Rate of Change of Quantities Questions in English

(D) Given the position function $S(t) = t^3 - 3t^2 + 4t - 2$.
Velocity $v(t)$ is the derivative of $S(t)$ with respect to $t$:
$v(t) = \frac{dS}{dt} = 3t^2 - 6t + 4$.
Acceleration $a(t)$ is the derivative of $v(t)$ with respect to $t$:
$a(t) = \frac{dv}{dt} = 6t - 6$.
Set acceleration to zero to find the time $t$:
$6t - 6 = 0 \implies t = 1 \text{ second}$.
Now,substitute $t = 1$ into the velocity function:
$v(1) = 3(1)^2 - 6(1) + 4 = 3 - 6 + 4 = 1 \text{ m/s}$.
Thus,the velocity of the particle when its acceleration is zero is $1 \text{ m/s}$.

(D) Given the curve $x^3 = 12y$. Differentiating both sides with respect to time $t$,we get:
$3x^2 \frac{dx}{dt} = 12 \frac{dy}{dt}$
$\frac{dy}{dt} = \frac{3x^2}{12} \frac{dx}{dt} = \frac{x^2}{4} \frac{dx}{dt}$
Given that the rate of change of $x$ is more than the rate of change of $y$,i.e.,$\frac{dx}{dt} > \frac{dy}{dt}$.
Substituting the value of $\frac{dy}{dt}$:
$\frac{dx}{dt} > \frac{x^2}{4} \frac{dx}{dt}$
Since $x > 0$,$\frac{dx}{dt}$ is positive (assuming $x$ is increasing),so we can divide by $\frac{dx}{dt}$:
$1 > \frac{x^2}{4}$
$x^2 < 4$
$|x| < 2$
Since the condition $x > 0$ is given,the interval for $x$ is $(0, 2)$.

(A) Let $V$ be the volume,$r$ be the radius,and $h$ be the height of the water in the inverted cone at any time $t$.
Given,$\frac{dV}{dt} = 3 \ m^3/min$.
The volume of the cone is $V = \frac{1}{3} \pi r^2 h$.
From the similarity of triangles in the cone,we have $\frac{r}{h} = \frac{4}{6} = \frac{2}{3}$,which implies $r = \frac{2}{3}h$.
Substituting $r$ in the volume formula:
$V = \frac{1}{3} \pi \left(\frac{2}{3}h\right)^2 h = \frac{1}{3} \pi \left(\frac{4}{9}h^2\right) h = \frac{4}{27} \pi h^3$.
Differentiating with respect to $t$:
$\frac{dV}{dt} = \frac{4}{27} \pi (3h^2) \frac{dh}{dt} = \frac{4}{9} \pi h^2 \frac{dh}{dt}$.
Given $\frac{dV}{dt} = 3$ and we need to find $\frac{dh}{dt}$ when $h = 3 \ m$:
$3 = \frac{4}{9} \pi (3)^2 \frac{dh}{dt}$
$3 = \frac{4}{9} \pi (9) \frac{dh}{dt}$
$3 = 4 \pi \frac{dh}{dt}$
$\frac{dh}{dt} = \frac{3}{4 \pi} \ m/min$.

(B) Let $V$ be the volume,$S$ be the surface area,and $r$ be the radius of the sphere.
It is given that $\frac{dV}{dt} = 1200 \text{ cm}^3/\text{s}$ and $r = 10 \text{ cm}$.
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: $1200 = 4 \pi (10)^2 \frac{dr}{dt} \implies 1200 = 400 \pi \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{3}{\pi} \text{ cm/s}$.
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 $r = 10 \text{ cm}$ and $\frac{dr}{dt} = \frac{3}{\pi} \text{ cm/s}$:
$\frac{dS}{dt} = 8 \pi (10) \left( \frac{3}{\pi} \right) = 80 \times 3 = 240 \text{ cm}^2/\text{s}$.

(B) Given that the rate of change of volume is $\frac{dV}{dt} = 2 \pi \text{ cm}^3/\text{s}$.
We know that the volume of a sphere is $V = \frac{4}{3} \pi r^3$.
Differentiating both sides with respect to $t$,we get $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$.
Given $V = 288 \pi$,we find the radius $r$ as:
$288 \pi = \frac{4}{3} \pi r^3 \Rightarrow 216 = r^3 \Rightarrow r = 6 \text{ cm}$.
Substituting the values into the derivative equation:
$2 \pi = 4 \pi (6)^2 \frac{dr}{dt}$
$2 \pi = 4 \pi (36) \frac{dr}{dt}$
$2 \pi = 144 \pi \frac{dr}{dt}$
$\frac{dr}{dt} = \frac{2 \pi}{144 \pi} = \frac{1}{72} \text{ cm}/\text{s}$.

(A) The area $A$ of a circular plate with radius $r$ is given by $A = \pi r^2$.
To find the rate at which the area increases,we differentiate $A$ with respect to time $t$:
$\frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = 2 \pi r \frac{dr}{dt}$.
Given that $\frac{dr}{dt} = 0.01 \text{ cm/s}$ and $r = 12 \text{ cm}$,we substitute these values into the equation:
$\frac{dA}{dt} = 2 \pi (12) (0.01) = 0.24 \pi \text{ cm}^2/\text{s}$.
Thus,the rate at which the area increases is $0.24 \pi \text{ cm}^2/\text{s}$.

(A) Given that,the rate of change of volume is $\frac{dV}{dt} = 30 \ ft^3 / \text{min}$ and the radius $r = 15 \ ft$.
The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$.
Differentiating both sides with respect to time $t$,we get $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$.
Substituting the given values,$30 = 4 \pi (15)^2 \frac{dr}{dt}$.
$30 = 4 \pi (225) \frac{dr}{dt} = 900 \pi \frac{dr}{dt}$.
Therefore,$\frac{dr}{dt} = \frac{30}{900 \pi} = \frac{1}{30 \pi} \ ft / \text{min}$.

(B) Given the displacement function: $s = \frac{1}{3} t^3 - 3 t^2 + 9 t + 17$.
The velocity $v$ is the derivative of displacement with respect to time: $v = \frac{ds}{dt} = t^2 - 6 t + 9$.
To find when the velocity decreases,we calculate the acceleration $a = \frac{dv}{dt}$.
$a = \frac{dv}{dt} = 2 t - 6$.
Velocity decreases when the acceleration is negative,i.e.,$\frac{dv}{dt} < 0$.
$2 t - 6 < 0 \implies 2 t < 6 \implies t < 3$.
Since time $t$ must be greater than $0$,the velocity decreases in the interval $0 < t < 3$.

(C) Let $a$ be the side of the equilateral triangle and $A$ be its area.
The area of an equilateral triangle is given by $A = \frac{\sqrt{3}}{4} a^2$.
Differentiating both sides with respect to time $t$,we get:
$\frac{dA}{dt} = \frac{\sqrt{3}}{4} \times 2a \times \frac{da}{dt} = \frac{\sqrt{3}}{2} a \frac{da}{dt}$.
Given that $\frac{da}{dt} = 2 \text{ cm/s}$ and $a = 20 \text{ cm}$.
Substituting these values into the equation:
$\frac{dA}{dt} = \frac{\sqrt{3}}{2} \times 20 \times 2 = 20 \sqrt{3} \text{ cm}^2/\text{s}$.
Thus,the rate at which the area is increasing is $20 \sqrt{3} \text{ cm}^2/\text{s}$.

(B) The growth function of the bacteria is given by $f(t) = t^4$.
To find the rate of growth,we calculate the derivative of $f(t)$ with respect to time $t$:
$f'(t) = \frac{d}{dt}(t^4) = 4t^3$.
We are given that the rate of growth at $t = t_0$ is $4000 \text{ units/second}$.
Therefore,$f'(t_0) = 4t_0^3 = 4000$.
Dividing both sides by $4$,we get $t_0^3 = 1000$.
Taking the cube root of both sides,$t_0 = \sqrt[3]{1000} = 10$.
Thus,$t_0 = 10$ seconds.

(A) Let $AB$ be the street light of height $H = 5 \frac{1}{3} = \frac{16}{3} \text{ m}$. Let $CD$ be the man of height $h = 2 \text{ m}$. Let $AC = x$ be the distance of the man from the light and $CE = y$ be the length of his shadow.
From similar triangles $\triangle ABE$ and $\triangle DCE$,we have:
$\frac{AB}{CD} = \frac{AE}{CE} \Rightarrow \frac{16/3}{2} = \frac{x+y}{y}$
$\frac{8}{3} = \frac{x}{y} + 1$ $\Rightarrow \frac{x}{y} = \frac{5}{3}$ $\Rightarrow y = \frac{3}{5}x$
Differentiating with respect to time $t$:
$\frac{dy}{dt} = \frac{3}{5} \frac{dx}{dt}$
Given that the man walks towards the light at $1 \frac{2}{3} \text{ m/s}$,so $\frac{dx}{dt} = -\frac{5}{3} \text{ m/s}$ (negative because $x$ is decreasing).
Therefore,$\frac{dy}{dt} = \frac{3}{5} \times (-\frac{5}{3}) = -1 \text{ m/s}$.
The length of the shadow is changing at the rate of $-1 \text{ m/s}$.

(A) Let $AB$ be the lamp-post and $PQ$ be the man. Let $C$ be the tip of the shadow. Let $AP = x$ be the distance of the man from the lamp-post and $PC = y$ be the length of his shadow.
Given: $AB = 9 \ m$,$PQ = 2 \ m$,and $\frac{dx}{dt} = 7 \ m/min$.
Since $\triangle CAB$ and $\triangle CPQ$ are similar triangles,we have:
$\frac{PC}{AC} = \frac{PQ}{AB}$
$\frac{y}{x+y} = \frac{2}{9}$
$9y = 2x + 2y$
$7y = 2x$
$x = \frac{7}{2}y$
Differentiating both sides with respect to time $t$:
$\frac{dx}{dt} = \frac{7}{2} \frac{dy}{dt}$
Substituting $\frac{dx}{dt} = 7$:
$7 = \frac{7}{2} \frac{dy}{dt}$
$\frac{dy}{dt} = 2 \ m/min$.
Thus,the length of his shadow increases at a rate of $2 \ m/min$.

(A) Let $A_1$ be the area of the first square and $A_2$ be the area of the second square.
Given the side of the first square is $x$,its area is $A_1 = x^2$.
Given the side of the second square is $y = x - x^2$,its area is $A_2 = y^2 = (x - x^2)^2$.
Expanding this,we get $A_2 = x^2 - 2 x^3 + x^4$.
We need to find the rate of change of $A_2$ with respect to $A_1$,which is $\frac{d A_2}{d A_1}$.
Using the chain rule,$\frac{d A_2}{d A_1} = \frac{d A_2 / d x}{d A_1 / d x}$.
First,calculate the derivatives with respect to $x$:
$\frac{d A_1}{d x} = \frac{d}{d x}(x^2) = 2 x$.
$\frac{d A_2}{d x} = \frac{d}{d x}(x^2 - 2 x^3 + x^4) = 2 x - 6 x^2 + 4 x^3$.
Now,substitute these into the chain rule formula:
$\frac{d A_2}{d A_1} = \frac{2 x - 6 x^2 + 4 x^3}{2 x} = \frac{2 x(1 - 3 x + 2 x^2)}{2 x} = 1 - 3 x + 2 x^2$.

(B) Let the radius of the circle be $r \text{ cm}$ and its area be $A \text{ cm}^2$.
Given that the rate of change of the radius is $\frac{dr}{dt} = 0.1 \text{ cm s}^{-1}$.
The area of a circle is given by $A = \pi r^2$.
Differentiating both sides with respect to time $t$,we get $\frac{dA}{dt} = 2 \pi r \frac{dr}{dt}$.
Substituting the given values $r = 5 \text{ cm}$ and $\frac{dr}{dt} = 0.1 \text{ cm s}^{-1}$,we have:
$\frac{dA}{dt} = 2 \pi (5) (0.1) = 10 \pi (0.1) = \pi \text{ cm}^2 \text{ s}^{-1}$.
Thus,the rate of change of the area is $\pi \text{ cm}^2 \text{ s}^{-1}$.

(D) Given the curve $y = x^3 - 2x^2 + 3x - 2$.
The slope of the tangent $m$ is given by $\frac{dy}{dx} = 3x^2 - 4x + 3$.
We need to find the rate of change of the slope $m$ with respect to time $t$,which is $\frac{dm}{dt}$.
Differentiating $m$ with respect to $x$,we get $\frac{dm}{dx} = \frac{d}{dx}(3x^2 - 4x + 3) = 6x - 4$.
By the chain rule,$\frac{dm}{dt} = \frac{dm}{dx} \cdot \frac{dx}{dt} = (6x - 4) \frac{dx}{dt}$.
At the point $(2, 4)$,$x = 2$.
Substituting $x = 2$ into the expression,we get $\frac{dm}{dt} = (6(2) - 4) \frac{dx}{dt} = (12 - 4) \frac{dx}{dt} = 8 \frac{dx}{dt}$.
The problem states that $\frac{dm}{dt} = k \cdot \frac{dx}{dt}$.
Comparing the two expressions,we find $k = 8$.

(C) The distance of the particle is given by $S = f(t) = t^3 - 5t^2 + 8t$.
Velocity $v(t)$ is the first derivative of distance with respect to time $t$:
$v(t) = \frac{dS}{dt} = \frac{d}{dt}(t^3 - 5t^2 + 8t) = 3t^2 - 10t + 8$.
Acceleration $a(t)$ is the derivative of velocity with respect to time $t$:
$a(t) = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 10t + 8) = 6t - 10$.
To find the acceleration at $t = 5 \text{ sec}$,substitute $t = 5$ into the acceleration equation:
$a(5) = 6(5) - 10 = 30 - 10 = 20 \text{ cm/sec}^2$.
Thus,the acceleration of the particle at $t = 5 \text{ sec}$ is $20 \text{ cm/sec}^2$.

(D) Let the position of the observer be at the origin $(0, 0)$. The balloon is at a constant altitude $y = 30 \ m$. Let its horizontal position at time $t$ be $x(t)$. Given that the balloon moves horizontally at $1 \ m/s$,we have $x(t) = 1 \cdot t = t$ (assuming it starts at $x=0$ at $t=0$).
The distance $s$ of the balloon from the observer is given by $s = \sqrt{x^2 + y^2} = \sqrt{t^2 + 30^2}$.
To find the rate at which the balloon is moving away from the observer,we differentiate $s$ with respect to $t$:
$\frac{ds}{dt} = \frac{1}{2\sqrt{t^2 + 30^2}} \cdot 2t = \frac{t}{\sqrt{t^2 + 30^2}}$.
At $t = 40 \ s$:
$\frac{ds}{dt} = \frac{40}{\sqrt{40^2 + 30^2}} = \frac{40}{\sqrt{1600 + 900}} = \frac{40}{\sqrt{2500}} = \frac{40}{50} = 0.8 \ m/s$.
Thus,the rate is $0.8 \ m/s$.

(B) Let $H = 15 \text{ ft}$ be the height of the light and $h = 5 \text{ ft}$ be the height of the man.
Let $x$ be the distance of the man from the light source and $s$ be the length of his shadow.
By similar triangles,we have $\frac{s}{h} = \frac{x+s}{H}$.
Substituting the values,$\frac{s}{5} = \frac{x+s}{15} \implies 3s = x + s \implies 2s = x$.
Differentiating with respect to time $t$,we get $2 \frac{ds}{dt} = \frac{dx}{dt}$.
Given $\frac{ds}{dt} = \frac{11}{5} \text{ ft/sec}$,so $\frac{dx}{dt} = 2 \times \frac{11}{5} = \frac{22}{5} \text{ ft/sec}$.
To convert $\frac{dx}{dt}$ to $\text{miles/hour}$,we have $\frac{22}{5} \text{ ft/sec} = \frac{22}{5} \times 3600 \text{ ft/hour} = \frac{22 \times 3600}{5 \times 5280} \text{ miles/hour}$.
Calculating this,$\frac{79200}{26400} = 3 \text{ miles/hour}$.
Thus,$K = 3$.

(A) Let $x$ be the distance of end $A$ from the wall and $y$ be the height of end $B$ from the floor. The length of the rod is constant,so $x^2 + y^2 = 41^2 = 1681$.
Differentiating with respect to time $t$,we get $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$,which simplifies to $x \frac{dx}{dt} + y \frac{dy}{dt} = 0$.
Given $\frac{dx}{dt} = 3 \ ft/min$. When $y = 9$,$x^2 + 9^2 = 1681 \implies x^2 = 1681 - 81 = 1600 \implies x = 40 \ ft$.
Substituting these values: $40(3) + 9 \frac{dy}{dt} = 0 \implies 9 \frac{dy}{dt} = -120 \implies \frac{dy}{dt} = -\frac{120}{9} = -\frac{40}{3} \ ft/min$.
The area of the triangle is $A = \frac{1}{2}xy$.
Differentiating with respect to $t$: $\frac{dA}{dt} = \frac{1}{2} \left( x \frac{dy}{dt} + y \frac{dx}{dt} \right)$.
Substituting the values: $\frac{dA}{dt} = \frac{1}{2} \left( 40 \times (-\frac{40}{3}) + 9 \times 3 \right) = \frac{1}{2} \left( -\frac{1600}{3} + 27 \right) = \frac{1}{2} \left( \frac{-1600 + 81}{3} \right) = \frac{1}{2} \left( -\frac{1519}{3} \right) = -\frac{1519}{6} \ ft^2/min$.

(C) Let $h$ be the height and $r$ be the radius of the cone. The semi-vertical angle $\alpha = \pi / 3$.
We know that $\tan(\alpha) = r / h$,so $r = h \tan(\pi / 3) = h \sqrt{3}$.
The volume of the cone is $V = \frac{1}{3} \pi r^2 h$.
Substituting $r = h \sqrt{3}$,we get $V = \frac{1}{3} \pi (h \sqrt{3})^2 h = \pi h^3$.
Since the volume $V$ is fixed,differentiating with respect to time $t$ gives $\frac{dV}{dt} = 3 \pi h^2 \frac{dh}{dt} = 0$.
However,the problem implies $V$ is constant at a specific instant.
From $V = \frac{1}{3} \pi r^2 h$,differentiating with respect to $t$: $\frac{dV}{dt} = \frac{1}{3} \pi (2rh \frac{dr}{dt} + r^2 \frac{dh}{dt}) = 0$.
Thus,$2rh \frac{dr}{dt} = -r^2 \frac{dh}{dt}$,which simplifies to $\frac{dr}{dt} = -\frac{r}{2h} \frac{dh}{dt}$.
Given $\frac{dh}{dt} = 2$ and $r = h \sqrt{3}$,we have $\frac{dr}{dt} = -\frac{h \sqrt{3}}{2h} (2) = -\sqrt{3}$.
The rate at which the radius is decreasing is $\sqrt{3} \text{ units/min}$.

(C) The distance $S$ is given by the function $S(t) = t^3 - t^2 - t + 3$.
The velocity $v$ of the particle is the rate of change of distance with respect to time,given by $v = \frac{dS}{dt}$.
$v = \frac{d}{dt}(t^3 - t^2 - t + 3) = 3t^2 - 2t - 1$.
The particle comes to rest when its velocity $v = 0$.
Setting $v = 0$,we get $3t^2 - 2t - 1 = 0$.
Factoring the quadratic equation: $3t^2 - 3t + t - 1 = 0 \Rightarrow 3t(t - 1) + 1(t - 1) = 0 \Rightarrow (3t + 1)(t - 1) = 0$.
Since time $t$ cannot be negative,we take $t = 1$ second.
Now,substitute $t = 1$ into the distance equation $S(t)$:
$S(1) = (1)^3 - (1)^2 - (1) + 3 = 1 - 1 - 1 + 3 = 2$ metres.
Thus,the distance travelled by the particle when it comes to rest is $2$ metres.

(C) Let $r$ be the radius and $h$ be the height of the water level in the cone at any time $t$.
The vertical angle is $60^{\circ}$,so the semi-vertical angle is $30^{\circ}$.
From the geometry of the cone,$\frac{r}{h} = \tan 30^{\circ} = \frac{1}{\sqrt{3}}$,which implies $h = \sqrt{3}r$.
The volume $V$ of the water in the cone is given by $V = \frac{1}{3} \pi r^2 h$.
Substituting $h = \sqrt{3}r$,we get $V = \frac{1}{3} \pi r^2 (\sqrt{3}r) = \frac{\sqrt{3}}{3} \pi r^3 = \frac{1}{\sqrt{3}} \pi r^3$.
Differentiating with respect to $t$,we get $\frac{dV}{dt} = \frac{1}{\sqrt{3}} \pi (3r^2) \frac{dr}{dt} = \sqrt{3} \pi r^2 \frac{dr}{dt}$.
Given $\frac{dV}{dt} = \frac{1}{\sqrt{3}}$,we have $\sqrt{3} \pi r^2 \frac{dr}{dt} = \frac{1}{\sqrt{3}}$,so $\frac{dr}{dt} = \frac{1}{3 \pi r^2}$.
Since $h = \sqrt{3}r$,when $h = 3 \text{ m}$,$r = \frac{3}{\sqrt{3}} = \sqrt{3} \text{ m}$.
Substituting $r = \sqrt{3}$ into the expression for $\frac{dr}{dt}$,we get $\frac{dr}{dt} = \frac{1}{3 \pi (\sqrt{3})^2} = \frac{1}{3 \pi (3)} = \frac{1}{9 \pi} \text{ m/min}$.

(A) Given the curve equation: $x^3 y^4 = 2^7$.
Differentiating both sides with respect to time $t$:
$3x^2 y^4 \frac{dx}{dt} + 4x^3 y^3 \frac{dy}{dt} = 0$.
We are given $\frac{dx}{dt} = -8 \text{ units/sec}$ (since it is decreasing).
At the point $(2, 2)$,substitute $x = 2, y = 2$ and $\frac{dx}{dt} = -8$:
$3(2)^2 (2)^4 (-8) + 4(2)^3 (2)^3 \frac{dy}{dt} = 0$.
$3(4)(16)(-8) + 4(8)(8) \frac{dy}{dt} = 0$.
$-1536 + 256 \frac{dy}{dt} = 0$.
$256 \frac{dy}{dt} = 1536$.
$\frac{dy}{dt} = \frac{1536}{256} = 6$.
Since $\frac{dy}{dt} > 0$,the $y$-coordinate increases at the rate of $6 \text{ units per second}$.

(C) Using the Law of Cosines: $a^2 = b^2 + c^2 - 2bc \cos A$.
Differentiating both sides with respect to $A$,we get: $2a \delta a = 2bc \sin A \delta A$.
Thus,$\delta a = \frac{bc \sin A \delta A}{a}$.
We know that the area of the triangle is $\Delta = \frac{1}{2} bc \sin A$,so $bc \sin A = 2\Delta$.
Also,by the Sine Rule,$\frac{a}{\sin A} = 2R$,which implies $a = 2R \sin A$.
Substituting these into the expression for $\delta a$: $\delta a = \frac{2\Delta \delta A}{2R \sin A} = \frac{\Delta \delta A}{R \sin A}$.
The percentage error is $\frac{\delta a}{a} \times 100 = \frac{\Delta \delta A}{R \sin A \cdot 2R \sin A} \times 100 = \frac{\Delta \delta A}{2R^2 \sin^2 A} \times 100$.

(C) The volume $V$ of a cylinder is given by $V = \pi r^2 h$,where $r$ is the radius and $h$ is the height (water level).
Given $r = 3.5 \ ft = \frac{7}{2} \ ft$ and the rate of change of volume $\frac{dV}{dt} = 1 \ ft^3/min$.
Differentiating with respect to $t$,we get $\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}$.
Substituting the values: $1 = \pi \times (\frac{7}{2})^2 \times \frac{dh}{dt}$.
$1 = \frac{22}{7} \times \frac{49}{4} \times \frac{dh}{dt}$.
$1 = \frac{11 \times 7}{2} \times \frac{dh}{dt} = \frac{77}{2} \times \frac{dh}{dt}$.
Therefore,$\frac{dh}{dt} = \frac{2}{77} \ ft/min$.

(C) Let $x$ be the distance of the lower end from the wall and $y$ be the height of the upper end from the ground. The ladder forms a right-angled triangle with the wall and the ground,so $x^2 + y^2 = 13^2 = 169$.
Differentiating both sides with respect to time $t$,we get $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$,which simplifies to $x \frac{dx}{dt} + y \frac{dy}{dt} = 0$.
Given that the lower end moves away from the wall at a speed of $\frac{dx}{dt} = 2 \ m/min$.
When $x = 5 \ m$,we find $y$ using $x^2 + y^2 = 169$: $5^2 + y^2 = 169 \Rightarrow 25 + y^2 = 169 \Rightarrow y^2 = 144 \Rightarrow y = 12 \ m$.
Substituting these values into the differentiated equation: $5(2) + 12 \frac{dy}{dt} = 0$.
$10 + 12 \frac{dy}{dt} = 0 \Rightarrow 12 \frac{dy}{dt} = -10 \Rightarrow \frac{dy}{dt} = -\frac{10}{12} = -\frac{5}{6} \ m/min$.
The negative sign indicates that the upper end is falling. Thus,the speed at which the upper end falls is $\frac{5}{6} \ m/min$.

(C) Given that the rate of change of the radius is $\frac{dr}{dt} = 5 \text{ inch/min}$.
When the radius $r = 10 \text{ inches}$.
The volume $V$ of a spherical balloon is given by $V = \frac{4}{3} \pi r^3$.
Differentiating both sides with respect to time $t$, we get:
$\frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3} \pi r^3 \right) = \frac{4}{3} \pi (3r^2) \frac{dr}{dt} = 4 \pi r^2 \frac{dr}{dt}$.
Substituting the given values $r = 10$ and $\frac{dr}{dt} = 5$:
$\frac{dV}{dt} = 4 \pi (10)^2 (5) = 4 \pi (100) (5) = 2000 \pi \text{ cubic inches/min}$.

(B) The displacement of the particle is given by $S(t) = t^3 - 4t^2 + 7t$.
Instantaneous velocity $v$ is the rate of change of displacement with respect to time,which is given by $v = \frac{dS}{dt}$.
Differentiating $S(t)$ with respect to $t$:
$v = \frac{d}{dt}(t^3 - 4t^2 + 7t) = 3t^2 - 8t + 7$.
To find the velocity at $t = 4 \ sec$,substitute $t = 4$ into the expression for $v$:
$v = 3(4)^2 - 8(4) + 7$
$v = 3(16) - 32 + 7$
$v = 48 - 32 + 7$
$v = 16 + 7 = 23 \ m/sec$.

(A) Let $A$ be the area and $P$ be the perimeter of the circle. We have,$\frac{dA}{dt} = \frac{1}{\sqrt{\pi}}$.
Since $A = \pi r^2$,differentiating with respect to $t$ gives $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$.
Substituting the given rate: $2\pi r \frac{dr}{dt} = \frac{1}{\sqrt{\pi}} \Rightarrow \frac{dr}{dt} = \frac{1}{2\pi r \sqrt{\pi}}$.
The perimeter is $P = 2\pi r$. Differentiating with respect to $t$ gives $\frac{dP}{dt} = 2\pi \frac{dr}{dt}$.
Substituting $\frac{dr}{dt}$: $\frac{dP}{dt} = 2\pi \times \frac{1}{2\pi r \sqrt{\pi}} = \frac{1}{r \sqrt{\pi}}$.
Given $P = \sqrt{\pi}$,we have $2\pi r = \sqrt{\pi} \Rightarrow r = \frac{\sqrt{\pi}}{2\pi} = \frac{1}{2\sqrt{\pi}}$.
Substituting $r$ into the expression for $\frac{dP}{dt}$: $\frac{dP}{dt} = \frac{1}{(\frac{1}{2\sqrt{\pi}}) \sqrt{\pi}} = \frac{1}{1/2} = 2$ units/sec.

(C) Let the volume of the rectangular parallelepiped be $V = 27 \ m^3$.
Let $A$ be the area of the base and $h$ be the height of the tank.
Then,$V = A \times h$.
Differentiating with respect to time $t$,we get $\frac{dV}{dt} = A \frac{dh}{dt}$.
According to the problem,the rate of change of the water level $\frac{dh}{dt}$ is thrice the rate of change of the water quantity $\frac{dV}{dt}$,i.e.,$\frac{dh}{dt} = 3 \frac{dV}{dt}$.
Substituting this into the derivative equation: $\frac{dV}{dt} = A \times (3 \frac{dV}{dt})$.
This implies $1 = 3A$,so $A = \frac{1}{3} \ m^2$.
Since $V = A \times h$,we have $27 = \frac{1}{3} \times h$.
Therefore,$h = 27 \times 3 = 81 \ m$.

(A) Given that,the rate of increase in volume is $\frac{dV}{dt} = 4 \pi \text{ cm}^3/\text{sec}$.
Volume of the sphere is $V = 288 \pi \text{ cm}^3$.
The formula for the volume of a sphere is $V = \frac{4}{3} \pi r^3$.
Substituting the given volume to find the radius $r$:
$288 \pi = \frac{4}{3} \pi r^3 \implies 288 = \frac{4}{3} r^3 \implies r^3 = 216 \implies r = 6 \text{ cm}$.
Differentiating the volume formula with respect to time $t$:
$\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$.
Substituting the known values $\frac{dV}{dt} = 4 \pi$ and $r = 6$:
$4 \pi = 4 \pi (6)^2 \frac{dr}{dt}$.
$1 = 36 \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{1}{36} \text{ cm/sec}$.

(B) Given that,at any instant of time,the rate of change in volume with respect to time is equal to the rate of change in surface area with respect to time,i.e.,$\frac{dV}{dt} = \frac{dS}{dt}$ $\ldots(i)$
The volume of a sphere of radius $r$ is $V = \frac{4}{3} \pi r^3$.
Differentiating with respect to $t$,we get $\frac{dV}{dt} = \frac{d}{dt}(\frac{4}{3} \pi r^3) = \frac{4}{3} \pi (3r^2) \frac{dr}{dt} = 4 \pi r^2 \frac{dr}{dt}$ $\ldots(ii)$
The surface area of a sphere of radius $r$ is $S = 4 \pi r^2$.
Differentiating with respect to $t$,we get $\frac{dS}{dt} = \frac{d}{dt}(4 \pi r^2) = 4 \pi (2r) \frac{dr}{dt} = 8 \pi r \frac{dr}{dt}$ $\ldots(iii)$
Substituting the values from equations $(ii)$ and $(iii)$ into equation $(i)$,we get $4 \pi r^2 \frac{dr}{dt} = 8 \pi r \frac{dr}{dt}$.
Assuming $\frac{dr}{dt} \neq 0$,we divide both sides by $4 \pi r \frac{dr}{dt}$ to obtain $r = 2$.

(B) Let $h$ be the height,$r$ be the radius,and $l$ be the slant height of the water at any time $t$. Given the semi-vertical angle $\alpha = 30^{\circ}$.
From the geometry of the cone,$r = h \tan 30^{\circ} = \frac{h}{\sqrt{3}}$.
The volume $V$ of the water is $V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \left(\frac{h}{\sqrt{3}}\right)^2 h = \frac{\pi h^3}{9}$.
Given $V = \frac{8 \pi}{\sqrt{3}}$,we have $\frac{\pi h^3}{9} = \frac{8 \pi}{\sqrt{3}} \Rightarrow h^3 = \frac{72}{\sqrt{3}} = 24 \sqrt{3} = (2 \sqrt{3})^3$,so $h = 2 \sqrt{3} \ ft$.
The slant height $l = \sqrt{h^2 + r^2} = \sqrt{h^2 + \frac{h^2}{3}} = \sqrt{\frac{4h^2}{3}} = \frac{2h}{\sqrt{3}}$.
Differentiating with respect to $t$,$\frac{dl}{dt} = \frac{2}{\sqrt{3}} \frac{dh}{dt}$.
Given $\frac{dl}{dt} = -\frac{1}{\sqrt{3}}$ (decreasing),we have $-\frac{1}{\sqrt{3}} = \frac{2}{\sqrt{3}} \frac{dh}{dt} \Rightarrow \frac{dh}{dt} = -\frac{1}{2} \ ft/min$.
The rate of change of volume is $\frac{dV}{dt} = \frac{d}{dt} \left(\frac{\pi h^3}{9}\right) = \frac{\pi}{3} h^2 \frac{dh}{dt}$.
Substituting $h = 2 \sqrt{3}$ and $\frac{dh}{dt} = -\frac{1}{2}$,we get $\frac{dV}{dt} = \frac{\pi}{3} (2 \sqrt{3})^2 \left(-\frac{1}{2}\right) = \frac{\pi}{3} (12) \left(-\frac{1}{2}\right) = -2 \pi \ cu. \ ft/min$.
Thus,the volume is decreasing at a rate of $2 \pi \ cu. \ ft/min$.

(C) Given the curve $y = 3x^5 + 15x - 8$.
The rate of change of $x$ is $\frac{dx}{dt} = \frac{1}{5} \text{ units/sec}$.
The rate of change of $y$ is $\frac{dy}{dt} = 6 \text{ units/sec}$.
Differentiating $y$ with respect to $t$:
$\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} = (15x^4 + 15) \cdot \frac{dx}{dt}$.
Substituting the given values:
$6 = (15x^4 + 15) \cdot \frac{1}{5} = 3(x^4 + 1)$.
$x^4 + 1 = 2 \Rightarrow x^4 = 1 \Rightarrow x = \pm 1$.
For $x = 1$,$y = 3(1)^5 + 15(1) - 8 = 10$. So,$A = (1, 10)$.
For $x = -1$,$y = 3(-1)^5 + 15(-1) - 8 = -3 - 15 - 8 = -26$. So,$B = (-1, -26)$.
The slope of $AB$ is $\frac{y_2 - y_1}{x_2 - x_1} = \frac{-26 - 10}{-1 - 1} = \frac{-36}{-2} = 18$.

(D) Given: $\frac{d(2r)}{dt} = 4 \text{ cm/s} \implies \frac{dr}{dt} = 2 \text{ cm/s}$.
Volume of cylinder $V = \pi r^2 h$. Since volume is constant, $\frac{dV}{dt} = 0$.
$\frac{dV}{dt} = \pi \left( r^2 \frac{dh}{dt} + 2rh \frac{dr}{dt} \right) = 0$.
$r^2 \frac{dh}{dt} + 2rh(2) = 0 \implies r \frac{dh}{dt} + 4h = 0 \implies \frac{dh}{dt} = -\frac{4h}{r}$.
At $r = 4 \text{ cm}$ and $h = 12 \text{ cm}$, $\frac{dh}{dt} = -\frac{4(12)}{4} = -12 \text{ cm/s}$.
Lateral surface area $S = 2 \pi rh$.
$\frac{dS}{dt} = 2 \pi \left( r \frac{dh}{dt} + h \frac{dr}{dt} \right)$.
Substituting the values: $\frac{dS}{dt} = 2 \pi \left( 4(-12) + 12(2) \right) = 2 \pi (-48 + 24) = 2 \pi (-24) = -48 \pi \text{ cm}^2/\text{s}$.

(D) Given: The rate of change of the radius is $\frac{dr}{dt} = 0.01 \text{ cm/sec}$.
We need to find the rate of change of the area $A$ when $r = 12 \text{ cm}$.
The area of a circular plate is given by $A = \pi r^2$.
Differentiating both sides with respect to time $t$,we get:
$\frac{dA}{dt} = 2 \pi r \frac{dr}{dt}$.
Substituting the given values $r = 12 \text{ cm}$ and $\frac{dr}{dt} = 0.01 \text{ cm/sec}$:
$\frac{dA}{dt} = 2 \pi (12) (0.01) = 24 \pi (0.01) = 0.24 \pi \text{ cm}^2/\text{sec}$.
Thus,the rate at which the area increases is $0.24 \pi \text{ cm}^2/\text{sec}$.

(A) Given the relation $p V^{1/4} = C$,where $C$ is a constant.
Taking the natural logarithm on both sides: $\ln p + \frac{1}{4} \ln V = \ln C$.
Differentiating both sides with respect to $V$: $\frac{dp}{p} + \frac{1}{4} \frac{dV}{V} = 0$.
This implies $\frac{dp}{p} = -\frac{1}{4} \frac{dV}{V}$.
Given that the percentage decrease in volume is $\frac{dV}{V} = -\frac{1}{2} \% = -0.5 \%$.
Substituting this into the equation: $\frac{dp}{p} = -\frac{1}{4} (-0.5 \%) = 0.125 \% = \frac{1}{8} \%$.
Thus,the percentage increase in pressure is $\frac{1}{8} \%$.

(D) Given the equation of the curve is $y = 4 - 2x^2$.
On differentiating both sides with respect to time $t$,we get:
$\frac{dy}{dt} = -4x \frac{dx}{dt}$.
We are given that the $x$-coordinate is decreasing at the rate of $5 \text{ units/s}$,so $\frac{dx}{dt} = -5 \text{ units/s}$.
At the point $(1, 2)$,we have $x = 1$.
Substituting these values into the derivative equation:
$\frac{dy}{dt} = -4(1)(-5) = 20 \text{ units/s}$.
Thus,the $y$-coordinate is changing at the rate of $20 \text{ units/s}$.

(C) Given the equation of the curve is $y = x^2 + 2x$.
Since the $x$ and $y$ coordinates of the particle change at the same rate,we have $\frac{dy}{dt} = \frac{dx}{dt}$.
Differentiating the equation of the curve with respect to $t$,we get:
$\frac{dy}{dt} = (2x + 2) \frac{dx}{dt}$.
Substituting $\frac{dy}{dt} = \frac{dx}{dt}$ into the equation,we get:
$\frac{dx}{dt} = (2x + 2) \frac{dx}{dt}$.
Assuming $\frac{dx}{dt} \neq 0$,we can divide both sides by $\frac{dx}{dt}$:
$1 = 2x + 2$
$2x = -1$
$x = -\frac{1}{2}$.
Now,substitute $x = -\frac{1}{2}$ into the curve equation to find $y$:
$y = \left(-\frac{1}{2}\right)^2 + 2\left(-\frac{1}{2}\right)$
$y = \frac{1}{4} - 1 = -\frac{3}{4}$.
Therefore,the required point is $\left(-\frac{1}{2}, -\frac{3}{4}\right)$.

(A) Let $V$ be the volume and $S$ be the surface area of the spherical balloon with radius $r$.
Given: $\frac{dV}{dt} = 2 \ cm^3/sec$.
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 the given values: $2 = 4 \pi r^2 \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{1}{2 \pi r^2}$.
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 $\frac{dr}{dt} = \frac{1}{2 \pi r^2}$ into the equation for $\frac{dS}{dt}$:
$\frac{dS}{dt} = 8 \pi r \left( \frac{1}{2 \pi r^2} \right) = \frac{4}{r}$.
At $r = 4 \ cm$,the rate of change of surface area is $\frac{dS}{dt} = \frac{4}{4} = 1 \ cm^2/sec$.
Thus,the correct option is $A$.

(C) The velocity of the particle is given by the derivative of position with respect to time:
$v = \frac{dx}{dt} = a \cos (\sqrt{\lambda} t + b) \cdot \sqrt{\lambda}$.
For the particle to come to rest,the velocity must be zero:
$v = 0 \Rightarrow a \sqrt{\lambda} \cos (\sqrt{\lambda} t + b) = 0$.
Since $a \neq 0$ and $\lambda > 0$,we have $\cos (\sqrt{\lambda} t + b) = 0$.
This occurs when the argument is $\pm \frac{\pi}{2}$.
Let the two points be $x_1$ and $x_2$:
$x_1 = a \sin (\frac{\pi}{2}) = a(1) = a$.
$x_2 = a \sin (-\frac{\pi}{2}) = a(-1) = -a$.
The distance between these two points is $|x_1 - x_2| = |a - (-a)| = |2a| = 2a$.

(A) Given the distance equation: $s = t^{2} + at - b + 17$.
Since the particle starts from rest,the velocity $v = \frac{ds}{dt}$ must be zero at $t = 5 \ s$.
Calculating the velocity: $v = \frac{ds}{dt} = 2t + a$.
Setting $v = 0$ at $t = 5$: $2(5) + a = 0 \implies 10 + a = 0 \implies a = -10$.
Now,given that at $t = 5 \ s$,the distance $s = 25$:
$25 = (5)^{2} + a(5) - b + 17$.
Substituting $a = -10$: $25 = 25 + (-10)(5) - b + 17$.
$25 = 25 - 50 - b + 17$.
$25 = -8 - b$.
$b = -8 - 25 = -33$.
Thus,the values are $a = -10$ and $b = -33$.

(C) Given the equation of motion: $x = \frac{1}{2} vt$.
Since velocity $v = \frac{dx}{dt}$,we substitute this into the equation:
$x = \frac{1}{2} \left( \frac{dx}{dt} \right) t$.
Rearranging the terms to separate variables:
$\frac{2 dt}{t} = \frac{dx}{x}$.
Integrating both sides:
$2 \int \frac{dt}{t} = \int \frac{dx}{x} \implies 2 \ln |t| + C' = \ln |x|$.
This simplifies to $\ln |t^2| + C' = \ln |x|$,which implies $x = c t^2$ for some constant $c$.
Now,find the velocity $v$ by differentiating $x$ with respect to $t$:
$v = \frac{dx}{dt} = \frac{d}{dt} (c t^2) = 2ct$.
Finally,find the acceleration $f$ by differentiating $v$ with respect to $t$:
$f = \frac{dv}{dt} = \frac{d}{dt} (2ct) = 2c$.
Since $2c$ is a constant,the acceleration $f$ is constant.

(A) Given that the ordinate $y$ decreases at the same rate at which the abscissa $x$ increases,we have $\frac{dy}{dt} = -\frac{dx}{dt}$.
The equation of the ellipse is $16x^{2} + 9y^{2} = 400$.
Differentiating both sides with respect to $t$,we get:
$16(2x) \frac{dx}{dt} + 9(2y) \frac{dy}{dt} = 0$
$32x \frac{dx}{dt} + 18y \frac{dy}{dt} = 0$
$16x \frac{dx}{dt} + 9y \frac{dy}{dt} = 0$.
Substituting $\frac{dy}{dt} = -\frac{dx}{dt}$ into the equation:
$16x \frac{dx}{dt} + 9y \left(-\frac{dx}{dt}\right) = 0$
$(16x - 9y) \frac{dx}{dt} = 0$.
Since $\frac{dx}{dt} \neq 0$ for the points to be moving,we have $16x - 9y = 0$,which implies $y = \frac{16}{9}x$.
Substituting $y = \frac{16}{9}x$ into the ellipse equation:
$16x^{2} + 9\left(\frac{16}{9}x\right)^{2} = 400$
$16x^{2} + 9 \cdot \frac{256}{81}x^{2} = 400$
$16x^{2} + \frac{256}{9}x^{2} = 400$
$\frac{144x^{2} + 256x^{2}}{9} = 400$
$\frac{400x^{2}}{9} = 400$
$x^{2} = 9 \Rightarrow x = \pm 3$.
If $x = 3$,then $y = \frac{16}{9}(3) = \frac{16}{3}$.
If $x = -3$,then $y = \frac{16}{9}(-3) = -\frac{16}{3}$.
Thus,the points are $\left(3, \frac{16}{3}\right)$ and $\left(-3, -\frac{16}{3}\right)$.