Rate of Change of Quantities Questions in English

(B) Let $r$ be the radius of the spherical balloon. Given that the volume $V = \frac{4}{3}\pi r^3$ is increasing at a rate of $\frac{dV}{dt} = 35 \, cc/min$.
We know that $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$.
Given diameter $d = 14 \, cm$,so radius $r = 7 \, cm$.
Substituting the values: $35 = 4\pi (7)^2 \frac{dr}{dt} = 196\pi \frac{dr}{dt}$.
Thus,$\frac{dr}{dt} = \frac{35}{196\pi} = \frac{5}{28\pi} \, cm/min$.
The surface area $S = 4\pi r^2$. The rate of increase of surface area is $\frac{dS}{dt} = 8\pi r \frac{dr}{dt}$.
Substituting $r = 7$ and $\frac{dr}{dt} = \frac{5}{28\pi}$:
$\frac{dS}{dt} = 8\pi (7) \left( \frac{5}{28\pi} \right) = 56\pi \left( \frac{5}{28\pi} \right) = 2 \times 5 = 10 \, cm^2/min$.

(D) Given that the displacement $s$ is proportional to the square of the velocity $v$,we can write:
$s = kv^2$,where $k$ is a constant of proportionality.
Rearranging for $v^2$,we get $v^2 = \frac{1}{k} s$.
Let $C = \frac{1}{k}$,so $v^2 = Cs$.
Differentiating both sides with respect to time $t$:
$\frac{d}{dt}(v^2) = \frac{d}{dt}(Cs)$
$2v \frac{dv}{dt} = C \frac{ds}{dt}$
Since $\frac{dv}{dt} = a$ (acceleration) and $\frac{ds}{dt} = v$ (velocity),we have:
$2v a = Cv$
Assuming $v \neq 0$,we divide by $v$:
$2a = C$
$a = \frac{C}{2}$
Since $C$ is a constant,the acceleration $a$ is a constant.

(A) The volume of a cylinder is given by $V = \pi r^2 h$.
By differentiating with respect to time $t$,we get:
$\frac{dV}{dt} = \pi \left( 2rh \frac{dr}{dt} + r^2 \frac{dh}{dt} \right)$.
Given: $r = 4 \, m$,$h = 6 \, m$,$\frac{dr}{dt} = 3 \, m/sec$,and $\frac{dh}{dt} = -4 \, m/sec$ (since altitude is decreasing).
Substituting these values:
$\frac{dV}{dt} = \pi \left[ 2(4)(6)(3) + (4)^2(-4) \right]$.
$\frac{dV}{dt} = \pi [144 - 64] = 80\pi \, m^3/sec$.

(B) 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 that the length of the ladder is $10 \, m$,we have the relation $x^2 + y^2 = 10^2 = 100$.
Differentiating both sides with respect to time $t$,we get:
$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$
$x \frac{dx}{dt} + y \frac{dy}{dt} = 0$
Given $\frac{dx}{dt} = 3 \, cm/sec$ and $\frac{dy}{dt} = -4 \, cm/sec$ (since the height is decreasing).
Substituting these values,we get:
$x(3) + y(-4) = 0$
$3x = 4y \implies x = \frac{4}{3}y$
Substituting $x = \frac{4}{3}y$ into the original equation $x^2 + y^2 = 100$:
$(\frac{4}{3}y)^2 + y^2 = 100$
$\frac{16}{9}y^2 + y^2 = 100$
$\frac{25}{9}y^2 = 100$
$y^2 = 100 \times \frac{9}{25} = 36$
$y = 6 \, m$.
Thus,the height of the upper end is $6 \, m$.

(C) Let $f(x) = x^3$ and $g(x) = 6x^2 + 15x + 5$.
For $f(x)$ to increase less rapidly than $g(x)$,the rate of change of $f(x)$ must be less than the rate of change of $g(x)$.
This implies $f'(x) < g'(x)$.
Calculating the derivatives: $f'(x) = 3x^2$ and $g'(x) = 12x + 15$.
Setting up the inequality: $3x^2 < 12x + 15$.
Rearranging the terms: $3x^2 - 12x - 15 < 0$.
Dividing by $3$: $x^2 - 4x - 5 < 0$.
Factoring the quadratic: $(x - 5)(x + 1) < 0$.
The inequality holds when $x$ lies between the roots of the quadratic equation $x^2 - 4x - 5 = 0$,which are $x = -1$ and $x = 5$.
Thus,the interval is $(-1, 5)$.

(C) Let $V$ be the volume and $S$ be the surface area of the spherical balloon with radius $r$.
Given that $\frac{dV}{dt} = 40 \ cm^3/\min$.
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: $40 = 4\pi (8)^2 \frac{dr}{dt} \implies 40 = 256\pi \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{40}{256\pi} = \frac{5}{32\pi} \ cm/\min$.
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 $r = 8$ and $\frac{dr}{dt} = \frac{5}{32\pi}$:
$\frac{dS}{dt} = 8\pi \times 8 \times \frac{5}{32\pi} = 64\pi \times \frac{5}{32\pi} = 2 \times 5 = 10 \ cm^2/\min$.

(B) Let $AB$ be the lamp post of height $4.5 \ m$ and $PQ$ be the man of height $1.8 \ m$. Let the man be at a distance $y$ from the lamp post,so $QB = y$. Let the length of the shadow be $x$,so $CQ = x$.
From similar triangles $\triangle AB C$ and $\triangle PQC$,we have:
$\frac{PQ}{AB} = \frac{CQ}{CB}$
$\frac{1.8}{4.5} = \frac{x}{x + y}$
$\frac{2}{5} = \frac{x}{x + y}$
$2(x + y) = 5x$
$2x + 2y = 5x$
$3x = 2y$
$x = \frac{2}{3}y$
Differentiating with respect to time $t$:
$\frac{dx}{dt} = \frac{2}{3} \frac{dy}{dt}$
Given that the man is moving away at $\frac{dy}{dt} = 1.2 \ m/sec$,we have:
$\frac{dx}{dt} = \frac{2}{3} \times 1.2 = 0.8 \ m/sec$.
Thus,the shadow is lengthening at a rate of $0.8 \ m/sec$.

(A) Given the distance function $s(t) = t^3 + 2t^2 + t$.
The speed $v$ of the particle is the rate of change of distance with respect to time,given by $v = \frac{ds}{dt}$.
Differentiating $s$ with respect to $t$:
$v = \frac{d}{dt}(t^3 + 2t^2 + t) = 3t^2 + 4t + 1$.
To find the speed after $1$ second,substitute $t = 1$ into the expression for $v$:
$v(1) = 3(1)^2 + 4(1) + 1 = 3 + 4 + 1 = 8 \, cm/sec$.
Thus,the speed of the particle after $1$ second is $8 \, cm/sec$.

(D) Let the side of the square be $x$ and its area be $A$.
Given that the rate of change of the side is $\frac{dx}{dt} = 4 \text{ cm/min}$.
The area of a square is given by $A = x^2$.
Differentiating both sides with respect to time $t$,we get $\frac{dA}{dt} = 2x \frac{dx}{dt}$.
When the side $x = 8 \text{ cm}$,the rate of change of the area is:
$\frac{dA}{dt} = 2 \times 8 \times 4 = 64 \text{ cm}^2/\text{min}$.

(C) Let $V$ be the volume and $S$ be the surface area of the spherical balloon with radius $r$.
Given: $\frac{dV}{dt} = 40 \text{ cm}^3/\text{min}$ and $r = 8 \text{ cm}$.
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: $40 = 4\pi (8)^2 \frac{dr}{dt} \Rightarrow 40 = 256\pi \frac{dr}{dt} \Rightarrow \frac{dr}{dt} = \frac{40}{256\pi} = \frac{5}{32\pi} \text{ cm/min}$.
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 $r = 8$ and $\frac{dr}{dt} = \frac{5}{32\pi}$:
$\frac{dS}{dt} = 8\pi \times 8 \times \frac{5}{32\pi} = 64\pi \times \frac{5}{32\pi} = 2 \times 5 = 10 \text{ cm}^2/\text{min}$.

(B) Given the volume of a sphere is $V = \frac{4}{3}\pi r^3$.
To find the rate of change of $V$ with respect to $t$,we differentiate both sides with respect to $t$:
$\frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3}\pi r^3 \right) = \frac{4}{3}\pi \cdot 3r^2 \cdot \frac{dr}{dt} = 4\pi r^2 \frac{dr}{dt}$.
Given $r = 10$ and $\frac{dr}{dt} = 0.01$.
Substituting these values:
$\frac{dV}{dt} = 4\pi (10)^2 (0.01) = 4\pi (100) (0.01) = 4\pi (1) = 4\pi$.

(A) 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:
$\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$.
Given that $\frac{dV}{dt} = 900 \ cm^3/sec$ and the radius $r = 15 \ cm$,we substitute these values into the equation:
$900 = 4 \pi (15)^2 \frac{dr}{dt}$.
$900 = 4 \pi (225) \frac{dr}{dt}$.
$900 = 900 \pi \frac{dr}{dt}$.
Solving for $\frac{dr}{dt}$:
$\frac{dr}{dt} = \frac{900}{900 \pi} = \frac{1}{\pi} \ cm/sec$.
Thus,the radius of the balloon is increasing at a rate of $1/\pi \ cm/sec$.

(C) Let $r$ be the radius,$h$ be the height,and $V$ be the volume of the cylinder at any time $t$.
Given: $\frac{dr}{dt} = 3 \text{ m/s}$ and $\frac{dh}{dt} = -4 \text{ m/s}$.
We need to find $\frac{dV}{dt}$ when $r = 4 \text{ m}$ and $h = 6 \text{ m}$.
The volume of a cylinder is given by $V = \pi r^2 h$.
Differentiating both sides with respect to $t$ using the product rule:
$\frac{dV}{dt} = \pi \left( r^2 \frac{dh}{dt} + h \cdot 2r \frac{dr}{dt} \right)$.
Substituting the given values:
$\frac{dV}{dt} = \pi \left( (4)^2(-4) + 6 \cdot 2(4)(3) \right)$.
$\frac{dV}{dt} = \pi \left( 16(-4) + 6(24) \right)$.
$\frac{dV}{dt} = \pi (-64 + 144) = 80\pi \text{ m}^3/\text{s}$.

(A) Let $v$ be the velocity and $s$ be the distance covered by the particle.
Given that $v \propto \sqrt{s}$,which implies $v = k\sqrt{s}$ for some constant $k$.
The acceleration $a$ is given by $a = \frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt} = v \frac{dv}{ds}$.
Since $v = k\sqrt{s}$,we have $\frac{dv}{ds} = k \cdot \frac{1}{2\sqrt{s}} = \frac{k}{2\sqrt{s}}$.
Substituting these into the expression for acceleration:
$a = (k\sqrt{s}) \cdot \left( \frac{k}{2\sqrt{s}} \right) = \frac{k^2}{2}$.
Since $k$ is a constant,$a = \frac{k^2}{2}$ is also a constant.
Therefore,the acceleration of the particle is constant.

(C) The displacement equation is given by $s = 13.8t - 4.9t^2$.
The velocity $v$ is the rate of change of displacement with respect to time,given by $v = \frac{ds}{dt}$.
Differentiating $s$ with respect to $t$:
$v = \frac{d}{dt}(13.8t - 4.9t^2) = 13.8 - 9.8t$.
To find the velocity at $t = 1$ second,substitute $t = 1$ into the velocity equation:
$v(1) = 13.8 - 9.8(1) = 13.8 - 9.8 = 4.0 \text{ } m/s$.
Thus,the velocity at $t = 1$ second is $4 \text{ } m/s$.

(B) Let the ladder be $PQ$ with length $5 \ m$. Let $P$ be the bottom of the ladder at a distance $x$ from the wall and $Q$ be the top of the ladder at a height $y$ from the ground.
By the Pythagorean theorem,we have $x^2 + y^2 = 5^2 = 25$.
Given that the bottom of the ladder is moving away from the wall at a rate of $\frac{dx}{dt} = 2 \ m/sec$.
We need to find the rate at which the height $y$ is decreasing,i.e.,$-\frac{dy}{dt}$,when $x = 4 \ m$.
When $x = 4$,substituting into $x^2 + y^2 = 25$ gives $4^2 + y^2 = 25$,so $y^2 = 25 - 16 = 9$,which means $y = 3 \ m$.
Differentiating $x^2 + y^2 = 25$ with respect to $t$,we get:
$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$
Substituting the known values $x = 4$,$y = 3$,and $\frac{dx}{dt} = 2$:
$2(4)(2) + 2(3) \frac{dy}{dt} = 0$
$16 + 6 \frac{dy}{dt} = 0$
$6 \frac{dy}{dt} = -16$
$\frac{dy}{dt} = -\frac{16}{6} = -\frac{8}{3} \ m/sec$.
Since the height is decreasing,the rate of decrease is $\frac{8}{3} \ m/sec$.

(B) Let $x$ be the side length and $A$ be the area of the equilateral triangle at time $t$.
The area of an equilateral triangle is given by $A = \frac{\sqrt{3}}{4} x^2$.
Differentiating both sides with respect to $t$,we get:
$\frac{dA}{dt} = \frac{\sqrt{3}}{4} \times 2x \times \frac{dx}{dt} = \frac{\sqrt{3}}{2} x \frac{dx}{dt}$.
Given that $\frac{dx}{dt} = 2 \ cm/s$ and $x = 10 \ cm$.
Substituting these values into the derivative formula:
$\frac{dA}{dt} = \frac{\sqrt{3}}{2} \times 10 \times 2 = 10\sqrt{3} \ cm^2/s$.

(D) Let $r$ be the thickness of the ice layer at any time $t$ and $V$ be the volume of the ice layer.
Given that the rate of change of volume is $\frac{dV}{dt} = -50 \text{ cm}^3/\text{min}$.
We need to find $\frac{dr}{dt}$ when $r = 5 \text{ cm}$.
The radius of the sphere including the ice layer is $(r + 10) \text{ cm}$.
The volume of the ice layer $V$ is the volume of the sphere with ice minus the volume of the iron sphere:
$V = \frac{4}{3}\pi(r + 10)^3 - \frac{4}{3}\pi(10)^3$.
Differentiating both sides with respect to $t$:
$\frac{dV}{dt} = 4\pi(r + 10)^2 \frac{dr}{dt}$.
Substituting the given values $\frac{dV}{dt} = -50$ and $r = 5$:
$-50 = 4\pi(5 + 10)^2 \frac{dr}{dt}$.
$-50 = 4\pi(15)^2 \frac{dr}{dt}$.
$-50 = 4\pi(225) \frac{dr}{dt}$.
$-50 = 900\pi \frac{dr}{dt}$.
$\frac{dr}{dt} = -\frac{50}{900\pi} = -\frac{1}{18\pi} \text{ cm/min}$.
Since the question asks for the rate at which the thickness is decreasing,the rate is $1/(18\pi) \text{ cm/min}$.

(A) Let $r$ be the radius of the ripple and $A$ be the area of the circle enclosed by the ripple.
Given,the rate of change of the radius is $\frac{dr}{dt} = 3.5 \text{ cm/sec}$.
The area of the circle is given by $A = \pi r^2$.
Differentiating both sides with respect to time $t$,we get $\frac{dA}{dt} = 2 \pi r \frac{dr}{dt}$.
At the instant when $r = 7.5 \text{ cm}$,we substitute the values:
$\frac{dA}{dt} = 2 \times \pi \times 7.5 \times 3.5$.
$\frac{dA}{dt} = 15 \times 3.5 \times \pi$.
$\frac{dA}{dt} = 52.5 \pi \text{ cm}^2/\text{sec}$.
Thus,the area is increasing at a rate of $52.5 \pi \text{ cm}^2/\text{sec}$.

(C) Let $y = \sqrt{x^2 + 16}$ and $z = \frac{x}{x - 1}$.
First,differentiate $y$ with respect to $x$:
$\frac{dy}{dx} = \frac{1}{2\sqrt{x^2 + 16}} \cdot (2x) = \frac{x}{\sqrt{x^2 + 16}}$.
Next,differentiate $z$ with respect to $x$ using the quotient rule:
$\frac{dz}{dx} = \frac{(x - 1)(1) - x(1)}{(x - 1)^2} = \frac{-1}{(x - 1)^2}$.
Now,find the rate of change of $y$ with respect to $z$:
$\frac{dy}{dz} = \frac{dy/dx}{dz/dx} = \frac{x}{\sqrt{x^2 + 16}} \cdot (-(x - 1)^2) = \frac{-x(x - 1)^2}{\sqrt{x^2 + 16}}$.
Substitute $x = 3$ into the expression:
$\left( \frac{dy}{dz} \right)_{x = 3} = \frac{-3(3 - 1)^2}{\sqrt{3^2 + 16}} = \frac{-3(2^2)}{\sqrt{9 + 16}} = \frac{-3(4)}{\sqrt{25}} = \frac{-12}{5}$.

(D) Let $x$ be the thickness of the ice. The total radius of the ball including the ice is $r = (10 + x) \ cm$.
The volume of the ice layer is $V = \frac{4}{3}\pi(10 + x)^3 - \frac{4}{3}\pi(10)^3$.
Differentiating with respect to time $t$,we get $\frac{dV}{dt} = 4\pi(10 + x)^2 \frac{dx}{dt}$.
Given $\frac{dV}{dt} = -50 \ cm^3/min$ (since the ice is melting,the volume is decreasing).
At $x = 5 \ cm$,we have $-50 = 4\pi(10 + 5)^2 \frac{dx}{dt}$.
$-50 = 4\pi(15)^2 \frac{dx}{dt} = 4\pi(225) \frac{dx}{dt} = 900\pi \frac{dx}{dt}$.
Therefore,$\frac{dx}{dt} = -\frac{50}{900\pi} = -\frac{1}{18\pi} \ cm/min$.
The rate at which the thickness decreases is $\frac{1}{18\pi} \ cm/min$.

(B) Given the displacement equation: $x = At^2 + Bt + C$.
Velocity $v$ is the rate of change of displacement with respect to time $t$:
$v = \frac{dx}{dt} = \frac{d}{dt}(At^2 + Bt + C) = 2At + B$.
Now,substitute $x$ and $v$ into the expression $4Ax - v^2$:
$4Ax - v^2 = 4A(At^2 + Bt + C) - (2At + B)^2$.
Expand the terms:
$4Ax - v^2 = (4A^2t^2 + 4ABt + 4AC) - (4A^2t^2 + 4ABt + B^2)$.
Simplify the expression by canceling common terms:
$4Ax - v^2 = 4A^2t^2 - 4A^2t^2 + 4ABt - 4ABt + 4AC - B^2$.
$4Ax - v^2 = 4AC - B^2$.

(B) Given the diameter $ d = \frac{3}{2}(2x + 3) $.
The radius $ r = \frac{d}{2} = \frac{3}{4}(2x + 3) $.
Differentiating $ r $ with respect to $ x $,we get $ \frac{dr}{dx} = \frac{3}{4} \times 2 = \frac{3}{2} $.
The volume of a sphere is $ V = \frac{4}{3}\pi r^3 $.
Differentiating $ V $ with respect to $ x $ using the chain rule:
$ \frac{dV}{dx} = \frac{dV}{dr} \times \frac{dr}{dx} $.
$ \frac{dV}{dx} = (4\pi r^2) \times \frac{dr}{dx} $.
Substituting the values of $ r $ and $ \frac{dr}{dx} $:
$ \frac{dV}{dx} = 4\pi \left[ \frac{3}{4}(2x + 3) \right]^2 \times \frac{3}{2} $.
$ \frac{dV}{dx} = 4\pi \times \frac{9}{16}(2x + 3)^2 \times \frac{3}{2} $.
$ \frac{dV}{dx} = \frac{27\pi}{8}(2x + 3)^2 $.

(A) Let $r$ be the radius and $V$ be the volume of the spherical bubble. The surface area $A$ is given by $A = 4\pi r^2$.
Differentiating with respect to time $t$,we get $\frac{dA}{dt} = 8\pi r \frac{dr}{dt}$.
Given $\frac{dA}{dt} = 2 \text{ cm}^2/\text{s}$ and $r = 6 \text{ cm}$,we substitute these values:
$2 = 8\pi(6) \frac{dr}{dt} \Rightarrow 2 = 48\pi \frac{dr}{dt} \Rightarrow \frac{dr}{dt} = \frac{1}{24\pi} \text{ cm/s}$.
The volume $V$ of the 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 $r = 6$ and $\frac{dr}{dt} = \frac{1}{24\pi}$:
$\frac{dV}{dt} = 4\pi(6)^2 \left(\frac{1}{24\pi}\right) = 4\pi(36) \left(\frac{1}{24\pi}\right) = \frac{144\pi}{24\pi} = 6 \text{ cm}^3/\text{s}$.

(A) The area of the triangle is given by $S = \frac{1}{2} ab \sin C$.
Taking the derivative with respect to $C$,we get $\frac{dS}{dC} = \frac{1}{2} ab \cos C$.
Let $\Delta S$ be the error in the area $S$ corresponding to an error $\Delta C = \alpha$ in angle $C$.
Then,$\Delta S \approx \frac{dS}{dC} \cdot \Delta C = \frac{1}{2} ab \cos C \cdot \alpha$.
The relative error in the area is given by $\frac{\Delta S}{S}$.
Substituting the values,we get $\frac{\Delta S}{S} = \frac{\frac{1}{2} ab \cos C \cdot \alpha}{\frac{1}{2} ab \sin C} = \alpha \cot C$.

(B) Given the displacement equation: $x = t^3 - 12t^2 + 6t + 8$.
Velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{dx}{dt} = 3t^2 - 24t + 6$.
Acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = 6t - 24$.
At the instant when acceleration is zero: $a = 0 \Rightarrow 6t - 24 = 0 \Rightarrow t = 4 \text{ s}$.
Now,substitute $t = 4$ into the velocity equation: $v = 3(4)^2 - 24(4) + 6$.
$v = 3(16) - 96 + 6 = 48 - 96 + 6 = -42 \text{ units/s}$.

(B) Let the side length of the square be $x$.
The diagonal $R$ is given by $R = \sqrt{x^2 + x^2} = x\sqrt{2}$.
The area $A$ is given by $A = x^2$.
From the area equation,we have $x = \sqrt{A}$.
Substituting this into the diagonal equation,we get $R = \sqrt{2} \cdot \sqrt{A} = \sqrt{2A}$.
To find the rate of change of $R$ with respect to $A$,we differentiate $R$ with respect to $A$:
$\frac{dR}{dA} = \frac{d}{dA}(\sqrt{2} \cdot A^{1/2}) = \sqrt{2} \cdot \frac{1}{2} A^{-1/2} = \frac{\sqrt{2}}{2\sqrt{A}} = \frac{1}{\sqrt{2A}}$.
Since $R = \sqrt{2A}$,we can substitute this back into the expression:
$\frac{dR}{dA} = \frac{1}{R}$.

(D) Let $y = x^3 - 5x^2 + 5x + 8$.
The rate of change of $y$ with respect to time $t$ is given by $\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$.
Given that $\frac{dy}{dt} = 2 \frac{dx}{dt}$,we have:
$(3x^2 - 10x + 5) \frac{dx}{dt} = 2 \frac{dx}{dt}$.
Assuming $\frac{dx}{dt} \neq 0$,we divide both sides by $\frac{dx}{dt}$:
$3x^2 - 10x + 5 = 2$.
$3x^2 - 10x + 3 = 0$.
Factoring the quadratic equation:
$(3x - 1)(x - 3) = 0$.
Therefore,$x = 3$ or $x = \frac{1}{3}$.

(D) Given the equation of the parabola is $y^2 = 18x$.
Differentiating both sides with respect to $t$,we get $2y \frac{dy}{dt} = 18 \frac{dx}{dt}$,which simplifies to $y \frac{dy}{dt} = 9 \frac{dx}{dt}$ (Equation $1$).
It is given that the rate of change of the $y$-coordinate is twice the rate of change of the $x$-coordinate,i.e.,$\frac{dy}{dt} = 2 \frac{dx}{dt}$.
Substituting this into Equation $1$,we get $y(2 \frac{dx}{dt}) = 9 \frac{dx}{dt}$.
Assuming $\frac{dx}{dt} \neq 0$,we have $2y = 9$,which implies $y = \frac{9}{2}$.
Substituting $y = \frac{9}{2}$ into the original equation $y^2 = 18x$,we get $\left( \frac{9}{2} \right)^2 = 18x$.
This gives $\frac{81}{4} = 18x$,so $x = \frac{81}{4 \times 18} = \frac{9}{8}$.
Thus,the required point is $\left( \frac{9}{8}, \frac{9}{2} \right)$.

(C) The surface area $A$ of a sphere with radius $r$ is given by $A = 4\pi r^2$.
Differentiating both sides with respect to time $t$,we get:
$\frac{dA}{dt} = 8\pi r \frac{dr}{dt}$.
Given that $\frac{dr}{dt} = 0.1 \ cm/s$ and $r = 200 \ cm$.
Substituting these values into the derivative formula:
$\frac{dA}{dt} = 8\pi \times 200 \times 0.1$.
$\frac{dA}{dt} = 160\pi \ cm^2/s$.

(B) Let $h$ be the height of the water and $V$ be the volume of the water at any time $t$.
The radius of the cylindrical vessel is given as $r = 2 \text{ cm}$.
The rate of change of volume is given as $\frac{dV}{dt} = 8 \text{ cm}^3/\text{s}$.
The volume of water in a cylinder is given by the formula $V = \pi r^2 h$.
Substituting $r = 2$,we get $V = \pi (2)^2 h = 4\pi h$.
Differentiating both sides with respect to $t$,we get $\frac{dV}{dt} = 4\pi \frac{dh}{dt}$.
Substituting the given value $\frac{dV}{dt} = 8$,we have $8 = 4\pi \frac{dh}{dt}$.
Therefore,$\frac{dh}{dt} = \frac{8}{4\pi} = \frac{2}{\pi} \text{ cm/s}$.

(A) Let the volume of the sphere be $V = \frac{4}{3}\pi r^3$ and the surface area be $S = 4\pi r^2$.
We need to find the rate of change of volume with respect to surface area,which is $\frac{dV}{dS}$.
First,differentiate $V$ with respect to $r$: $\frac{dV}{dr} = 4\pi r^2$.
Next,differentiate $S$ with respect to $r$: $\frac{dS}{dr} = 8\pi r$.
Using the chain rule,$\frac{dV}{dS} = \frac{dV/dr}{dS/dr} = \frac{4\pi r^2}{8\pi r}$.
Simplifying this,we get $\frac{dV}{dS} = \frac{r}{2}$.

(C) Given,$x = t^3 - 9t^2 + 24t + 6$.
The velocity $v$ is given by the first derivative of $x$ with respect to $t$:
$v = \frac{dx}{dt} = 3t^2 - 18t + 24$.
The acceleration $a$ is given by the second derivative of $x$ with respect to $t$:
$a = \frac{d^2x}{dt^2} = 6t - 18$.
Given that the acceleration $a = 6$:
$6t - 18 = 6$
$6t = 24$
$t = 4$.
Now,substitute $t = 4$ into the velocity equation:
$v = 3(4)^2 - 18(4) + 24$
$v = 3(16) - 72 + 24$
$v = 48 - 72 + 24$
$v = 0$ units.

(A) Let the position of the first cyclist be $A$ and the second cyclist be $B$ at time $t$. Let the distance between them be $x$.
Given $OA = 4t$ and $OB = 3t$.
Using the Law of Cosines in $\triangle OAB$:
$x^2 = (OA)^2 + (OB)^2 - 2(OA)(OB) \cos(120^{\circ})$
$x^2 = (4t)^2 + (3t)^2 - 2(4t)(3t) \left(-\frac{1}{2}\right)$
$x^2 = 16t^2 + 9t^2 + 12t^2$
$x^2 = 37t^2$
$x = \sqrt{37}t$
The rate at which they are separating is $\frac{dx}{dt}$.
$\frac{dx}{dt} = \frac{d}{dt}(\sqrt{37}t) = \sqrt{37} \text{ km/h}$.

(B) Let the ladder be represented by the hypotenuse of a right-angled triangle formed by the ladder,the wall,and the floor.
Let $x$ be the distance of the man from the wall along the ladder.
Let $y$ be the horizontal distance of the man from the wall.
The angle between the ladder and the wall is $\theta = 30^\circ$.
From trigonometry,the horizontal distance $y$ is given by $y = x \sin(\theta)$.
Here,$\theta = 30^\circ$,so $y = x \sin(30^\circ) = x \times \frac{1}{2}$.
Differentiating with respect to time $t$,we get $\frac{dy}{dt} = \frac{dx}{dt} \times \frac{1}{2}$.
Given that the rate of climbing the ladder is $\frac{dx}{dt} = 3 \text{ ft/sec}$.
Therefore,the rate at which he approaches the wall is $\frac{dy}{dt} = 3 \times \frac{1}{2} = \frac{3}{2} \text{ ft/sec}$.

(B) Given the equation of the curve is $y = x^2 + 2x$.
We are given that the $x$ and $y$ coordinates change at the same rate,which means $\frac{dx}{dt} = \frac{dy}{dt}$.
This implies $\frac{dy/dt}{dx/dt} = 1$.
Differentiating the equation $y = x^2 + 2x$ with respect to $t$,we get:
$\frac{dy}{dt} = (2x + 2) \frac{dx}{dt}$.
Dividing both sides by $\frac{dx}{dt}$,we get:
$\frac{dy/dt}{dx/dt} = 2x + 2$.
Since $\frac{dy/dt}{dx/dt} = 1$,we have:
$1 = 2x + 2$.
$2x = 1 - 2 = -1$.
$x = -\frac{1}{2}$.
Now,substitute $x = -\frac{1}{2}$ into the original equation to find $y$:
$y = (-\frac{1}{2})^2 + 2(-\frac{1}{2}) = \frac{1}{4} - 1 = -\frac{3}{4}$.
Thus,the required point is $\left( -\frac{1}{2}, -\frac{3}{4} \right)$.

(A) Let the man be at a distance $x$ from the pole at any time $t$,and let the height of his shadow on the wall be $y$. The total distance from the pole to the wall is $12 \ m$. When the man is $8 \ m$ from the wall,his distance from the pole is $x = 12 - 8 = 4 \ m$. Given $dx/dt = 1/2 \ m/s$.
Using similar triangles,the ratio of the height of the shadow to the height of the man relative to the pole is:
$\frac{y}{2} = \frac{12}{x}$
$y = \frac{24}{x}$
Differentiating with respect to $t$:
$\frac{dy}{dt} = -\frac{24}{x^2} \cdot \frac{dx}{dt}$
At $x = 4$ and $dx/dt = 1/2$:
$\frac{dy}{dt} = -\frac{24}{16} \cdot \frac{1}{2} = -\frac{3}{4} \ m/s$.
The negative sign indicates that the length of the shadow is decreasing. Thus,the rate at which the shadow is decreasing is $3/4 \ m/s$.

(D) Let $x$ be the length of the edge of the cube and $A$ be its surface area. The surface area of a cube is given by $A = 6x^2$.
Given that $\frac{dA}{dt} = 2 \ cm^2/sec$.
Differentiating $A$ with respect to $t$,we get $\frac{dA}{dt} = 12x \frac{dx}{dt}$.
Substituting the given values: $2 = 12 \times 90 \times \frac{dx}{dt}$.
Thus,$\frac{dx}{dt} = \frac{2}{12 \times 90} = \frac{1}{540} \ cm/sec$.
The volume of the cube is $V = x^3$.
Differentiating $V$ with respect to $t$,we get $\frac{dV}{dt} = 3x^2 \frac{dx}{dt}$.
Substituting $x = 90$ and $\frac{dx}{dt} = \frac{1}{540}$:
$\frac{dV}{dt} = 3 \times (90)^2 \times \frac{1}{540} = 3 \times 8100 \times \frac{1}{540} = \frac{24300}{540} = 45 \ cm^3/sec$.

(C) Let $x$ be the distance of the foot of the ladder from the wall and $y$ be the height of the top of the ladder on the wall.
By the Pythagorean theorem,we have $x^2 + y^2 = 10^2 = 100$.
Differentiating both sides with respect to time $t$,we get $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$.
Given that $x = 6 \ m$ and $\frac{dx}{dt} = 2 \ m/min$.
When $x = 6$,$y = \sqrt{100 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8 \ m$.
Substituting these values into the differentiated equation:
$2(6)(2) + 2(8) \frac{dy}{dt} = 0$
$24 + 16 \frac{dy}{dt} = 0$
$16 \frac{dy}{dt} = -24$
$\frac{dy}{dt} = -\frac{24}{16} = -\frac{3}{2} \ m/min$.
Thus,the height is decreasing at the rate of $3/2 \ m/min$.

(C) Let the radius of the sphere be $r$ and its surface area be $S$.
The surface area of a sphere is given by $S = 4\pi r^2$.
We are given that the rate of change of the radius is $\frac{dr}{dt} = 2 \text{ cm/s}$.
Differentiating $S$ with respect to time $t$,we get:
$\frac{dS}{dt} = \frac{d}{dt}(4\pi r^2) = 4\pi \times 2r \times \frac{dr}{dt}$.
Substituting the given value of $\frac{dr}{dt}$:
$\frac{dS}{dt} = 8\pi r \times 2 = 16\pi r$.
Since $16\pi$ is a constant,we have $\frac{dS}{dt} \propto r$.
Therefore,the rate of change of the surface area is proportional to $r$.

(C) The initial volume of the balloon is $V_i = 4500\pi \text{ m}^3$.
The rate of change of volume is $\frac{dV}{dt} = -72\pi \text{ m}^3/\text{min}$.
After $t = 49$ minutes,the volume $V$ is:
$V = 4500\pi - (72\pi \times 49) = 4500\pi - 3528\pi = 972\pi \text{ m}^3$.
The volume of a sphere is $V = \frac{4}{3}\pi r^3$.
At $t = 49$ minutes,$972\pi = \frac{4}{3}\pi r^3$,which implies $r^3 = \frac{972 \times 3}{4} = 729$.
Thus,$r = \sqrt[3]{729} = 9 \text{ m}$.
Differentiating $V = \frac{4}{3}\pi r^3$ with respect to $t$,we get $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$.
Substituting the values: $-72\pi = 4\pi (9)^2 \frac{dr}{dt}$.
$-72\pi = 4\pi (81) \frac{dr}{dt} = 324\pi \frac{dr}{dt}$.
$\frac{dr}{dt} = -\frac{72\pi}{324\pi} = -\frac{2}{9}$.
The rate at which the radius decreases is $\frac{2}{9} \text{ m/min}$.

(C) Given the rate of change of production: $\frac{dP}{dx} = 100 - 12\sqrt{x}$.
Integrating both sides with respect to $x$:
$\int dP = \int (100 - 12x^{1/2}) dx$
$P = 100x - 12 \times \frac{x^{3/2}}{3/2} + C$
$P = 100x - 8x^{3/2} + C$
Initially,when $x = 0$,the production $P = 2000$. Substituting these values:
$2000 = 100(0) - 8(0)^{3/2} + C \Rightarrow C = 2000$.
So,the production function is $P = 100x - 8x^{3/2} + 2000$.
For $x = 25$ additional workers,the new production level is:
$P = 100(25) - 8(25)^{3/2} + 2000$
$P = 2500 - 8(125) + 2000$
$P = 2500 - 1000 + 2000$
$P = 3500$.

(B) The height of the stone is given by the function $s(t) = 24t - 0.8t^2$.
To find the velocity $v(t)$,we differentiate the position function with respect to time $t$:
$v(t) = \frac{ds}{dt} = \frac{d}{dt}(24t - 0.8t^2) = 24 - 1.6t$.
To find the acceleration $a(t)$,we differentiate the velocity function with respect to time $t$:
$a(t) = \frac{dv}{dt} = \frac{d}{dt}(24 - 1.6t) = -1.6 \ m/s^2$.
The negative sign indicates that the acceleration is directed downwards (opposing the upward motion).
Therefore,the magnitude of the acceleration due to gravity on the surface of the moon is $1.6 \ m/s^2$.

(C) Given the distance $s = \sqrt{at^2 + bt + c}$,we have $s^2 = at^2 + bt + c$.
Differentiating with respect to $t$:
$2s \frac{ds}{dt} = 2at + b$
$\frac{ds}{dt} = \frac{2at + b}{2s}$
Differentiating again with respect to $t$ to find acceleration $f = \frac{d^2s}{dt^2}$:
$2 \left( \frac{ds}{dt} \right)^2 + 2s \frac{d^2s}{dt^2} = 2a$
$s \frac{d^2s}{dt^2} = a - \left( \frac{ds}{dt} \right)^2$
$s \frac{d^2s}{dt^2} = a - \frac{(2at + b)^2}{4s^2} = \frac{4as^2 - (2at + b)^2}{4s^2}$
Substituting $s^2 = at^2 + bt + c$:
$s \frac{d^2s}{dt^2} = \frac{4a(at^2 + bt + c) - (4a^2t^2 + 4abt + b^2)}{4s^2}$
$s \frac{d^2s}{dt^2} = \frac{4a^2t^2 + 4abt + 4ac - 4a^2t^2 - 4abt - b^2}{4s^2} = \frac{4ac - b^2}{4s^2}$
$\frac{d^2s}{dt^2} = \frac{4ac - b^2}{4s^3}$
Since $4ac - b^2$ is a constant,the acceleration $f \propto s^{-3}$.

(D) Let $r$ be the radius of the circular ripple and $A$ be its area.
Given that the rate of change of the radius is $\frac{dr}{dt} = 6 \text{ cm/sec}$.
The area of a circle is given by $A = \pi r^2$.
Differentiating both sides with respect to time $t$,we get:
$\frac{dA}{dt} = 2 \pi r \frac{dr}{dt}$.
Substitute the given values $r = 10 \text{ cm}$ and $\frac{dr}{dt} = 6 \text{ cm/sec}$:
$\frac{dA}{dt} = 2 \pi \times 10 \times 6 = 120 \pi \text{ cm}^2/\text{sec}$.
Thus,the rate of increase in the area is $120 \pi \text{ cm}^2/\text{sec}$.

(A) The focus $S$ of the parabola $y^2 = 4x$ is $(1, 0)$. Let $P$ be a point on the parabola,so $P = (T^2, 2T)$.
The vector $\vec{SP} = (T^2 - 1)\hat{i} + 2T\hat{j}$.
The line is $x + y - 1 = 0$. The normal vector to the line is $\vec{n} = \hat{i} + \hat{j}$. The unit normal vector is $\hat{u} = \frac{\hat{i} + \hat{j}}{\sqrt{2}}$.
The projection of $\vec{SP}$ on the line $x + y = 1$ is the component of $\vec{SP}$ along the direction parallel to the line. The direction vector of the line is $\vec{v} = \hat{i} - \hat{j}$.
The projection $L$ is given by $\frac{\vec{SP} \cdot \vec{v}}{|\vec{v}|} = \frac{(T^2 - 1) - 2T}{\sqrt{2}} = \frac{T^2 - 2T - 1}{\sqrt{2}}$.
Differentiating with respect to $t$: $\frac{dL}{dt} = \frac{1}{\sqrt{2}} (2T - 2) \frac{dT}{dt}$.
Given $\frac{dx}{dt} = 4$. Since $x = T^2$,$\frac{dx}{dt} = 2T \frac{dT}{dt} = 4$. At $P(4, 4)$,$T = 2$,so $4 \frac{dT}{dt} = 4 \implies \frac{dT}{dt} = 1$.
Substituting $\frac{dT}{dt} = 1$ and $T = 2$ into the derivative of $L$:
$\frac{dL}{dt} = \frac{1}{\sqrt{2}} (2(2) - 2)(1) = \frac{2}{\sqrt{2}} = \sqrt{2}$.

(D) The volume of a right circular cylinder is given by $V = \pi r^2h$.
Differentiating both sides with respect to time $t$,we get:
$\frac{dV}{dt} = \pi \left( r^2 \frac{dh}{dt} + 2rh \frac{dr}{dt} \right)$.
Given that $\frac{dr}{dt} = 0.1 = \frac{1}{10} \text{ cm/min}$ and $\frac{dh}{dt} = -0.2 = -\frac{2}{10} \text{ cm/min}$.
Substituting these values into the derivative expression:
$\frac{dV}{dt} = \pi \left( r^2 \left( -\frac{2}{10} \right) + 2rh \left( \frac{1}{10} \right) \right)$.
$\frac{dV}{dt} = \frac{\pi r}{10} (-2r + 2h) = \frac{2\pi r}{10} (h - r) = \frac{\pi r}{5} (h - r)$.
At $r = 2 \text{ cm}$ and $h = 3 \text{ cm}$:
$\frac{dV}{dt} = \frac{\pi (2)}{5} (3 - 2) = \frac{2\pi}{5} \text{ cm}^3/\text{min}$.

(D) The volume of coffee in the cylindrical pot is given by $V = \pi r^2 h$,where $r$ is the radius and $h$ is the height of the coffee level.
Given the diameter of the pot is $15 \, cm$,the radius $r = \frac{15}{2} \, cm$.
Since the pot is cylindrical,the radius $r$ remains constant as the level rises,so $\frac{dr}{dt} = 0$.
Differentiating the volume with respect to time $t$,we get:
$\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}$.
We are given $\frac{dV}{dt} = 100 \, cm^3/min$.
Substituting the values:
$100 = \pi \left( \frac{15}{2} \right)^2 \frac{dh}{dt}$
$100 = \pi \left( \frac{225}{4} \right) \frac{dh}{dt}$
$\frac{dh}{dt} = \frac{100 \times 4}{225 \pi} = \frac{400}{225 \pi} = \frac{16}{9 \pi} \, cm/min$.
Thus,the rate at which the level in the pot is rising is $\frac{16}{9 \pi} \, cm/min$.

(B) Let $r$ be the radius of the circle and $v = 20 \, km/hr$ be the speed of the horse.
Let $\theta$ be the angle subtended by the horse at the centre of the circle at any time $t$.
The position of the shadow on the fence is given by $x = r \tan \theta$.
Differentiating with respect to $t$,we get $\frac{dx}{dt} = r \sec^2 \theta \frac{d\theta}{dt}$.
Since the speed of the horse is $v = r \frac{d\theta}{dt}$,we have $\frac{d\theta}{dt} = \frac{v}{r}$.
Substituting this into the expression for $\frac{dx}{dt}$,we get $\frac{dx}{dt} = r \sec^2 \theta \left(\frac{v}{r}\right) = v \sec^2 \theta$.
The horse covers $1/8$ of the circle,which corresponds to an angle $\theta = \frac{1}{8} \times 360^\circ = 45^\circ$.
At $\theta = 45^\circ$,$\sec^2(45^\circ) = (\sqrt{2})^2 = 2$.
Therefore,the speed of the shadow is $\frac{dx}{dt} = 20 \times 2 = 40 \, km/hr$.

(D) The area $A$ of an isosceles triangle with equal sides $x$ and included angle $\theta$ is given by $A = \frac{1}{2} x^2 \sin\theta$.
To find the rate of change of area with respect to time $t$,we differentiate with respect to $t$:
$\frac{dA}{dt} = \frac{1}{2} \left( 2x \frac{dx}{dt} \sin\theta + x^2 \cos\theta \frac{d\theta}{dt} \right) = x \frac{dx}{dt} \sin\theta + \frac{1}{2} x^2 \cos\theta \frac{d\theta}{dt}$.
Given values: $x = 12 \ m$,$\theta = \pi/4$,$\frac{dx}{dt} = 1/12 \ m/hr$,and $\frac{d\theta}{dt} = \pi/180 \ \text{rad/hr}$.
Substituting these values:
$\frac{dA}{dt} = (12) \left( \frac{1}{12} \right) \sin(\pi/4) + \frac{1}{2} (12)^2 \cos(\pi/4) \left( \frac{\pi}{180} \right)$
$\frac{dA}{dt} = (1) \left( \frac{1}{\sqrt{2}} \right) + \frac{1}{2} (144) \left( \frac{1}{\sqrt{2}} \right) \left( \frac{\pi}{180} \right)$
$\frac{dA}{dt} = \frac{1}{\sqrt{2}} + \frac{72}{\sqrt{2}} \left( \frac{\pi}{180} \right) = \frac{1}{\sqrt{2}} + \frac{2\pi}{5\sqrt{2}} = \frac{1}{\sqrt{2}} \left( 1 + \frac{2\pi}{5} \right) = \frac{\sqrt{2}}{2} \left( 1 + \frac{2\pi}{5} \right)$.
Wait,checking the options,the expression is $2^{1/2} (1/2 + \pi/5) = \sqrt{2} (1/2 + \pi/5) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}\pi}{5} = \frac{1}{\sqrt{2}} + \frac{2\pi}{5\sqrt{2}}$. This matches option $D$.