate of change of displacement with respect to time,given by $v = \frac{ds}{dt}$.
Given $s = t^3 - 6t^2 + 9t$.
Differentiating with respect to $t$:
$v = \frac{d}{dt}(t^3 - 6t^2 + 9t) = 3t^2 - 12t + 9$.
To find the velocity at $t = 2 \ s$,substitute $t = 2$ into the velocity equation:
$v(2) = 3(2)^2 - 12(2) + 9$
$v(2) = 3(4) - 24 + 9$
$v(2) = 12 - 24 + 9 = -3 \ m/s$.
Therefore,the velocity at $t = 2 \ s$ is $-3 \ m/s$.

(B) The volume $V$ of a sphere with radius $r$ is given by the formula: $V = \frac{4}{3} \pi r^3$.
To find the rate of change of the volume with respect to its radius,we differentiate $V$ with respect to $r$:
$\frac{dV}{dr} = \frac{d}{dr} (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi (3r^2) = 4 \pi r^2$.
Now,we evaluate this derivative at $r = 7 \ cm$:
$\frac{dV}{dr} \Big|_{r=7} = 4 \pi (7)^2 = 4 \pi (49) = 196 \pi$.
The unit of the rate of change of volume with respect to radius is $\frac{cm^3}{cm} = cm^2$.
Therefore,the rate of change is $196 \pi \ cm^2$.

(B) The marginal revenue is defined as the derivative of the total revenue function $R(x)$ with respect to $x$,denoted as $MR = \frac{dR}{dx}$.
Given $R(x) = 10x^2 + 20x + 1500$.
Differentiating with respect to $x$:
$\frac{dR}{dx} = \frac{d}{dx}(10x^2 + 20x + 1500) = 20x + 20$.
To find the marginal revenue when $x = 1500$:
$MR = 20(1500) + 20 = 30000 + 20 = 30020$.
Thus,the correct option is $B$.

(B) Let the radius of the sphere be $r$.
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 $S = 4 \pi r^2$.
From the surface area formula,we have $r^2 = \frac{S}{4 \pi}$,which implies $r = \sqrt{\frac{S}{4 \pi}} = \frac{1}{2} \sqrt{\frac{S}{\pi}}$.
We need to find $\frac{dV}{dS}$.
Using the chain rule,$\frac{dV}{dS} = \frac{dV/dr}{dS/dr}$.
$\frac{dV}{dr} = \frac{d}{dr} (\frac{4}{3} \pi r^3) = 4 \pi r^2$.
$\frac{dS}{dr} = \frac{d}{dr} (4 \pi r^2) = 8 \pi r$.
Therefore,$\frac{dV}{dS} = \frac{4 \pi r^2}{8 \pi r} = \frac{r}{2}$.
Substituting the value of $r$,we get $\frac{dV}{dS} = \frac{1}{2} (\frac{1}{2} \sqrt{\frac{S}{\pi}}) = \frac{1}{4} \sqrt{\frac{S}{\pi}}$.
Thus,the correct option is $B$.

(C) Let $V$ be the volume and $r$ be the radius of the sphere.
Given that the rate of change of volume is $\frac{dV}{dt} = \pi \text{ cm}^3/\text{s}$.
The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$.
Differentiating both sides with respect to $t$,we get $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$.
Substituting the given values $\frac{dV}{dt} = \pi$ and $r = 2 \text{ cm}$:
$\pi = 4 \pi (2)^2 \frac{dr}{dt}$
$\pi = 16 \pi \frac{dr}{dt}$
$\frac{dr}{dt} = \frac{\pi}{16 \pi} = \frac{1}{16} \text{ cm/s}$.
Thus,the rate at which the radius increases is $\frac{1}{16} \text{ cm/s}$.

(A) The volume of a sphere is given by $V = \frac{4}{3} \Pi r^3$.
Differentiating with respect to time $t$,we get $\frac{dV}{dt} = 4 \Pi r^2 \frac{dr}{dt}$.
Given $\frac{dV}{dt} = 10 \text{ cm}^3 \text{s}^{-1}$ and $r = 15 \text{ cm}$.
Substituting these values: $10 = 4 \Pi (15)^2 \frac{dr}{dt}$.
$\frac{dr}{dt} = \frac{10}{4 \Pi (225)} = \frac{10}{900 \Pi} = \frac{1}{90 \Pi} \text{ cm s}^{-1}$.

(B) Let the radius of the sphere be $r = 4 \text{ cm}$.
The volume $V$ of a sphere is given by $V = \frac{4}{3} \pi r^3$.
The surface area $S$ of a sphere is given by $S = 4 \pi r^2$.
Differentiating $V$ and $S$ with respect to $r$:
$\frac{dV}{dr} = 4 \pi r^2$
$\frac{dS}{dr} = 8 \pi r$
Using the chain rule,the rate of change of volume with respect to surface area is:
$\frac{dV}{dS} = \frac{dV/dr}{dS/dr} = \frac{4 \pi r^2}{8 \pi r} = \frac{r}{2}$.
At $r = 4 \text{ cm}$,we have:
$\frac{dV}{dS} = \frac{4}{2} = 2 \text{ cm}^3 \text{ cm}^{-2}$.

(B) Given that the rate of change of the radius is $ \frac{dr}{dt} = 5 \text{ cm s}^{-1} $.
The area of a circle is given by $ A = \pi r^2 $.
Differentiating both sides with respect to time $ t $,we get:
$ \frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = 2\pi r \frac{dr}{dt} $.
Substituting the given values $ r = 8 \text{ cm} $ and $ \frac{dr}{dt} = 5 \text{ cm s}^{-1} $:
$ \frac{dA}{dt} = 2 \pi (8) (5) = 80 \pi \text{ cm}^2 \text{ s}^{-1} $.
Thus,the enclosed area is increasing at a rate of $ 80 \pi \text{ cm}^2 \text{ s}^{-1} $.

(B) Given,$s = \frac{2 t^3}{3} - 18 t + \frac{5}{3}$.
Velocity $v = \frac{ds}{dt} = \frac{d}{dt} (\frac{2 t^3}{3} - 18 t + \frac{5}{3}) = 2 t^2 - 18$.
Acceleration $a = \frac{dv}{dt} = \frac{d}{dt} (2 t^2 - 18) = 4 t$.
The particle comes to rest when $v = 0$.
$2 t^2 - 18 = 0 \Rightarrow 2 t^2 = 18 \Rightarrow t^2 = 9 \Rightarrow t = 3 \ s$ (since $t > 0$).
Now,substitute $t = 3$ in the expression for acceleration:
$a = 4(3) = 12 \ m/s^2$.

(A) Given the curve equation: $\frac{x^2}{16} + \frac{y^2}{4} = 1$.
Differentiating both sides with respect to $t$:
$\frac{2x}{16} \frac{dx}{dt} + \frac{2y}{4} \frac{dy}{dt} = 0$.
$\frac{x}{8} \frac{dx}{dt} + \frac{y}{2} \frac{dy}{dt} = 0$.
Given that the rate of change of abscissa $(\frac{dx}{dt})$ is $4$ times the rate of change of ordinate $(\frac{dy}{dt})$,i.e.,$\frac{dx}{dt} = 4 \frac{dy}{dt}$.
Substituting this into the differentiated equation:
$\frac{x}{8} (4 \frac{dy}{dt}) + \frac{y}{2} \frac{dy}{dt} = 0$.
$(\frac{x}{2} + \frac{y}{2}) \frac{dy}{dt} = 0$.
Since $\frac{dy}{dt} \neq 0$,we have $\frac{x}{2} + \frac{y}{2} = 0$,which implies $x = -y$.
Substituting $x = -y$ into the original curve equation:
$\frac{(-y)^2}{16} + \frac{y^2}{4} = 1$.
$\frac{y^2}{16} + \frac{4y^2}{16} = 1 \Rightarrow \frac{5y^2}{16} = 1 \Rightarrow y^2 = \frac{16}{5}$.
Thus,$y = \pm \frac{4}{\sqrt{5}}$.
Since $x = -y$,if $y = \frac{4}{\sqrt{5}}$,then $x = -\frac{4}{\sqrt{5}}$ (Quadrant $II$).
If $y = -\frac{4}{\sqrt{5}}$,then $x = \frac{4}{\sqrt{5}}$ (Quadrant $IV$).
Therefore,the particle lies in the $II$ or $IV$ quadrant.

(C) Let $r$ be the radius and $A$ be the area of the circular plate at any time $t$.
We know that the area of a 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 that the rate of change of radius is $\frac{dr}{dt} = 0.05 \text{ cm/s}$.
We need to find the rate of change of area when $r = 5.2 \text{ cm}$.
Substituting the values into the derivative formula:
$\frac{dA}{dt} = 2 \times \pi \times 5.2 \times 0.05$.
$\frac{dA}{dt} = 10.4 \times 0.05 \times \pi = 0.52 \pi \text{ cm}^2/\text{s}$.
Thus,the rate at which the area is increasing is $0.52 \pi \text{ cm}^2/\text{s}$.

(A) Let the side of the equilateral triangle be $x$ and its area be $A$.
Given that the rate of change of the side is $\frac{dx}{dt} = 4 \text{ cm/sec}$.
The area of an equilateral triangle is given by $A = \frac{\sqrt{3}}{4} x^2$.
Differentiating both sides with respect to time $t$,we get:
$\frac{dA}{dt} = \frac{\sqrt{3}}{4} \cdot 2x \cdot \frac{dx}{dt}$
$\frac{dA}{dt} = \frac{\sqrt{3}}{2} \cdot x \cdot \frac{dx}{dt}$
Substituting the given values $x = 14 \text{ cm}$ and $\frac{dx}{dt} = 4 \text{ cm/sec}$:
$\frac{dA}{dt} = \frac{\sqrt{3}}{2} \cdot 14 \cdot 4$
$\frac{dA}{dt} = \sqrt{3} \cdot 7 \cdot 4$
$\frac{dA}{dt} = 28 \sqrt{3} \text{ cm}^2/\text{sec}$.
Thus,the area is increasing at the rate of $28 \sqrt{3} \text{ cm}^2/\text{sec}$.

(D) The area of a circle $A$ is given by the formula:
$A = \pi r^2$
To find the rate of change of the area with respect to the radius $r$,we differentiate $A$ with respect to $r$:
$\frac{dA}{dr} = \frac{d}{dr}(\pi r^2) = 2\pi r$
Now,we evaluate this derivative at $r = 2 \text{ cm}$:
$\left(\frac{dA}{dr}\right)_{r=2} = 2\pi(2) = 4\pi \text{ cm}^2/\text{cm}$
Thus,the rate of change is $4\pi$.

(A) Given the rate of increase in the volume of the sphere is $\frac{dV}{dt} = \pi \text{ cm}^3/\text{s}$.
We know that the volume of a sphere is $V = \frac{4}{3} \pi r^3$.
Differentiating with respect to $t$,we get $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$.
Substituting $\frac{dV}{dt} = \pi$,we have $\pi = 4 \pi r^2 \frac{dr}{dt}$,which implies $\frac{dr}{dt} = \frac{1}{4r^2}$.
The surface area of a sphere is $S = 4 \pi r^2$.
Differentiating with respect to $t$,we get $\frac{dS}{dt} = 8 \pi r \frac{dr}{dt}$.
Substituting $\frac{dr}{dt} = \frac{1}{4r^2}$,we get $\frac{dS}{dt} = 8 \pi r \left( \frac{1}{4r^2} \right) = \frac{2 \pi}{r}$.
When $r = 1 \text{ cm}$,the rate of increase in surface area is $\frac{dS}{dt} = \frac{2 \pi}{1} = 2 \pi \text{ cm}^2/\text{s}$.

(A) Let $O$ be the origin $(0,0)$. Let the position of $X$ at time $t$ be $x(t) = 4t$ along $OA$ and the position of $Y$ at time $t$ be $y(t) = 3t$ along $OB$.
Let $A$ be the shortest distance between $X$ and $Y$ at time $t$.
Using the Law of Cosines in $\triangle OXY$:
$A^2 = (4t)^2 + (3t)^2 - 2(4t)(3t) \cos(120^{\circ})$
Since $\cos(120^{\circ}) = -\frac{1}{2}$,we have:
$A^2 = 16t^2 + 9t^2 - 24t^2 \left(-\frac{1}{2}\right)$
$A^2 = 25t^2 + 12t^2 = 37t^2$
$A = \sqrt{37}t$
Differentiating with respect to $t$:
$2A \frac{dA}{dt} = 37(2t)$
$\frac{dA}{dt} = \frac{37t}{A}$
Substituting $A = \sqrt{37}t$:
$\frac{dA}{dt} = \frac{37t}{\sqrt{37}t} = \sqrt{37} \text{ km/h}$.
Thus,the rate at which the distance is increasing is $\sqrt{37} \text{ km/h}$.

(D) Let the side of the cube be $x$. The surface area $S$ of the cube is given by $S = 6x^2$.
When the side $x$ is increased by $5 \%$,the new side becomes $x' = x + 0.05x = 1.05x$.
The new surface area $S'$ is given by $S' = 6(1.05x)^2$.
$S' = 6(1.1025x^2) = 1.1025(6x^2) = 1.1025S$.
The percentage increase in surface area is $\frac{S' - S}{S} \times 100 \%$.
$= \frac{1.1025S - S}{S} \times 100 \% = 0.1025 \times 100 \% = 10.25 \%$.

(C) Given that,$f(x) = 2x^{2}$.
We need to evaluate the expression $\frac{f(3.8) - f(4)}{3.8 - 4}$.
Substitute the values into the function:
$\frac{f(3.8) - f(4)}{3.8 - 4} = \frac{2(3.8)^{2} - 2(4)^{2}}{3.8 - 4}$.
Factor out $2$ from the numerator:
$= \frac{2(3.8^{2} - 4^{2})}{3.8 - 4}$.
Using the algebraic identity $a^{2} - b^{2} = (a - b)(a + b)$,where $a = 3.8$ and $b = 4$:
$= \frac{2(3.8 - 4)(3.8 + 4)}{3.8 - 4}$.
Cancel the common term $(3.8 - 4)$ from the numerator and denominator:
$= 2(3.8 + 4)$.
$= 2(7.8) = 15.6$.

(A) Given the function $y = 5x^2 + 6x + 6$.
First,we find the derivative of $y$ with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(5x^2 + 6x + 6) = 10x + 6$.
By definition,the differential $dy$ is given by $dy = (\frac{dy}{dx}) dx$.
Here,$dx = \Delta x = 0.001$ and $x = 2$.
Substituting these values into the expression for $dy$:
$dy = (10(2) + 6) \times 0.001$.
$dy = (20 + 6) \times 0.001$.
$dy = 26 \times 0.001$.
$dy = 0.026$.

(B) Given,the displacement of a particle at time $t$ is $s = \frac{t^3}{3} - 6t$.
Velocity $v$ is the rate of change of displacement with respect to time: $v = \frac{ds}{dt} = \frac{d}{dt}(\frac{t^3}{3} - 6t) = t^2 - 6$.
When the velocity vanishes,$v = 0$,so $t^2 - 6 = 0$,which gives $t = \sqrt{6} \text{ s}$ (considering $t > 0$).
Acceleration $a$ is the rate of change of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(t^2 - 6) = 2t$.
Substituting $t = \sqrt{6} \text{ s}$ into the acceleration equation,we get $a = 2(\sqrt{6}) = 2\sqrt{6} \text{ m/s}^2$.

(D) Given displacement $s = 5 \sin(2t)$.
Velocity $v$ is the rate of change of displacement with respect to time $t$,given by $v = \frac{ds}{dt}$.
$v = \frac{d}{dt} [5 \sin(2t)] = 5 \cdot \cos(2t) \cdot 2 = 10 \cos(2t)$.
Now,substitute $t = \frac{\pi}{3}$ into the velocity equation:
$v = 10 \cos(2 \cdot \frac{\pi}{3}) = 10 \cos(\frac{2\pi}{3})$.
Since $\cos(\frac{2\pi}{3}) = \cos(120^{\circ}) = -\frac{1}{2}$,
$v = 10 \cdot (-\frac{1}{2}) = -5$.
Thus,the velocity is $-5$. The correct option is $D$.

(C) Given the distance function: $s = t^2 - 2t + 5$.
Velocity $v$ is the first derivative of distance with respect to time $t$:
$v = \frac{ds}{dt} = \frac{d}{dt}(t^2 - 2t + 5) = 2t - 2$.
Acceleration $a$ is the derivative of velocity with respect to time $t$:
$a = \frac{dv}{dt} = \frac{d}{dt}(2t - 2) = 2$.
Therefore,the acceleration of the particle is $2$ units.

(A) The velocity $(v)$ of a particle is the rate of change of distance $(s)$ with respect to time $(t)$,which is given by the derivative $v = \frac{ds}{dt}$.
Given the distance equation $s = 4t^2 + 2t + 3$.
Differentiating with respect to $t$:
$v = \frac{d}{dt}(4t^2 + 2t + 3) = 8t + 2$.
To find the velocity at $t = 3$ seconds,substitute $t = 3$ into the velocity equation:
$v = 8(3) + 2 = 24 + 2 = 26 \text{ unit/sec}$.

(B) Given that the velocity $v$ is proportional to the cube root of displacement $x$:
$v = k x^{1/3}$,where $k$ is a constant.
We know that acceleration $a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx}$.
First,find $\frac{dv}{dx}$:
$\frac{dv}{dx} = k \cdot \frac{1}{3} x^{-2/3} = \frac{k}{3 x^{2/3}}$.
Now,substitute $v$ and $\frac{dv}{dx}$ into the acceleration formula:
$a = (k x^{1/3}) \cdot \left( \frac{k}{3 x^{2/3}} \right) = \frac{k^2}{3 x^{1/3}}$.
Since $v = k x^{1/3}$,we can write $x^{1/3} = \frac{v}{k}$.
Substituting this into the expression for $a$:
$a = \frac{k^2}{3 (v/k)} = \frac{k^3}{3v}$.
Therefore,$a \propto \frac{1}{v}$,which means the acceleration is inversely proportional to its velocity.

(C) Let $V$ be the volume and $S$ be the surface area of the sphere with radius $r$.
Given,$\frac{dV}{dt} = 12 \text{ cm}^3/\text{sec}$.
The volume of a sphere is $V = \frac{4}{3} \pi r^3$.
Differentiating with respect to $t$,we get $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$.
Given diameter $d = 12 \text{ cm}$,so radius $r = 6 \text{ cm}$.
Substituting the values: $12 = 4 \pi (6)^2 \frac{dr}{dt} \implies 12 = 144 \pi \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{12}{144 \pi} = \frac{1}{12 \pi} \text{ cm/sec}$.
The surface area of a sphere is $S = 4 \pi r^2$.
Differentiating with respect to $t$,we get $\frac{dS}{dt} = 8 \pi r \frac{dr}{dt}$.
Substituting $r = 6$ and $\frac{dr}{dt} = \frac{1}{12 \pi}$:
$\frac{dS}{dt} = 8 \pi (6) \left( \frac{1}{12 \pi} \right) = 48 \pi \times \frac{1}{12 \pi} = 4 \text{ cm}^2/\text{sec}$.

(D) Let $r$ be the radius,$S$ be the surface area,and $V$ be the volume of the spherical bubble.
Given: $\frac{dS}{dt} = 4 \text{ cm}^2/\text{sec}$.
The surface area of a sphere is $S = 4\pi r^2$.
Differentiating with respect to $t$: $\frac{dS}{dt} = 8\pi r \frac{dr}{dt}$.
Substituting the given values: $4 = 8\pi(8) \frac{dr}{dt} \implies 4 = 64\pi \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{1}{16\pi} \text{ cm/sec}$.
The volume of a sphere is $V = \frac{4}{3}\pi r^3$.
Differentiating with respect to $t$: $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$.
Substituting $r = 8$ and $\frac{dr}{dt} = \frac{1}{16\pi}$:
$\frac{dV}{dt} = 4\pi(8)^2 \left(\frac{1}{16\pi}\right) = 4\pi(64) \left(\frac{1}{16\pi}\right) = 4(4) = 16 \text{ cm}^3/\text{sec}$.

(A) The velocity $v$ of the particle is the rate of change of displacement with respect to time $t$,given by $v = \frac{dS}{dt}$.
Given $S = 2t^3 + 2t^2 - 2t - 3$.
Differentiating with respect to $t$,we get $v = \frac{d}{dt}(2t^3 + 2t^2 - 2t - 3) = 6t^2 + 4t - 2$.
$A$ particle changes its direction when its velocity becomes zero.
Setting $v = 0$,we have $6t^2 + 4t - 2 = 0$.
Dividing by $2$,we get $3t^2 + 2t - 1 = 0$.
Factoring the quadratic equation: $3t^2 + 3t - t - 1 = 0 \implies 3t(t + 1) - 1(t + 1) = 0$.
$(3t - 1)(t + 1) = 0$.
This gives $t = \frac{1}{3}$ or $t = -1$.
Since time cannot be negative,we take $t = \frac{1}{3}$ seconds.

(A) Given the curve equation: $y = x^3 - 3x^2 + 2x - 1$.
Differentiating both sides with respect to time $t$,we get:
$\frac{dy}{dt} = (3x^2 - 6x + 2) \frac{dx}{dt}$.
We are given that $\frac{dy}{dt} = 6$ units/sec and the point is $(2, -1)$,so $x = 2$.
Substituting these values into the derivative equation:
$6 = (3(2)^2 - 6(2) + 2) \frac{dx}{dt}$.
$6 = (12 - 12 + 2) \frac{dx}{dt}$.
$6 = 2 \frac{dx}{dt}$.
Therefore,$\frac{dx}{dt} = \frac{6}{2} = 3$ units/sec.

(C) Given $y = x - x^2$.
We need to find the rate of change of $y^2$ with respect to $x^2$.
Let $v = y^2 = (x - x^2)^2 = x^2 + x^4 - 2x^3$.
Let $u = x^2$.
We want to calculate $\frac{dv}{du} = \frac{dv/dx}{du/dx}$.
First,differentiate $v$ with respect to $x$: $\frac{dv}{dx} = \frac{d}{dx}(x^2 + x^4 - 2x^3) = 2x + 4x^3 - 6x^2$.
Next,differentiate $u$ with respect to $x$: $\frac{du}{dx} = \frac{d}{dx}(x^2) = 2x$.
Now,$\frac{dv}{du} = \frac{2x + 4x^3 - 6x^2}{2x} = 1 + 2x^2 - 3x$.
At $x = 2$,the rate of change is $1 + 2(2)^2 - 3(2) = 1 + 8 - 6 = 3$.

(C) Given displacement $s = 2t^3 - 9t$.
Velocity $v = \frac{ds}{dt} = \frac{d}{dt}(2t^3 - 9t) = 6t^2 - 9$.
The velocity vanishes when $v = 0$,so $6t^2 - 9 = 0 \Rightarrow t^2 = \frac{9}{6} = \frac{3}{2} \Rightarrow t = \sqrt{\frac{3}{2}}$.
Acceleration $a = \frac{dv}{dt} = \frac{d}{dt}(6t^2 - 9) = 12t$.
Substituting $t = \sqrt{\frac{3}{2}}$ into the acceleration equation:
$a = 12 \times \sqrt{\frac{3}{2}} = 12 \times \frac{\sqrt{3}}{\sqrt{2}} = 6 \times \sqrt{2} \times \sqrt{3} = 6\sqrt{6}$.

(C) Let $OA$ be the light pole of height $6 \ m$ and $FG$ be the man of height $1.8 \ m$. Let $x$ be the distance of the man from the pole and $y$ be the length of his shadow.
From the similar triangles $\triangle BGF$ and $\triangle BOA$,we have:
$\frac{FG}{OA} = \frac{BG}{BO}$
$\frac{1.8}{6} = \frac{y}{x+y}$
$1.8(x+y) = 6y$
$1.8x + 1.8y = 6y$
$1.8x = 4.2y$
$y = \frac{1.8}{4.2}x = \frac{18}{42}x = \frac{3}{7}x$
Differentiating both sides with respect to time $t$:
$\frac{dy}{dt} = \frac{3}{7} \frac{dx}{dt}$
Given that the man is walking away from the pole at a speed of $7 \ km/h$,so $\frac{dx}{dt} = 7 \ km/h$.
Therefore,$\frac{dy}{dt} = \frac{3}{7} \times 7 = 3 \ km/h$.
The rate of change of the length of his shadow is $3 \ km/h$.

(B) Given that the radius of the circle is $OP = 8$.
$M$ is the foot of the perpendicular from $P$ on $OA$,so $\triangle OMP$ is a right-angled triangle at $M$.
In $\triangle OMP$,we have $\cos \theta = \frac{OM}{OP}$.
Given $OM = 4$ and $OP = 8$,so $\cos \theta = \frac{4}{8} = \frac{1}{2}$.
Also,$PM = OP \sin \theta = 8 \sin \theta$.
To find the rate of change of $PM$ with respect to time $t$,we differentiate $PM$ with respect to $t$:
$\frac{d(PM)}{dt} = \frac{d}{dt}(8 \sin \theta) = 8 \cos \theta \frac{d\theta}{dt}$.
Given $\frac{d\theta}{dt} = 6 \text{ rad/sec}$ and $\cos \theta = \frac{1}{2}$,we substitute these values:
$\frac{d(PM)}{dt} = 8 \times \frac{1}{2} \times 6 = 24 \text{ units/sec}$.

(D) The area of a circle is given by $A = \pi r^2$.
Taking the natural logarithm on both sides,we get $\ln A = \ln \pi + 2 \ln r$.
Differentiating both sides with respect to $r$,we obtain $\frac{dA}{A} = 2 \frac{dr}{r}$.
For small errors,this can be written as $\frac{\Delta A}{A} = 2 \times \frac{\Delta r}{r}$.
Given that the relative error in the radius is $\frac{\Delta r}{r} = 0.05 \%$,the relative error in the area is $\frac{\Delta A}{A} = 2 \times 0.05 \% = 0.1 \%$.
Thus,the corresponding error in calculating the area is $0.1 \%$.

(D) Let $D$ be the diameter,$r$ be the radius,and $h$ be the altitude of the right circular cone.
Given $D = 10 \ cm$,so $r = 5 \ cm$.
Given $h = 20 \ cm$.
The rate of change of diameter is $\frac{dD}{dt} = 2 \ cm/s$.
Since $D = 2r$,we have $\frac{dr}{dt} = \frac{1}{2} \frac{dD}{dt} = \frac{1}{2} \times 2 = 1 \ cm/s$.
The volume of the cone is $V = \frac{1}{3} \pi r^2 h$.
To keep the volume constant,$\frac{dV}{dt} = 0$.
Using the product rule,$\frac{dV}{dt} = \frac{1}{3} \pi (2rh \frac{dr}{dt} + r^2 \frac{dh}{dt}) = 0$.
Substituting the values: $2(5)(20)(1) + (5)^2 \frac{dh}{dt} = 0$.
$200 + 25 \frac{dh}{dt} = 0$.
$25 \frac{dh}{dt} = -200$.
$\frac{dh}{dt} = -8 \ cm/s$.
Thus,the altitude must change at a rate of $-8 \ cm/s$.

(D) Let the radius of the iron ball be $r_0 = 10 \ cm$ and the thickness of the ice layer be $x \ cm$. The total radius of the sphere (iron ball + ice) is $R = 10 + x \ cm$.
The volume of the ice layer is $V = \frac{4}{3}\pi (10+x)^3 - \frac{4}{3}\pi (10)^3$.
Since the iron ball is constant,the rate of change of the total volume is equal to the rate of change of the ice volume.
$V = \frac{4}{3}\pi (10+x)^3$.
Differentiating with respect to time $t$:
$\frac{dV}{dt} = 4\pi (10+x)^2 \frac{dx}{dt}$.
Given $\frac{dV}{dt} = -50 \ cm^3/min$ (since it melts) and $x = 5 \ cm$:
$-50 = 4\pi (10+5)^2 \frac{dx}{dt}$.
$-50 = 4\pi (15)^2 \frac{dx}{dt}$.
$-50 = 4\pi (225) \frac{dx}{dt}$.
$-50 = 900\pi \frac{dx}{dt}$.
$\frac{dx}{dt} = -\frac{50}{900\pi} = -\frac{1}{18\pi} \ cm/min$.
The negative sign indicates a decrease in thickness.
Therefore,the rate at which the thickness decreases is $\frac{1}{18\pi} \ cm/min$.
Thus,option $(D)$ is correct.

(C) Let $v$ be the volume and $s$ be the surface area of the spherical balloon with radius $r$.
Given: $\frac{dv}{dt} = 30 \ cm^3/min$.
The volume of a sphere is $v = \frac{4}{3} \pi r^3$.
Differentiating with respect to $t$: $\frac{dv}{dt} = 4 \pi r^2 \frac{dr}{dt}$.
Substituting the given values: $30 = 4 \pi r^2 \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{30}{4 \pi r^2} = \frac{15}{2 \pi r^2}$.
The surface area of a sphere is $s = 4 \pi r^2$.
Differentiating with respect to $t$: $\frac{ds}{dt} = 8 \pi r \frac{dr}{dt}$.
Substituting $\frac{dr}{dt}$: $\frac{ds}{dt} = 8 \pi r \left( \frac{15}{2 \pi r^2} \right) = \frac{60}{r}$.
For $r = 6 \ cm$: $\frac{ds}{dt} = \frac{60}{6} = 10 \ cm^2/min$.

(B) Given,$\frac{dV}{dt} = 4 \pi \text{ cm}^3/\text{s}$ ...$(i)$
We need to find $\frac{dr}{dt}$ when $V = 288 \pi \text{ cm}^3$.
Let $r$ be the radius of the spherical ball.
The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$.
Differentiating both sides with respect to $t$,we get:
$\frac{dV}{dt} = \frac{4}{3} \pi (3r^2) \frac{dr}{dt} = 4 \pi r^2 \frac{dr}{dt}$.
Substituting the given rate $\frac{dV}{dt} = 4 \pi$:
$4 \pi = 4 \pi r^2 \frac{dr}{dt} \Rightarrow \frac{dr}{dt} = \frac{1}{r^2}$ ...(ii)
When $V = 288 \pi$,we have:
$288 \pi = \frac{4}{3} \pi r^3 \Rightarrow r^3 = 288 \times \frac{3}{4} = 216$.
Thus,$r = \sqrt[3]{216} = 6 \text{ cm}$.
Substituting $r = 6$ into equation (ii):
$\frac{dr}{dt} = \frac{1}{6^2} = \frac{1}{36} \text{ cm/s}$.

(C) Given the curve $y = 5x - 2x^3$.
The slope of the curve is given by $m = \frac{dy}{dx} = 5 - 6x^2$.
We need to find the rate of change of the slope with respect to time $t$,which is $\frac{dm}{dt}$.
Differentiating $m$ with respect to $t$ using the chain rule:
$\frac{dm}{dt} = \frac{d}{dt}(5 - 6x^2) = -12x \cdot \frac{dx}{dt}$.
Given that $\frac{dx}{dt} = 2 \text{ units/sec}$ and we need to evaluate this at $x = 3$:
$\left(\frac{dm}{dt}\right)_{x=3} = -12(3) \times 2 = -72$.
Thus,the rate of change of the slope at $x = 3$ is $-72 \text{ units/sec}^2$.

(D) Given the distance function $s = 60t - 5t^2$.
The velocity $v$ is the rate of change of distance with respect to time $t$,given by $v = \frac{ds}{dt}$.
$\frac{ds}{dt} = \frac{d}{dt}(60t - 5t^2) = 60 - 10t$.
The particle comes to rest when its velocity $v = 0$.
Setting $60 - 10t = 0$,we get $10t = 60$,which implies $t = 6 \text{ s}$.
Now,we calculate the distance covered at $t = 6 \text{ s}$:
$s(6) = 60(6) - 5(6)^2$
$s(6) = 360 - 5(36)$
$s(6) = 360 - 180 = 180 \text{ units}$.
Thus,the distance covered before coming to rest is $180 \text{ units}$.

(B) Let $N(t)$ be the number of bacteria at time $t$. The growth is given by $N(t) = t^3$.
The rate of growth of the bacteria is given by the derivative $\frac{dN}{dt} = \frac{d}{dt}(t^3) = 3t^2$.
We are given that the rate of growth is $1200 \ \text{per } s$.
Therefore,$3t^2 = 1200$.
Dividing by $3$,we get $t^2 = 400$.
Taking the square root,$t = \sqrt{400} = 20 \ s$.
Thus,the time taken is $20 \ s$.

(A) The area $A$ of an equilateral triangle with side $a$ is given by:
$A = \frac{\sqrt{3}}{4} a^2$
Differentiating both sides with respect to time $t$,we get:
$\frac{dA}{dt} = \frac{\sqrt{3}}{4} \cdot 2a \cdot \frac{da}{dt}$
$\frac{dA}{dt} = \frac{\sqrt{3}}{2} a \cdot \frac{da}{dt}$
Given that the rate of increase of the side is $\frac{da}{dt} = 2 \text{ cm s}^{-1}$ and the side length is $a = 10 \text{ cm}$,we substitute these values into the equation:
$\frac{dA}{dt} = \frac{\sqrt{3}}{2} \times 10 \times 2$
$\frac{dA}{dt} = 10 \sqrt{3} \text{ cm}^2 \text{ s}^{-1}$
Thus,the area of the triangle increases at the rate of $10 \sqrt{3} \text{ cm}^2 \text{ s}^{-1}$.

(D) Let $r$ be the radius of the sphere.
Given,the rate of change in radius is $\frac{dr}{dt} = 0.04 \text{ cm/sec}$.
The volume of the sphere is $V = \frac{4}{3} \pi r^3$.
Differentiating with respect to $t$,we get:
$\frac{dV}{dt} = \frac{4}{3} \pi (3r^2) \cdot \frac{dr}{dt} = 4 \pi r^2 \cdot \frac{dr}{dt} \quad \dots(i)$
The surface area of the sphere is $S = 4 \pi r^2$.
Differentiating with respect to $t$,we get:
$\frac{dS}{dt} = 4 \pi (2r) \cdot \frac{dr}{dt} = 8 \pi r \cdot \frac{dr}{dt} \quad \dots(ii)$
Dividing equation $(i)$ by equation $(ii)$,we get:
$\frac{dV/dt}{dS/dt} = \frac{dV}{dS} = \frac{4 \pi r^2 \cdot \frac{dr}{dt}}{8 \pi r \cdot \frac{dr}{dt}}$
$\frac{dV}{dS} = \frac{r}{2}$
Given $r = 10 \text{ cm}$,we substitute the value:
$\frac{dV}{dS} = \frac{10}{2} = 5 \text{ cm}$.
Thus,the rate of increase in volume with respect to surface area is $5 \text{ cm}$.
Hence,option $(D)$ is correct.

(C) Let $x$ be the distance of the lower end of the ladder from the wall and $y$ be the height of the upper end of the ladder from the ground.
Given 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 $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$,which simplifies to $x \frac{dx}{dt} + y \frac{dy}{dt} = 0$.
We are given $\frac{dx}{dt} = 3 \ m/sec$ and the ladder is descending at $\frac{dy}{dt} = -4 \ m/sec$.
Substituting these values into the differentiated equation: $x(3) + y(-4) = 0$,which gives $3x = 4y$,or $x = \frac{4}{3}y$.
Substitute $x = \frac{4}{3}y$ into the original equation $x^2 + y^2 = 25$:
$(\frac{4}{3}y)^2 + y^2 = 25$
$\frac{16}{9}y^2 + y^2 = 25$
$\frac{25}{9}y^2 = 25$
$y^2 = 9$
$y = 3 \ m$ (since height must be positive).
Thus,the height of the upper end is $3 \ m$.

(B) Let $V$ be the volume and $S$ be the surface area of the spherical balloon with radius $r$.
Given, the rate of discharge of air is $\frac{dV}{dt} = -4 \,m^3 / min$ (since air is discharging, volume is decreasing).
The volume of a sphere is $V = \frac{4}{3} \pi r^3$.
Differentiating with respect to $t$, we get $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$.
Substituting the given values at $r = 8 \,m$:
$-4 = 4 \pi (8)^2 \frac{dr}{dt} \Rightarrow -4 = 256 \pi \frac{dr}{dt} \Rightarrow \frac{dr}{dt} = -\frac{1}{64 \pi} \,m / min$.
The surface area of a sphere is $S = 4 \pi r^2$.
Differentiating with respect to $t$, we get $\frac{dS}{dt} = 8 \pi r \frac{dr}{dt}$.
Substituting $r = 8 \,m$ and $\frac{dr}{dt} = -\frac{1}{64 \pi} \,m / min$:
$\frac{dS}{dt} = 8 \pi (8) \left( -\frac{1}{64 \pi} \right) = -1 \,m^2 / min$.
Since the surface area is shrinking at a rate of $1 \,m^2 / min$, the rate of shrinking is $1 \,m^2 / min$.

(B) Let $x$ be the length of the edge of the cube and $V$ be its volume.
Given that the rate of change of the edge length is $\frac{dx}{dt} = 1 \text{ cm/sec}$.
The volume of a cube is given by $V = x^3$.
Differentiating both sides with respect to time $t$,we get $\frac{dV}{dt} = 3x^2 \frac{dx}{dt}$.
When the edge length $x = 5 \text{ cm}$,we substitute the values into the derivative:
$\frac{dV}{dt} = 3(5)^2(1) = 3 \times 25 \times 1 = 75 \text{ cc/sec}$.
Thus,the rate of change of volume is $75 \text{ cc/sec}$.