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 $126$.