Rate of Change of Quantities Questions in English

(C) Let the height of the balloon when the stone is dropped be $h$ and its upward velocity be $v$.
When the stone is dropped, its initial velocity is $u = v$ (upwards).
Using the equation of motion for the stone: $s = ut + \frac{1}{2}at^2$.
Here, $s = -h$ (displacement is downwards), $u = v$, $a = -g = -32 \ ft/sec^2$, and $t = 4 \ sec$.
$-h = v(4) + \frac{1}{2}(-32)(4)^2$.
$-h = 4v - 16(16)$.
$-h = 4v - 256$.
$h = 256 - 4v$.
In $4 \ sec$, the balloon rises further by a distance $d = v \times t = v \times 4 = 4v$.
The total height of the balloon from the ground when the stone hits the ground is $H = h + d$.
$H = (256 - 4v) + 4v = 256 \ ft$.

(C) Let the final velocity be $v$. For particle $B$, $v = f't$, so $t = \frac{v}{f'}$. The distance covered is $s = \frac{1}{2}f't^2 = \frac{1}{2}f'\left(\frac{v}{f'}\right)^2 = \frac{v^2}{2f'}$.
For particle $A$, $v = f(t+m)$, so $t+m = \frac{v}{f}$, which means $t = \frac{v}{f} - m$. The distance covered is $s+n = \frac{1}{2}f(t+m)^2 = \frac{1}{2}f\left(\frac{v}{f}\right)^2 = \frac{v^2}{2f}$.
From the velocity equations: $f't = f(t+m) \implies t(f'-f) = fm \implies t = \frac{fm}{f'-f}$.
Substituting $t$ into the velocity equation $v = f't$: $v = \frac{f'fm}{f'-f}$.
Now, $n = (s+n) - s = \frac{v^2}{2f} - \frac{v^2}{2f'} = \frac{v^2}{2} \left(\frac{f'-f}{ff'}\right)$.
Substituting $v^2 = \left(\frac{ff'm}{f'-f}\right)^2$:
$n = \frac{1}{2} \left(\frac{ff'm}{f'-f}\right)^2 \left(\frac{f'-f}{ff'}\right) = \frac{1}{2} \frac{(ff')^2 m^2}{(f'-f)^2} \cdot \frac{f'-f}{ff'} = \frac{ff'm^2}{2(f'-f)}$.
Rearranging gives: $(f'-f)n = \frac{1}{2}ff'm^2$.

(B) Let $r = 10 \ m$ be the radius of the circular track.
The speed of the man along the track is $v = r \frac{d\theta}{dt} = 10 \ m/sec$.
Since $r = 10 \ m$, we have $10 \frac{d\theta}{dt} = 10$, which implies $\frac{d\theta}{dt} = 1 \ rad/sec$.
Let $y$ be the position of the shadow on the wall at an angular distance $\theta$ from $P$.
From the geometry, we have $\tan \theta = \frac{y}{r}$, so $y = r \tan \theta$.
Differentiating with respect to time $t$, we get $\frac{dy}{dt} = r \sec^2 \theta \cdot \frac{d\theta}{dt}$.
At $\theta = 60^{\circ}$, we have $\sec(60^{\circ}) = 2$, so $\sec^2(60^{\circ}) = 4$.
Substituting the values, $\frac{dy}{dt} = 10 \times 4 \times 1 = 40 \ m/sec$.
Thus, the speed of the shadow is $40 \ m/sec$.

(C) Given the equation of the hyperbola is $y = \frac{10}{x}$.
The rate of change of the abscissa is given as $\frac{dx}{dt} = 1 \text{ unit/s}$.
To find the rate of change of the ordinate $\frac{dy}{dt}$, we differentiate $y$ with respect to $t$ using the chain rule:
$\frac{dy}{dt} = \frac{d}{dt} \left( \frac{10}{x} \right) = -\frac{10}{x^2} \cdot \frac{dx}{dt}$.
Substitute the given values $x = 5$ and $\frac{dx}{dt} = 1$ into the derivative:
$\frac{dy}{dt} = -\frac{10}{(5)^2} \cdot (1) = -\frac{10}{25} = -\frac{2}{5} \text{ unit/s}$.
Since the result is negative, the ordinate $y$ decreases at the rate of $\frac{2}{5} \text{ unit/s}$.

(A) Let $x$ be the distance of the lower end of the ladder from the wall and $y$ be the height of the top end of the ladder from the floor.
Given the length of the ladder is $20 \ ft$, by the Pythagorean theorem, we have:
$x^2 + y^2 = 20^2 = 400$
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 that the top end slides downwards at a rate of $2 \ ft/sec$, we have $\frac{dy}{dt} = -2 \ ft/sec$.
When $x = 12 \ ft$, we find $y$ using the equation $x^2 + y^2 = 400$:
$12^2 + y^2 = 400$
$144 + y^2 = 400$
$y^2 = 256 \Rightarrow y = 16 \ ft$
Now, substitute $x = 12$, $y = 16$, and $\frac{dy}{dt} = -2$ into the differentiated equation:
$12 \left(\frac{dx}{dt}\right) + 16(-2) = 0$
$12 \left(\frac{dx}{dt}\right) - 32 = 0$
$12 \left(\frac{dx}{dt}\right) = 32$
$\frac{dx}{dt} = \frac{32}{12} = \frac{8}{3} \ ft/sec$
Thus, the lower end moves away from the wall at a rate of $\frac{8}{3} \ ft/sec$.

(D) Given the curve $12 y = x^{3}$.
Let the rate of change of the ordinate be $\frac{dy}{dt}$ and the rate of change of the abscissa be $\frac{dx}{dt}$.
We are given that the rate of change of the ordinate exceeds that of the abscissa, so $\frac{dy}{dt} > \frac{dx}{dt}$.
Differentiating the equation of the curve with respect to $t$, we get $12 \frac{dy}{dt} = 3x^{2} \frac{dx}{dt}$, which implies $\frac{dy}{dt} = \frac{x^{2}}{4} \frac{dx}{dt}$.
Substituting this into the inequality $\frac{dy}{dt} > \frac{dx}{dt}$, we have $\frac{x^{2}}{4} \frac{dx}{dt} > \frac{dx}{dt}$.
Assuming $\frac{dx}{dt} > 0$, we get $\frac{x^{2}}{4} > 1$, which means $x^{2} > 4$.
This inequality $x^{2} - 4 > 0$ factors as $(x - 2)(x + 2) > 0$.
The solution to this inequality is $x \in (-\infty, -2) \cup (2, \infty)$.
Since the question asks for the condition where the rate of change of the ordinate exceeds the abscissa, and considering the provided options, the condition $x > 2$ is a valid subset of the solution set.

(A) Given: Initial velocity $u = 0$, acceleration $a = 8 \text{ m/s}^2$.
Using the equation of motion $S = ut + \frac{1}{2}at^2$, since $u = 0$, we have $S = \frac{1}{2}at^2$.
$1$. Time taken to cover the first metre $(S_1 = 1 \text{ m})$:
$1 = \frac{1}{2} \times 8 \times t_1^2 \implies 1 = 4t_1^2 \implies t_1^2 = \frac{1}{4} \implies t_1 = \frac{1}{2} \text{ s}$.
$2$. Time taken to cover the first two metres $(S_2 = 2 \text{ m})$:
$2 = \frac{1}{2} \times 8 \times t_2^2 \implies 2 = 4t_2^2 \implies t_2^2 = \frac{1}{2} \implies t_2 = \frac{1}{\sqrt{2}} \text{ s}$.
$3$. Time taken to move the second metre is the difference between the time taken to cover $2 \text{ m}$ and $1 \text{ m}$:
$\Delta t = t_2 - t_1 = \frac{1}{\sqrt{2}} - \frac{1}{2} = \frac{\sqrt{2}-1}{2} \text{ s}$.

(B) We know that velocity $v = \frac{dx}{dt}$, so $\frac{dt}{dx} = \frac{1}{v}$.
Now, differentiate with respect to $x$:
$\frac{d^2 t}{dx^2} = \frac{d}{dx} \left( \frac{1}{v} \right) = -\frac{1}{v^2} \frac{dv}{dx}$.
Using the chain rule, $\frac{dv}{dx} = \frac{dv}{dt} \times \frac{dt}{dx} = f \times \frac{1}{v} = \frac{f}{v}$.
Substituting this back into the equation:
$\frac{d^2 t}{dx^2} = -\frac{1}{v^2} \times \frac{f}{v} = -\frac{f}{v^3}$.
Rearranging for $f$, we get $f = -v^3 \frac{d^2 t}{dx^2}$.

(B) Given the displacement equation: $x = At^2 + Bt + C$.
Velocity $v$ is the rate of change of displacement with respect to time $t$: $v = \frac{dx}{dt} = 2At + B$.
Now, calculate $v^2$: $v^2 = (2At + B)^2 = 4A^2t^2 + 4ABt + B^2$.
Next, calculate $4Ax$: $4Ax = 4A(At^2 + Bt + C) = 4A^2t^2 + 4ABt + 4AC$.
Now, subtract $v^2$ from $4Ax$: $4Ax - v^2 = (4A^2t^2 + 4ABt + 4AC) - (4A^2t^2 + 4ABt + B^2)$.
Simplifying the expression: $4Ax - v^2 = 4AC - B^2$.

(A) The distance $x$ covered by the particle is given by the equation: $x = 3 + 8t - 4t^2$.
Velocity $v$ is defined as the rate of change of displacement with respect to time, which is given by the derivative of $x$ with respect to $t$:
$v = \frac{dx}{dt} = \frac{d}{dt}(3 + 8t - 4t^2)$.
Applying the power rule for differentiation:
$v = 0 + 8(1) - 4(2t) = 8 - 8t$.
To find the velocity after $1$ second, substitute $t = 1$ into the velocity equation:
$v = 8 - 8(1) = 8 - 8 = 0$ unit/second.

(B) Let $r$ be the radius and $A$ be the area of the circle.
Given that the rate of change of the radius is $\frac{dr}{dt} = 5 \text{ cm/sec}$.
The area of a circle is given by $A = \pi r^2$.
Differentiating both sides with respect to time $t$, we get:
$\frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = 2 \pi r \frac{dr}{dt}$.
Substituting the given values $r = 20 \text{ cm}$ and $\frac{dr}{dt} = 5 \text{ cm/sec}$:
$\frac{dA}{dt} = 2 \pi (20)(5) = 200 \pi \text{ cm}^2/\text{sec}$.
Thus, the rate of increase of the area is $200 \pi \text{ cm}^2/\text{sec}$.

(B) Given the displacement function: $x = t^3 - 6t^2 + t$.
Velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt}(t^3 - 6t^2 + t) = 3t^2 - 12t + 1$.
Acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 12t + 1) = 6t - 12$.
For acceleration to be zero: $a = 0$.
$6t - 12 = 0$.
$6t = 12$.
$t = 2$ unit time.

(A) $1$. Initial motion of the balloon: $u = 0$, $a = 4 \ ft/sec^2$, $t = 5 \ sec$.
Height of the balloon at $t = 5 \ sec$: $h_0 = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times 4 \times 5^2 = 50 \ ft$.
Velocity of the balloon at $t = 5 \ sec$: $v_0 = u + at = 0 + 4 \times 5 = 20 \ ft/sec$.
$2$. Motion of the stone after being dropped: The stone has an initial upward velocity $v_0 = 20 \ ft/sec$ and acceleration $g = -32 \ ft/sec^2$.
Using $s = ut + \frac{1}{2}at^2$ for the stone to reach the ground $(s = -50 \ ft)$:
$-50 = 20T + \frac{1}{2}(-32)T^2$
$-50 = 20T - 16T^2$ $\Rightarrow 16T^2 - 20T - 50 = 0$ $\Rightarrow 8T^2 - 10T - 25 = 0$.
Solving for $T$: $T = \frac{10 \pm \sqrt{100 - 4(8)(-25)}}{2(8)} = \frac{10 \pm \sqrt{900}}{16} = \frac{10 \pm 30}{16}$.
Since $T > 0$, $T = \frac{40}{16} = 2.5 \ sec = 5/2 \ sec$.
$3$. Height of the balloon when the stone reaches the ground: The balloon continues to ascend with $a = 4 \ ft/sec^2$ for an additional $T = 2.5 \ sec$.
$H = h_0 + v_0 T + \frac{1}{2}aT^2 = 50 + 20(2.5) + \frac{1}{2}(4)(2.5)^2 = 50 + 50 + 12.5 = 112.5 \ ft$.

(A) The marginal cost function $MC$ is the derivative of the cost function $C(x)$ with respect to $x$.
$MC = \frac{dC}{dx} = \frac{d}{dx}(0.05x^3 - 0.2x^2 + 3x + 500)$.
Applying the power rule, we get $MC = 0.15x^2 - 0.4x + 3$.
To find the marginal cost at $x = 3$, we substitute $x = 3$ into the $MC$ function:
$MC(3) = 0.15(3)^2 - 0.4(3) + 3$.
$MC(3) = 0.15(9) - 1.2 + 3$.
$MC(3) = 1.35 - 1.2 + 3$.
$MC(3) = 0.15 + 3 = 3.15$.
Thus, the marginal cost at $x = 3$ is $3.15$ Rupees.

(D) Marginal Revenue $(MR)$ is defined as the derivative of the total revenue function $R(x)$ with respect to $x$.
$MR = \frac{dR}{dx} = \frac{d}{dx}(3x^2 + 36x + 5)$
Applying the power rule of differentiation:
$MR = 6x + 36$
Now, substitute $x = 15$ into the expression for $MR$:
$MR = 6(15) + 36$
$MR = 90 + 36 = 126$
Therefore, the marginal revenue when $x = 15$ is $126$.

(B) Given the parabola equation: $y^2 = \frac{36}{5}x$.
Differentiating both sides with respect to time $t$: $2y \frac{dy}{dt} = \frac{36}{5} \frac{dx}{dt}$.
We are given that the ordinate $(y)$ increases at thrice the rate of the abscissa $(x)$, so $\frac{dy}{dt} = 3 \frac{dx}{dt}$.
Substituting this into the differentiated equation: $2y(3 \frac{dx}{dt}) = \frac{36}{5} \frac{dx}{dt}$.
Assuming $\frac{dx}{dt} \neq 0$, we get $6y = \frac{36}{5}$, which simplifies to $y = \frac{6}{5}$.
Now, substitute $y = \frac{6}{5}$ into the original parabola equation: $(\frac{6}{5})^2 = \frac{36}{5}x$.
$\frac{36}{25} = \frac{36}{5}x$.
Solving for $x$: $x = \frac{36}{25} \times \frac{5}{36} = \frac{1}{5}$.
Thus, the required point is $(1/5, 6/5)$.

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

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

(D) The area $s$ of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by $s = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Given vertices $A(a, 0)$, $B(0, b)$, and $C(x, x)$, we have:
$s = \frac{1}{2} |a(b - x) + 0(x - 0) + x(0 - b)|$
$s = \frac{1}{2} |ab - ax - bx| = \frac{1}{2} |ab - x(a + b)|$.
Since $s$ is the absolute value, $s = \pm \frac{1}{2} (ab - x(a + b))$.
Differentiating with respect to $x$:
$\frac{ds}{dx} = \pm \frac{1}{2} (-(a + b)) = \mp \frac{a + b}{2}$.
Given the options, the magnitude is $\frac{a + b}{2}$.

(B) The volume of a sphere is $V = \frac{4}{3}\pi r^3$.
Differentiating with respect to time $t$, we get $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$.
Given $\frac{dV}{dt} = -8 \text{ cm}^3/\text{s}$ (since volume is decreasing) and $r = 2 \text{ cm}$.
Substituting the values: $-8 = 4\pi(2)^2 \frac{dr}{dt}$.
$-8 = 16\pi \frac{dr}{dt}$.
$\frac{dr}{dt} = -\frac{8}{16\pi} = -\frac{1}{2\pi} \text{ cm/s}$.
The negative sign indicates that the radius is decreasing at the rate of $\frac{1}{2\pi} \text{ cm/s}$.

(A) Let the side of the equilateral triangle be $x$.
The area $A$ of an equilateral triangle is given by $A = \frac{\sqrt{3}}{4} x^2$.
Differentiating both sides with respect to time $t$, we get $\frac{dA}{dt} = \frac{\sqrt{3}}{4} \cdot 2x \cdot \frac{dx}{dt} = \frac{\sqrt{3}}{2} x \frac{dx}{dt}$.
Given $\frac{dx}{dt} = \sqrt{3} \text{ cm/sec}$ and $x = 12 \text{ cm}$.
Substituting these values, $\frac{dA}{dt} = \frac{\sqrt{3}}{2} (12) (\sqrt{3}) = 6 \times 3 = 18 \text{ cm}^2/\text{sec}$.

(D) Given that $s \propto v^2$, we can write $s = kv^2$, where $k$ is a constant.
Rearranging for velocity, $v^2 = s/k$, which implies $v = \frac{1}{\sqrt{k}} s^{1/2}$.
Acceleration $a$ is given by $a = v \frac{dv}{ds}$.
Differentiating $v = \frac{1}{\sqrt{k}} s^{1/2}$ with respect to $s$, we get $\frac{dv}{ds} = \frac{1}{\sqrt{k}} \cdot \frac{1}{2} s^{-1/2}$.
Substituting these into the acceleration formula: $a = (\frac{1}{\sqrt{k}} s^{1/2}) \cdot (\frac{1}{2\sqrt{k}} s^{-1/2})$.
Simplifying, $a = \frac{1}{2k} \cdot s^{1/2 - 1/2} = \frac{1}{2k}$.
Since $k$ is a constant, $a$ is a constant.

(A) The particle hits the ground when the height $s(t) = 0$.
$128t - 16t^2 = 0 \implies 16t(8 - t) = 0$.
Since $t > 0$, the particle hits the ground at $t = 8 \text{ s}$.
The velocity $v(t)$ is the derivative of the position function $s(t)$ with respect to time $t$: $v(t) = s'(t) = \frac{d}{dt}(128t - 16t^2) = 128 - 32t$.
Substituting $t = 8$ into the velocity equation: $v(8) = 128 - 32(8) = 128 - 256 = -128 \text{ ft/sec}$.

(D) Let $r$ be the radius of the spherical balloon.
Surface area $S = 4\pi r^2$, so $\frac{dS}{dt} = 8\pi r \frac{dr}{dt} = 5$.
Volume $V = \frac{4}{3}\pi r^3$, so $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} = 10$.
Dividing the volume rate equation by the surface area rate equation:
$\frac{4\pi r^2 \frac{dr}{dt}}{8\pi r \frac{dr}{dt}} = \frac{10}{5}$.
$\frac{r}{2} = 2$.
$r = 4 \text{ cm}$.

(A) Let $x$ be the thickness of the ice layer. The total radius of the ball including the ice is $R = 10 + x$.
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 $\frac{dV}{dt} = -50 \text{ cm}^3/\text{min}$ (since it melts) and $x = 5 \text{ cm}$.
Substituting these values: $-50 = 4\pi(10+5)^2 \frac{dx}{dt}$.
$-50 = 4\pi(15)^2 \frac{dx}{dt} \implies -50 = 4\pi(225) \frac{dx}{dt}$.
$-50 = 900\pi \frac{dx}{dt} \implies \frac{dx}{dt} = -\frac{50}{900\pi} = -\frac{1}{18\pi} \text{ cm/min}$.
Thus, the rate at which the thickness decreases is $\frac{1}{18\pi} \text{ cm/min}$.

(A) Let $x$ be the horizontal distance and $z$ be the straight-line distance from the observer to the aeroplane.
Given altitude $h = 1 \text{ km}$. By Pythagoras theorem, $z^2 = x^2 + h^2$, so $z^2 = x^2 + 1^2$.
Differentiating with respect to time $t$: $2z \frac{dz}{dt} = 2x \frac{dx}{dt} \implies z \frac{dz}{dt} = x \frac{dx}{dt}$.
Given $z = 1250 \text{ m} = 1.25 \text{ km}$ and $\frac{dx}{dt} = 600 \text{ km/hr}$.
Calculate $x$ when $z = 1.25 \text{ km}$: $x = \sqrt{z^2 - 1^2} = \sqrt{1.25^2 - 1^2} = \sqrt{1.5625 - 1} = \sqrt{0.5625} = 0.75 \text{ km}$.
Substitute values into the derivative equation: $1.25 \frac{dz}{dt} = 0.75 \times 600$.
$\frac{dz}{dt} = \frac{450}{1.25} = 360 \text{ km/hr}$.

(C) Given the equation of the path: $3y = x^3 - 3$.
Differentiating both sides with respect to time $t$: $3 \frac{dy}{dt} = 3x^2 \frac{dx}{dt}$.
Given the condition: $\frac{dy}{dt} = 9 \frac{dx}{dt}$.
Substituting the condition into the differentiated equation: $3(9 \frac{dx}{dt}) = 3x^2 \frac{dx}{dt}$.
This simplifies to $27 \frac{dx}{dt} = 3x^2 \frac{dx}{dt}$, which implies $x^2 = 9$, so $x = \pm 3$.
If $x = 3$, then $3y = (3)^3 - 3 = 27 - 3 = 24$, so $y = 8$. The point is $(3, 8)$.
If $x = -3$, then $3y = (-3)^3 - 3 = -27 - 3 = -30$, so $y = -10$. The point is $(-3, -10)$.
Comparing with the given options, $(3, 8)$ is the correct position.