Rate of Change of Quantities Questions in English

(C) Given the curve equation $y^3 = 27x$.
Differentiating both sides with respect to time $t$,we get:
$3y^2 \frac{dy}{dt} = 27 \frac{dx}{dt}$
$\frac{dx}{dt} = \frac{y^2}{9} \frac{dy}{dt}$.
The problem states that the abscissa $(x)$ changes at a slower rate than the ordinate $(y)$,which means:
$\left| \frac{dx}{dt} \right| < \left| \frac{dy}{dt} \right|$
Substituting the expression for $\frac{dx}{dt}$:
$\left| \frac{y^2}{9} \frac{dy}{dt} \right| < \left| \frac{dy}{dt} \right|$
$\frac{y^2}{9} < 1$
$y^2 < 9$
$-3 < y < 3$.
Since $y^3 = 27x$,when $y = -3$,$x = -1$,and when $y = 3$,$x = 1$.
Thus,the interval for the abscissa $x$ is $(-1, 1)$.

(D) Let $\frac{dr}{dt} = c$ and $h = ar + b$.
Given $\frac{dh}{dt} = 3 \frac{dr}{dt}$,so $a \frac{dr}{dt} = 3 \frac{dr}{dt}$,which implies $a = 3$.
Thus,$h = 3r + b$.
When $r = 1, h = 6$,so $6 = 3(1) + b$,which gives $b = 3$.
Therefore,$h = 3r + 3 = 3(r + 1)$.
The volume $V = \pi r^2 h = \pi r^2 (3r + 3) = 3\pi (r^3 + r^2)$.
Differentiating with respect to $t$,$\frac{dV}{dt} = 3\pi (3r^2 + 2r) \frac{dr}{dt}$.
When $r = 6$,$\frac{dV}{dt} = 1$,so $1 = 3\pi (3(36) + 2(6)) \frac{dr}{dt} = 3\pi (108 + 12) \frac{dr}{dt} = 360\pi \frac{dr}{dt}$.
Thus,$\frac{dr}{dt} = \frac{1}{360\pi}$.
When $r = 36$,$\frac{dV}{dt} = n = 3\pi (3(36)^2 + 2(36)) \frac{dr}{dt}$.
$n = 3\pi (3888 + 72) \frac{1}{360\pi} = 3\pi (3960) \frac{1}{360\pi} = 3 \times 11 = 33$.

(A) Let the distance of the particle from the origin be $r$. Then $r^2 = x^2 + y^2$.
Given $y = x^{3/2}$,so $y^2 = x^3$. Thus,$r^2 = x^2 + x^3$.
Differentiating with respect to $t$,we get $2r \frac{dr}{dt} = (2x + 3x^2) \frac{dx}{dt}$.
Given $\frac{dr}{dt} = 11$ and $x = 3$. Then $y = 3^{3/2} = 3\sqrt{3}$.
The distance $r = \sqrt{x^2 + y^2} = \sqrt{3^2 + (3\sqrt{3})^2} = \sqrt{9 + 27} = \sqrt{36} = 6$.
Substituting the values into the derivative equation:
$2(6)(11) = (2(3) + 3(3^2)) \frac{dx}{dt}$
$132 = (6 + 27) \frac{dx}{dt}$
$132 = 33 \frac{dx}{dt}$
$\frac{dx}{dt} = \frac{132}{33} = 4$.

(B) Let $x$ be the side length of the cube. The volume $V = x^3$ and surface area $S = 6x^2$.
Given $\frac{dV}{dt} = -4 \, cm^3/min$ (since the ice is melting,the volume is decreasing).
We know $\frac{dV}{dt} = 3x^2 \frac{dx}{dt}$.
Substituting the values: $-4 = 3x^2 \frac{dx}{dt} \implies \frac{dx}{dt} = -\frac{4}{3x^2}$.
We need to find $\frac{dS}{dt}$.
$\frac{dS}{dt} = \frac{d}{dt}(6x^2) = 12x \frac{dx}{dt}$.
Substituting $\frac{dx}{dt}$: $\frac{dS}{dt} = 12x \left( -\frac{4}{3x^2} \right) = -\frac{16}{x}$.
Given $V = 125 \, cm^3$,so $x^3 = 125 \implies x = 5 \, cm$.
Therefore,$\frac{dS}{dt} = -\frac{16}{5} \, cm^2/min$.

(C) We know that in any triangle,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,where $R$ is the circumradius.
Since $R$ is constant,we have $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.
Differentiating these with respect to the angles,we get $da = 2R \cos A \, dA$,$db = 2R \cos B \, dB$,and $dc = 2R \cos C \, dC$.
Substituting these into the given expression:
$\frac{da}{\cos A} + \frac{db}{\cos B} + \frac{dc}{\cos C} = \frac{2R \cos A \, dA}{\cos A} + \frac{2R \cos B \, dB}{\cos B} + \frac{2R \cos C \, dC}{\cos C} = 2R(dA + dB + dC)$.
Since the sum of angles in a triangle is $A + B + C = \pi$,differentiating gives $dA + dB + dC = 0$.
Therefore,the expression equals $2R(0) = 0$.

(B) Let the side of the equilateral triangle be $x$. Given that the rate of change of the side is $\frac{dx}{dt} = 2 \, cm/s$.
The area $A$ of an equilateral triangle with side $x$ is given by $A = \frac{\sqrt{3}}{4} x^2$.
To find the rate of increase of the area,we differentiate $A$ with respect to time $t$:
$\frac{dA}{dt} = \frac{d}{dt} \left( \frac{\sqrt{3}}{4} x^2 \right) = \frac{\sqrt{3}}{4} \cdot 2x \cdot \frac{dx}{dt} = \frac{\sqrt{3}}{2} x \frac{dx}{dt}$.
Given $x = 10 \, cm$ and $\frac{dx}{dt} = 2 \, cm/s$,we substitute these values into the expression:
$\frac{dA}{dt} = \frac{\sqrt{3}}{2} \times 10 \times 2 = 10 \sqrt{3} \, cm^2/s$.
Thus,the rate of increase of the area is $10 \sqrt{3} \, cm^2/s$.

(C) The diameter of the spherical balloon is $D = 3x + \frac{9}{2} = \frac{6x + 9}{2} = \frac{3}{2}(2x + 3)$.
The radius $r$ is given by $r = \frac{D}{2} = \frac{3}{4}(2x + 3)$.
The volume $V$ of a sphere is $V = \frac{4}{3}\pi r^3$.
Substituting $r$,we get $V = \frac{4}{3}\pi \left[ \frac{3}{4}(2x + 3) \right]^3 = \frac{4}{3}\pi \cdot \frac{27}{64} (2x + 3)^3 = \frac{9\pi}{16} (2x + 3)^3$.
Now,differentiating $V$ with respect to $x$ using the chain rule:
$\frac{dV}{dx} = \frac{9\pi}{16} \cdot 3(2x + 3)^2 \cdot \frac{d}{dx}(2x + 3)$.
$\frac{dV}{dx} = \frac{27\pi}{16} (2x + 3)^2 \cdot 2$.
$\frac{dV}{dx} = \frac{27\pi}{8} (2x + 3)^2$.

(A) Let $AB$ be the lamp post and $CD$ be the man at a particular time $t$.
Let $AC = x$ be the distance of the man from the lamp post and $CE = y$ be the length of his shadow.
Given,the speed of the man is $\frac{dx}{dt} = 5 \ km/hr$.
We need to find the rate at which the shadow increases,i.e.,$\frac{dy}{dt}$.
Since $\Delta ABE \sim \Delta CDE$,we have:
$\frac{AB}{CD} = \frac{AE}{CE}$
$\frac{6}{2} = \frac{x + y}{y}$
$3 = \frac{x + y}{y}$
$3y = x + y$
$2y = x$
Differentiating both sides with respect to $t$:
$2 \frac{dy}{dt} = \frac{dx}{dt}$
$2 \frac{dy}{dt} = 5$
$\frac{dy}{dt} = 2.5 \ km/hr$.
Thus,the length of the shadow increases at the rate of $2.5 \ km/hr$.

(A) Let $AB$ be the lamp post of height $6 \ m$ and $CD$ be the man of height $2 \ m$ at a particular time $t$.
Let $AC = x$ be the distance of the man from the lamp post and $CE = y$ be the length of his shadow.
Given that the man walks at a speed of $5 \ km/hr$,so $\frac{dx}{dt} = 5 \ km/hr$.
We need to find the rate at which the shadow increases,i.e.,$\frac{dy}{dt}$.
Since $\triangle ABE \sim \triangle CDE$,we have:
$\frac{AB}{CD} = \frac{AE}{CE}$
$\frac{6}{2} = \frac{x + y}{y}$
$3 = \frac{x + y}{y}$
$3y = x + y$
$2y = x$
Differentiating both sides with respect to $t$:
$2 \frac{dy}{dt} = \frac{dx}{dt}$
$2 \frac{dy}{dt} = 5$
$\frac{dy}{dt} = \frac{5}{2} = 2.5 \ km/hr$.
Thus,the length of the shadow increases at the rate of $2.5 \ km/hr$.

(B) The volume $V$ of a spherical balloon is given by $V = \frac{4}{3} \pi r^3$,where $r$ is the radius.
Differentiating both sides with respect to time $t$,we get:
$\frac{dV}{dt} = \frac{4}{3} \pi (3r^2) \frac{dr}{dt} = 4 \pi r^2 \frac{dr}{dt}$.
According to the problem,the rate of increase of volume is $16$ times the rate of increase of the radius:
$\frac{dV}{dt} = 16 \frac{dr}{dt}$.
Substituting this into the derivative equation:
$16 \frac{dr}{dt} = 4 \pi r^2 \frac{dr}{dt}$.
Assuming $\frac{dr}{dt} \neq 0$,we can divide both sides by $\frac{dr}{dt}$:
$16 = 4 \pi r^2$.
$r^2 = \frac{16}{4 \pi} = \frac{4}{\pi}$.
Taking the square root of both sides:
$r = \sqrt{\frac{4}{\pi}} = \frac{2}{\sqrt{\pi}}$.

(A) The volume of a cylinder is given by $V = \pi r^2 h$. Since the volume is constant,its derivative with respect to time $t$ is zero: $\frac{dV}{dt} = 0$.
Applying the product rule to $V = \pi r^2 h$,we get: $\frac{dV}{dt} = \pi (2rh \frac{dr}{dt} + r^2 \frac{dh}{dt}) = 0$.
Dividing by $\pi r$,we have $2h \frac{dr}{dt} + r \frac{dh}{dt} = 0$.
Given $\frac{dr}{dt} = 5 \ cm/min$,$r = 5 \ cm$,and $h = 3 \ cm$,we substitute these values:
$2(3)(5) + 5 \frac{dh}{dt} = 0$.
$30 + 5 \frac{dh}{dt} = 0$.
$5 \frac{dh}{dt} = -30$.
$\frac{dh}{dt} = -6 \ cm/min$.
The negative sign indicates a decrease. Therefore,the rate of decrease of the height is $6 \ cm/min$.

(C) Let the side length of the square and the cube be $x$.
For the square $S$,the area $A = x^2$.
Given that the rate of change of the area is equal to its side length,we have $\frac{dA}{dt} = x$.
Since $\frac{dA}{dt} = \frac{d}{dt}(x^2) = 2x \frac{dx}{dt}$,we equate the two expressions:
$2x \frac{dx}{dt} = x$.
Assuming $x \neq 0$,we get $\frac{dx}{dt} = \frac{1}{2}$.
For the cube $C$,the volume $V = x^3$.
The rate of change of the volume is $\frac{dV}{dt} = 3x^2 \frac{dx}{dt}$.
Given that the rate of change of the side of the cube is the same as that of the square,we use $\frac{dx}{dt} = \frac{1}{2}$.
At the time when the side length $x = 2$ units:
$\frac{dV}{dt} = 3(2)^2 \times \frac{1}{2} = 3 \times 4 \times \frac{1}{2} = 6$ $units^3/sec$.

(C) Given the equation of the circle is $x^2 + y^2 = 25$.
Differentiating both sides with respect to time $t$,we get:
$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$
$\Rightarrow x \frac{dx}{dt} + y \frac{dy}{dt} = 0$.
Given that the ordinate $y$ decreases at the rate of $1 \, cm/sec$,so $\frac{dy}{dt} = -1 \, cm/sec$.
When $y = 3 \, cm$,we find $x$ using the circle equation:
$x^2 + 3^2 = 25 \Rightarrow x^2 = 25 - 9 = 16 \Rightarrow x = 4 \, cm$ (since $x > 0$).
Substituting the values into the derivative equation:
$4 \frac{dx}{dt} + 3(-1) = 0$
$4 \frac{dx}{dt} = 3$
$\frac{dx}{dt} = \frac{3}{4} \, cm/sec$.
Thus,the abscissa increases at the rate of $\frac{3}{4} \, cm/sec$.

(A) Let $OA = x \ km$,$OB = y \ km$,and $AB = R \ km$.
Using the Law of Cosines in $\triangle AOB$:
$R^2 = x^2 + y^2 - 2xy \cos(120^o)$
Since $\cos(120^o) = -\frac{1}{2}$,we have:
$R^2 = x^2 + y^2 - 2xy(-\frac{1}{2}) = x^2 + y^2 + xy \quad \dots(1)$
At the given instance,$x = 8 \ km$ and $y = 6 \ km$:
$R^2 = 8^2 + 6^2 + (8 \times 6) = 64 + 36 + 48 = 148$
$R = \sqrt{148} = 2\sqrt{37} \ km$.
Differentiating equation $(1)$ with respect to time $t$:
$2R \frac{dR}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt} + (x \frac{dy}{dt} + y \frac{dx}{dt})$
Given $\frac{dx}{dt} = 20 \ km/hr$ and $\frac{dy}{dt} = 30 \ km/hr$:
$2(2\sqrt{37}) \frac{dR}{dt} = 2(8)(20) + 2(6)(30) + (8 \times 30 + 6 \times 20)$
$4\sqrt{37} \frac{dR}{dt} = 320 + 360 + (240 + 120)$
$4\sqrt{37} \frac{dR}{dt} = 680 + 360 = 1040$
$\frac{dR}{dt} = \frac{1040}{4\sqrt{37}} = \frac{260}{\sqrt{37}} \ km/hr$.

(C) 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:
$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$.
Given that $\frac{dV}{dt} = 4\pi \, cc/sec$,we substitute this into the equation:
$4\pi = 4\pi r^2 \frac{dr}{dt} \Rightarrow \frac{dr}{dt} = \frac{1}{r^2}$.
We are given that the volume $V = 288\pi \, cc$.
Using the volume formula:
$288\pi = \frac{4}{3}\pi r^3
\Rightarrow r^3 = \frac{288 \times 3}{4} = 216
\Rightarrow r = 6 \, cm$.
Substituting $r = 6$ into the expression for $\frac{dr}{dt}$:
$\frac{dr}{dt} = \frac{1}{6^2} = \frac{1}{36} \, cm/sec$.

(D) The volume of a sphere is given by $V = \frac{4}{3}\pi r^3$.
Differentiating with respect to time $t$,we get $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \dots (i)$.
The surface area of a sphere is $S = 4\pi r^2$.
Differentiating with respect to time $t$,we get $\frac{dS}{dt} = 8\pi r \frac{dr}{dt}$.
Given that $\frac{dS}{dt} = 8 \, cm^2/s$,we have $8 = 8\pi r \frac{dr}{dt}$,which implies $\frac{dr}{dt} = \frac{1}{\pi r}$.
Substituting the value of $\frac{dr}{dt}$ into equation $(i)$,we get $\frac{dV}{dt} = 4\pi r^2 \times \frac{1}{\pi r} = 4r$.
Thus,the rate of change of volume $\frac{dV}{dt}$ is proportional to $r$.

(A) Let $r$ be the radius of the spherical balloon. The volume $V$ of the 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 $\frac{dV}{dt} = 35 \, cm^3/min$,we have $35 = 4\pi r^2 \frac{dr}{dt}$,which implies $\frac{dr}{dt} = \frac{35}{4\pi r^2} \quad (1)$.
The surface area $S$ of the sphere is $S = 4\pi r^2$.
Differentiating with respect to $t$,we get $\frac{dS}{dt} = 8\pi r \frac{dr}{dt}$.
Substituting the value of $\frac{dr}{dt}$ from $(1)$ into the expression for $\frac{dS}{dt}$:
$\frac{dS}{dt} = 8\pi r \left( \frac{35}{4\pi r^2} \right) = \frac{70}{r}$.
Given the diameter is $14 \, cm$,the radius $r = 7 \, cm$.
Substituting $r = 7$ into the expression for $\frac{dS}{dt}$:
$\frac{dS}{dt} = \frac{70}{7} = 10 \, cm^2/min$.

(B) Let the radius of the circular sheet be $r$ and its area be $A$.
Given that $A = \pi r^2$.
Differentiating both sides with respect to time $t$,we get:
$\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$.
We are given that $\frac{dA}{dt} = 6\pi \, cm^2/hr$ and $r = 30 \, cm$.
Substituting these values into the equation:
$6\pi = 2\pi (30) \frac{dr}{dt}$.
$6\pi = 60\pi \frac{dr}{dt}$.
$\frac{dr}{dt} = \frac{6\pi}{60\pi} = \frac{1}{10} = 0.1 \, cm/hr$.
Thus,the rate at which the radius of the circular sheet increases is $0.1 \, cm/hr$.

(D) Let $A$ be the area,$b$ be the breadth,and $\ell$ be the length of the rectangle.
Given: $\frac{dA}{dt} = -5$,$\frac{d\ell}{dt} = 2$,and $\frac{db}{dt} = -3$.
We know that the area of a rectangle is $A = \ell \times b$.
Differentiating both sides with respect to time $t$,we get:
$\frac{dA}{dt} = \ell \cdot \frac{db}{dt} + b \cdot \frac{d\ell}{dt}$.
Substituting the given values:
$-5 = \ell(-3) + b(2)$.
$-5 = -3\ell + 2b$.
When the breadth $b = 2 \, m$,we substitute this into the equation:
$-5 = -3\ell + 2(2)$.
$-5 = -3\ell + 4$.
$3\ell = 4 + 5$.
$3\ell = 9$.
$\ell = 3 \, m$.
Thus,the length of the rectangle is $3 \, m$.

(C) Given $W = nw$.
Using the product rule for differentiation,we have:
$\frac{dW}{dt} = n \frac{dw}{dt} + w \frac{dn}{dt}$
Given $n = 2t^2 + 3$ and $w = t^2 - t + 2$.
Calculating the derivatives with respect to $t$:
$\frac{dn}{dt} = 4t$
$\frac{dw}{dt} = 2t - 1$
At $t = 1$:
$n = 2(1)^2 + 3 = 5$
$w = (1)^2 - 1 + 2 = 2$
$\frac{dn}{dt} = 4(1) = 4$
$\frac{dw}{dt} = 2(1) - 1 = 1$
Substituting these values into the derivative formula:
$\frac{dW}{dt} = (5)(1) + (2)(4)$
$\frac{dW}{dt} = 5 + 8 = 13$

(B) Let $A$ be the area of the circular plate and $r$ be its radius. The area is given by $A = \pi r^2$.
Given that the radius $r = 50 \, cm$ and the rate of change of the radius is $\frac{dr}{dt} = 1 \, mm/hour = 0.1 \, cm/hour = \frac{1}{10} \, cm/hour$.
Differentiating the area formula with respect to time $t$,we get $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$.
Substituting the given values,$\frac{dA}{dt} = 2 \times \pi \times 50 \times \frac{1}{10} = 10\pi \, cm^2/hour$.
Thus,the rate at which the area of the plate increases is $10\pi \, cm^2/hour$.

(B) Let $h$ be the depth of water and $r$ be the radius of the water surface at time $t$.
Given the semi-vertical angle $\theta = \tan^{-1}(1/2)$,we have $\tan \theta = \frac{r}{h} = \frac{1}{2}$,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 both sides with respect to time $t$,we get $\frac{dV}{dt} = \frac{\pi}{12} (3h^2) \frac{dh}{dt} = \frac{\pi h^2}{4} \frac{dh}{dt}$.
Given $\frac{dV}{dt} = 5 \ m^3/min$ and $h = 10 \ m$,we substitute these values:
$5 = \frac{\pi (10)^2}{4} \frac{dh}{dt}$
$5 = \frac{100\pi}{4} \frac{dh}{dt}$
$5 = 25\pi \frac{dh}{dt}$
$\frac{dh}{dt} = \frac{5}{25\pi} = \frac{1}{5\pi} \ m/min$.

(D) Let $r = 10 \, cm$ be the radius of the iron ball and $h$ be the thickness of the ice layer.
The total radius of the sphere including the ice is $R = 10 + h$.
The volume of the ice layer $V$ is given by the difference between the volume of the sphere with ice and the volume of the iron ball:
$V = \frac{4}{3}\pi (10 + h)^3 - \frac{4}{3}\pi (10)^3$
Differentiating both sides with respect to time $t$:
$\frac{dV}{dt} = \frac{4}{3}\pi \cdot 3(10 + h)^2 \cdot \frac{dh}{dt} = 4\pi (10 + h)^2 \frac{dh}{dt}$
Given that the ice melts at a rate of $50 \, cm^3/min$,we have $\frac{dV}{dt} = -50 \, cm^3/min$.
Substituting $h = 5 \, cm$ and $\frac{dV}{dt} = -50$ into the equation:
$-50 = 4\pi (10 + 5)^2 \frac{dh}{dt}$
$-50 = 4\pi (15)^2 \frac{dh}{dt}$
$-50 = 4\pi (225) \frac{dh}{dt}$
$-50 = 900\pi \frac{dh}{dt}$
$\frac{dh}{dt} = -\frac{50}{900\pi} = -\frac{1}{18\pi} \, cm/min$.
The rate at which the thickness decreases is $\frac{1}{18\pi} \, cm/min$.

(D) Let $x$ be the distance of the bottom of the ladder from the wall and $y$ be the height of the top of the ladder from the ground. The length of the ladder is $L = 2 \ m = 200 \ cm$.
By Pythagoras theorem,we have $x^2 + y^2 = 200^2$.
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 that the top slides down at $25 \ cm/sec$,we have $\frac{dy}{dt} = -25 \ cm/sec$.
When $y = 1 \ m = 100 \ cm$,we find $x$ using $x^2 + 100^2 = 200^2$,so $x^2 = 40000 - 10000 = 30000$,which gives $x = \sqrt{30000} = 100\sqrt{3} \ cm$.
Substituting these values into the differentiated equation: $(100\sqrt{3}) \frac{dx}{dt} + (100)(-25) = 0$.
$(100\sqrt{3}) \frac{dx}{dt} = 2500$.
$\frac{dx}{dt} = \frac{2500}{100\sqrt{3}} = \frac{25}{\sqrt{3}} \ cm/sec$.

(B) The area $A$ of a circle with radius $r$ is given by $A = \pi r^2$.
To find the rate of change of the area with respect to the radius,we differentiate $A$ with respect to $r$:
$\frac{dA}{dr} = \frac{d}{dr}(\pi r^2) = 2\pi r$.
Given that $r = 5 \text{ cm}$,we substitute this value into the derivative:
$\frac{dA}{dr} = 2 \times \pi \times 5 = 10\pi$.
Therefore,the rate of change of the area of the circle with respect to its radius when $r = 5 \text{ cm}$ is $10\pi \text{ cm}^2/\text{cm}$.

(A) Let $x$ be the length of an edge of the cube. Let $V$ be the volume and $S$ be the surface area of the cube. Then $V = x^3$ and $S = 6x^2$, where $x$ is a function of time $t$.
Given that the rate of change of volume is $\frac{dV}{dt} = 9 \text{ cm}^3/\text{s}$.
Using the chain rule, $\frac{dV}{dt} = \frac{d}{dt}(x^3) = 3x^2 \cdot \frac{dx}{dt}$.
Substituting the given value, $9 = 3x^2 \cdot \frac{dx}{dt}$, which gives $\frac{dx}{dt} = \frac{3}{x^2}$.
Now, we need to find the rate of change of the surface area, $\frac{dS}{dt}$.
Using the chain rule, $\frac{dS}{dt} = \frac{d}{dt}(6x^2) = 12x \cdot \frac{dx}{dt}$.
Substituting $\frac{dx}{dt} = \frac{3}{x^2}$ into the equation, we get $\frac{dS}{dt} = 12x \cdot \left(\frac{3}{x^2}\right) = \frac{36}{x}$.
When the edge length $x = 10 \text{ cm}$, the rate of change of the surface area is $\frac{dS}{dt} = \frac{36}{10} = 3.6 \text{ cm}^2/\text{s}$.

(A) The area $A$ of a circle with radius $r$ is given by $A = \pi r^2$.
Taking the derivative with respect to time $t$,we get:
$\frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = 2 \pi r \frac{dr}{dt}$.
Given that the rate of change of the radius is $\frac{dr}{dt} = 4 \text{ cm/s}$.
At the instant when $r = 10 \text{ cm}$,we substitute the values into the derivative formula:
$\frac{dA}{dt} = 2 \pi (10 \text{ cm}) (4 \text{ cm/s}) = 80 \pi \text{ cm}^2/\text{s}$.
Thus,the enclosed area is increasing at the rate of $80 \pi \text{ cm}^2/\text{s}$.

(A) Given that the length $x$ is decreasing at the rate of $3 \text{ cm/min}$,so $\frac{dx}{dt} = -3 \text{ cm/min}$.
Given that the width $y$ is increasing at the rate of $2 \text{ cm/min}$,so $\frac{dy}{dt} = 2 \text{ cm/min}$.
The area $A$ of the rectangle is given by $A = x \cdot y$.
Differentiating both sides with respect to $t$,we get $\frac{dA}{dt} = \frac{dx}{dt} \cdot y + x \cdot \frac{dy}{dt}$.
Substituting the given values $x = 10 \text{ cm}$,$y = 6 \text{ cm}$,$\frac{dx}{dt} = -3 \text{ cm/min}$,and $\frac{dy}{dt} = 2 \text{ cm/min}$:
$\frac{dA}{dt} = (-3)(6) + (10)(2) = -18 + 20 = 2 \text{ cm}^2/\text{min}$.
Thus,the rate of change of the area is $2 \text{ cm}^2/\text{min}$.

(A) The marginal cost $(MC)$ is the derivative of the total cost function $C(x)$ with respect to $x$:
$MC = \frac{dC}{dx} = \frac{d}{dx}(0.005x^{3} - 0.02x^{2} + 30x + 5000)$
$MC = 0.005(3x^{2}) - 0.02(2x) + 30$
$MC = 0.015x^{2} - 0.04x + 30$
To find the marginal cost when $x = 3$ units are produced,substitute $x = 3$ into the expression for $MC$:
$MC = 0.015(3)^{2} - 0.04(3) + 30$
$MC = 0.015(9) - 0.12 + 30$
$MC = 0.135 - 0.12 + 30$
$MC = 0.015 + 30 = 30.015$
Thus,the marginal cost is $Rs. 30.015$.

(A) 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 $\frac{dR}{dx}$.
Given $R(x) = 3x^2 + 36x + 5$.
Marginal Revenue $(MR) = \frac{dR}{dx} = \frac{d}{dx}(3x^2 + 36x + 5) = 6x + 36$.
To find the marginal revenue when $x = 5$,we substitute $x = 5$ into the expression for $MR$:
$MR = 6(5) + 36 = 30 + 36 = 66$.
Thus,the marginal revenue is $Rs. 66$.

(A) The area of a circle $(A)$ with radius $(r)$ is given by the formula:
$A = \pi r^2$
To find the rate of change of the area with respect to its radius, we differentiate $A$ with respect to $r$:
$\frac{dA}{dr} = \frac{d}{dr}(\pi r^2) = 2 \pi r$
Now, substitute the value $r = 3 \text{ cm}$ into the expression:
$\frac{dA}{dr} = 2 \pi (3) = 6 \pi \text{ cm}$
Thus, the area of the circle is changing at the rate of $6 \pi \text{ cm}$ when its radius is $3 \text{ cm}$.

(A) The area of a circle $(A)$ with radius $(r)$ is given by the formula:
$A = \pi r^{2}$
To find the rate of change of the area with respect to its radius,we differentiate $A$ with respect to $r$:
$\frac{dA}{dr} = \frac{d}{dr}(\pi r^{2}) = 2 \pi r$
Now,substitute the given value $r = 4 \, cm$ into the derivative:
$\frac{dA}{dr} = 2 \pi (4) = 8 \pi \, cm$
Thus,the rate of change of the area of the circle with respect to its radius when $r = 4 \, cm$ is $8 \pi \, cm$.

(A) Let $x$ be the length of an edge,$V$ be the volume,and $S$ be the surface area of the cube.
We have $V = x^{3}$ and $S = 6x^{2}$.
Given that the rate of change of volume is $\frac{dV}{dt} = 8 \, cm^{3}/s$.
Using the chain rule,$\frac{dV}{dt} = \frac{d}{dt}(x^{3}) = 3x^{2} \cdot \frac{dx}{dt}$.
Substituting the given values: $8 = 3x^{2} \cdot \frac{dx}{dt} \implies \frac{dx}{dt} = \frac{8}{3x^{2}}$.
Now,the rate of change of surface area is $\frac{dS}{dt} = \frac{d}{dt}(6x^{2}) = 12x \cdot \frac{dx}{dt}$.
Substituting $\frac{dx}{dt}$ from the previous step: $\frac{dS}{dt} = 12x \cdot \left(\frac{8}{3x^{2}}\right) = \frac{32}{x}$.
When the edge length $x = 12 \, cm$,the rate of change of surface area is $\frac{dS}{dt} = \frac{32}{12} = \frac{8}{3} \, cm^{2}/s$.

(A) The area of a circle $(A)$ with radius $(r)$ is given by $A = \pi r^2$.
Differentiating both sides with respect to time $(t)$,we get:
$\frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = 2 \pi r \frac{dr}{dt}$ (using the chain rule).
Given that the rate of change of the radius is $\frac{dr}{dt} = 3 \text{ cm/s}$.
Substituting the values,we get:
$\frac{dA}{dt} = 2 \pi r (3) = 6 \pi r$.
When the radius $r = 10 \text{ cm}$,the rate of change of the area is:
$\frac{dA}{dt} = 6 \pi (10) = 60 \pi \text{ cm}^2/\text{s}$.
Thus,the area of the circle is increasing at the rate of $60 \pi \text{ cm}^2/\text{s}$.

(A) Let $x$ be the length of the edge and $V$ be the volume of the cube.
Then,the volume is given by $V = x^3$.
Differentiating both sides with respect to time $t$,we get:
$\frac{dV}{dt} = \frac{d}{dt}(x^3) = 3x^2 \cdot \frac{dx}{dt}$ (by chain rule).
Given that the rate of increase of the edge is $\frac{dx}{dt} = 3 \, cm/s$.
Substituting this value,we get:
$\frac{dV}{dt} = 3x^2(3) = 9x^2$.
When the edge length $x = 10 \, cm$,the rate of increase of the volume is:
$\frac{dV}{dt} = 9(10)^2 = 9(100) = 900 \, cm^3/s$.
Thus,the volume of the cube is increasing at the rate of $900 \, cm^3/s$.

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

(A) The circumference $C$ of a circle with radius $r$ is given by $C = 2 \pi r$.
Differentiating both sides with respect to time $t$,we get the rate of change of circumference as:
$\frac{dC}{dt} = \frac{d}{dt}(2 \pi r) = 2 \pi \frac{dr}{dt}$.
Given that the rate of increase of the radius is $\frac{dr}{dt} = 0.7 \, cm/s$.
Substituting this value into the equation,we get:
$\frac{dC}{dt} = 2 \pi (0.7) = 1.4 \pi \, cm/s$.
Thus,the rate of increase of the circumference is $1.4 \pi \, cm/s$.

(A) Given that the length $x$ is decreasing at the rate of $5 \text{ cm/min}$,we have $\frac{dx}{dt} = -5 \text{ cm/min}$.
Given that the width $y$ is increasing at the rate of $4 \text{ cm/min}$,we have $\frac{dy}{dt} = 4 \text{ cm/min}$.
The perimeter $P$ of a rectangle is given by the formula $P = 2(x + y)$.
To find the rate of change of the perimeter with respect to time $t$,we differentiate $P$ with respect to $t$:
$\frac{dP}{dt} = 2 \left( \frac{dx}{dt} + \frac{dy}{dt} \right)$.
Substituting the given values:
$\frac{dP}{dt} = 2(-5 + 4) = 2(-1) = -2 \text{ cm/min}$.
Thus,the perimeter is decreasing at the rate of $2 \text{ cm/min}$.

(A) Given that the length $x$ is decreasing at the rate of $5 \text{ cm/min}$,we have $\frac{dx}{dt} = -5 \text{ cm/min}$.
Given that the width $y$ is increasing at the rate of $4 \text{ cm/min}$,we have $\frac{dy}{dt} = 4 \text{ cm/min}$.
The area $A$ of a rectangle is given by $A = x \times y$.
Differentiating both sides with respect to $t$,we get $\frac{dA}{dt} = \frac{dx}{dt} \cdot y + x \cdot \frac{dy}{dt}$.
Substituting the given values $x = 8 \text{ cm}$,$y = 6 \text{ cm}$,$\frac{dx}{dt} = -5 \text{ cm/min}$,and $\frac{dy}{dt} = 4 \text{ cm/min}$:
$\frac{dA}{dt} = (-5)(6) + (8)(4) = -30 + 32 = 2 \text{ cm}^2/\text{min}$.
Thus,the area of the rectangle is increasing at the rate of $2 \text{ cm}^2/\text{min}$.

(A) The volume of a sphere $(V)$ with radius $(r)$ is given by $V = \frac{4}{3} \pi r^3$.
Differentiating with respect to time $(t)$,we get the rate of change of volume:
$\frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3} \pi r^3 \right) = 4 \pi r^2 \frac{dr}{dt}$.
Given that $\frac{dV}{dt} = 900 \, cm^3/s$,we substitute this into the equation:
$900 = 4 \pi r^2 \frac{dr}{dt}$.
Solving for $\frac{dr}{dt}$:
$\frac{dr}{dt} = \frac{900}{4 \pi r^2} = \frac{225}{\pi r^2}$.
When the radius $r = 15 \, cm$:
$\frac{dr}{dt} = \frac{225}{\pi (15)^2} = \frac{225}{225 \pi} = \frac{1}{\pi} \, cm/s$.
Thus,the rate at which the radius increases is $\frac{1}{\pi} \, cm/s$.

(A) 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 at which the volume is increasing with respect to the radius,we calculate the derivative $\frac{dV}{dr}$:
$\frac{dV}{dr} = \frac{d}{dr} \left( \frac{4}{3} \pi r^3 \right) = \frac{4}{3} \pi (3r^2) = 4 \pi r^2$.
We are asked to find this rate when the radius $r = 10 \text{ cm}$.
Substituting $r = 10$ into the derivative:
$\frac{dV}{dr} = 4 \pi (10)^2 = 4 \pi (100) = 400 \pi$.
Thus,the rate of change of volume with respect to the radius is $400 \pi \text{ cm}^3/\text{cm}$.

(A) Let $x$ be the distance of the foot of the ladder from the wall and $y$ be the height of the ladder on the wall.
Given the length of the ladder is $5 \ m$,by Pythagoras theorem: $x^2 + y^2 = 5^2 = 25$.
Differentiating with respect to time $t$: $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$,which simplifies to $\frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt}$.
Given $\frac{dx}{dt} = 2 \ cm/s = 0.02 \ m/s$.
When $x = 4 \ m$,$y = \sqrt{25 - 4^2} = \sqrt{9} = 3 \ m$.
Substituting the values: $\frac{dy}{dt} = -\frac{4}{3} \times 2 = -\frac{8}{3} \ cm/s$.
The negative sign indicates that the height is decreasing.
Thus,the height is decreasing at the rate of $\frac{8}{3} \ cm/s$.

(A) The equation of the curve is given as $6y = x^3 + 2$.
Differentiating both sides with respect to time $t$,we get:
$6 \frac{dy}{dt} = 3x^2 \frac{dx}{dt}$
Dividing by $3$,we get:
$2 \frac{dy}{dt} = x^2 \frac{dx}{dt}$
Given that the $y$-coordinate is changing $8$ times as fast as the $x$-coordinate,i.e.,$\frac{dy}{dt} = 8 \frac{dx}{dt}$.
Substituting this into the equation:
$2(8 \frac{dx}{dt}) = x^2 \frac{dx}{dt}$
$16 \frac{dx}{dt} = x^2 \frac{dx}{dt}$
$(x^2 - 16) \frac{dx}{dt} = 0$
Assuming $\frac{dx}{dt} \neq 0$,we have $x^2 = 16$,which gives $x = 4$ or $x = -4$.
For $x = 4$,$6y = 4^3 + 2 = 64 + 2 = 66$,so $y = 11$. The point is $(4, 11)$.
For $x = -4$,$6y = (-4)^3 + 2 = -64 + 2 = -62$,so $y = -62/6 = -31/3$. The point is $(-4, -31/3)$.
Thus,the required points are $(4, 11)$ and $(-4, -31/3)$.

(A) The air bubble is in the shape of a sphere.
The volume $V$ of a sphere with radius $r$ is given by $V = \frac{4}{3} \pi r^3$.
Differentiating both sides with respect to time $t$,we get:
$\frac{dV}{dt} = \frac{4}{3} \pi \cdot 3r^2 \cdot \frac{dr}{dt} = 4 \pi r^2 \frac{dr}{dt}$.
Given that $\frac{dr}{dt} = \frac{1}{2} \text{ cm/s}$ and $r = 1 \text{ cm}$.
Substituting these values into the equation:
$\frac{dV}{dt} = 4 \pi (1)^2 \left( \frac{1}{2} \right) = 2 \pi \text{ cm}^3/\text{s}$.
Thus,the volume of the bubble is increasing at the rate of $2 \pi \text{ cm}^3/\text{s}$.

(B) The volume of a sphere $(V)$ with radius $(r)$ is given by $V = \frac{4}{3} \pi r^3$.
Given the diameter $d = \frac{3}{2}(2x+1)$,the radius $r$ is $r = \frac{d}{2} = \frac{3}{4}(2x+1)$.
Substituting $r$ into the volume formula:
$V = \frac{4}{3} \pi \left( \frac{3}{4}(2x+1) \right)^3 = \frac{4}{3} \pi \cdot \frac{27}{64} (2x+1)^3 = \frac{9}{16} \pi (2x+1)^3$.
To find the rate of change of volume with respect to $x$,we differentiate $V$ with respect to $x$:
$\frac{dV}{dx} = \frac{d}{dx} \left( \frac{9}{16} \pi (2x+1)^3 \right) = \frac{9}{16} \pi \cdot 3(2x+1)^2 \cdot \frac{d}{dx}(2x+1)$.
$\frac{dV}{dx} = \frac{27}{16} \pi (2x+1)^2 \cdot 2 = \frac{27}{8} \pi (2x+1)^2$.

(A) The volume of a cone $V$ with radius $r$ and height $h$ is given by $V = \frac{1}{3} \pi r^2 h$.
Given that $h = \frac{1}{6} r$,we have $r = 6h$.
Substituting $r$ in the volume formula: $V = \frac{1}{3} \pi (6h)^2 h = \frac{1}{3} \pi (36h^2) h = 12 \pi h^3$.
Differentiating with respect to time $t$: $\frac{dV}{dt} = 12 \pi (3h^2) \frac{dh}{dt} = 36 \pi h^2 \frac{dh}{dt}$.
Given $\frac{dV}{dt} = 12 \text{ cm}^3/\text{s}$,we set $12 = 36 \pi h^2 \frac{dh}{dt}$.
At $h = 4 \text{ cm}$,$12 = 36 \pi (4)^2 \frac{dh}{dt} = 36 \pi (16) \frac{dh}{dt} = 576 \pi \frac{dh}{dt}$.
Thus,$\frac{dh}{dt} = \frac{12}{576 \pi} = \frac{1}{48 \pi} \text{ cm/s}$.

(A) Marginal cost is the rate of change of total cost with respect to output.
Marginal cost $(MC) = \frac{dC}{dx} = 0.007(3x^{2}) - 0.003(2x) + 15$
$= 0.021x^{2} - 0.006x + 15$
When $x = 17$,$MC = 0.021(17^{2}) - 0.006(17) + 15$
$= 0.021(289) - 0.102 + 15$
$= 6.069 - 0.102 + 15$
$= 20.967$
Hence,when $17$ units are produced,the marginal cost is $Rs. 20.967$.

(A) 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 $\frac{dR}{dx}$.
Given the total revenue function $R(x) = 13x^2 + 26x + 15$.
To find the marginal revenue $(MR)$,we differentiate $R(x)$ with respect to $x$:
$MR = \frac{dR}{dx} = \frac{d}{dx}(13x^2 + 26x + 15)$
$MR = 13(2x) + 26(1) + 0$
$MR = 26x + 26$
Now,we evaluate the marginal revenue at $x = 7$:
$MR = 26(7) + 26$
$MR = 182 + 26$
$MR = 208$
Thus,the marginal revenue when $x = 7$ is $Rs. 208$.

(B) The area of a circle $A$ with radius $r$ is given by the formula:
$A = \pi r^2$
To find the rate of change of the area with respect to its 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 $r = 6 \text{ cm}$:
$\frac{dA}{dr} = 2 \pi (6) = 12 \pi \text{ cm}^2/\text{cm}$
Thus,the rate of change of the area of the circle with respect to its radius at $r = 6 \text{ cm}$ is $12 \pi \text{ cm}^2/\text{cm}$.
The correct option is $B$.