5.15 Related RatesHL ONLY

Solution:
Let $x$ be the side of the square and $A$ be the area. We are given $\dfrac{dx}{dt} = 2\text{ m/sec}$.
The relation between $A$ and $x$ is $ A=x^2 $. Differentiating with respect to time $t$ gives $$ \dfrac{dA}{dt} = \dfrac{d}{dt}(x^2) = 2x \dfrac{dx}{dt} $$ Therefore, when $x=10\text{ m}$, $$ \dfrac{dA}{dt} = 2 \times 10 \times 2 = \mathbf{40\text{ m}^2/\text{sec}} $$

Solution:
We are given $\dfrac{dV}{dt} = 5\text{ cm}^3/\text{sec}$, and we need to find $\dfrac{dr}{dt}$.
The relation between $V$ and $r$ is given by the volume of a sphere: $$ V = \dfrac{4}{3} \pi r^3 $$ Differentiating with respect to time $t$: $$ \dfrac{dV}{dt} = 4\pi r^2 \dfrac{dr}{dt} $$ (i) When $r = 3$: $$ \begin{aligned} 5 &= 4\pi (3^2) \dfrac{dr}{dt} \\ 5 &= 36\pi \dfrac{dr}{dt} \\ \dfrac{dr}{dt} &= \mathbf{\dfrac{5}{36\pi}\text{ cm/sec}} \end{aligned} $$ (ii) When $V=36\pi$, we must first find $r$. Using the original relation: $$ \begin{aligned} 36\pi &= \dfrac{4}{3} \pi r^3 \\ 108\pi &= 4\pi r^3 \\ 27 &= r^3 \implies r = 3\text{ cm} \end{aligned} $$ Since $r = 3\text{ cm}$, the answer is identically the same as above: $\mathbf{\dfrac{dr}{dt} = \dfrac{5}{36\pi}\text{ cm/sec}}$.

Solution:
Let $x$ be the distance $OA$ and $y$ be the distance $OB$. The car $A$ moves away from $O$, so $\dfrac{dx}{dt} = 50\text{ km/h}$. Car $B$ moves towards $O$, so $\dfrac{dy}{dt} = -70\text{ km/h}$.
At the given instant, $x = 5$ and $y = 5$.

(a) Let $s$ be the distance between the cars. By Pythagoras: $$ s^2 = x^2 + y^2 $$ Differentiating with respect to $t$: $$ 2s \dfrac{ds}{dt} = 2x \dfrac{dx}{dt} + 2y \dfrac{dy}{dt} \implies s \dfrac{ds}{dt} = x \dfrac{dx}{dt} + y \dfrac{dy}{dt} $$ At this instant, $s = \sqrt{5^2 + 5^2} = \sqrt{50} = 5\sqrt{2}$.
$$ \begin{aligned} (5\sqrt{2}) \dfrac{ds}{dt} &= (5)(50) + (5)(-70) \\ 5\sqrt{2} \dfrac{ds}{dt} &= 250 - 350 \\ 5\sqrt{2} \dfrac{ds}{dt} &= -100 \\ \dfrac{ds}{dt} &= -\dfrac{100}{5\sqrt{2}} = -\dfrac{20}{\sqrt{2}} = \mathbf{-10\sqrt{2}\text{ km/h}} \approx -14.1\text{ km/h} \end{aligned} $$ (b) The relation between $x$, $y$, and $\theta$ is: $$ \tan\theta = \dfrac{y}{x} \implies y = x \tan\theta $$ Differentiating both sides with respect to $t$ (using the product rule on the right): $$ \dfrac{dy}{dt} = \dfrac{dx}{dt}\tan\theta + x \sec^2\theta \dfrac{d\theta}{dt} $$ At the instant $x=5$ and $y=5$, we have $\tan\theta = \dfrac{5}{5} = 1 \implies \theta = \dfrac{\pi}{4}$ radians.
Substituting the known values: $$ \begin{aligned} -70 &= 50 \tan\left(\dfrac{\pi}{4}\right) + 5 \sec^2\left(\dfrac{\pi}{4}\right) \dfrac{d\theta}{dt} \\ -70 &= 50(1) + 5(\sqrt{2})^2 \dfrac{d\theta}{dt} \\ -70 &= 50 + 10 \dfrac{d\theta}{dt} \\ -120 &= 10 \dfrac{d\theta}{dt} \\ \dfrac{d\theta}{dt} &= \mathbf{-12\text{ rad/h}} = \mathbf{-0.2\text{ rad/min}} \end{aligned} $$

Solution:
(a) Since $h=2r$, the original relation becomes: $$ A = \dfrac{1}{3}r^2(2r) + 2r^3 = \dfrac{2}{3}r^3 + 2r^3 = \dfrac{8}{3}r^3 $$ Differentiating with respect to $t$: $$ \dfrac{dA}{dt} = \dfrac{8}{3}(3r^2) \dfrac{dr}{dt} = 8r^2 \dfrac{dr}{dt} $$ Therefore, when $r=3$ and $\dfrac{dA}{dt} = 30$: $$ \begin{aligned} 30 &= 8(3^2) \dfrac{dr}{dt} \\ 30 &= 72 \dfrac{dr}{dt} \\ \dfrac{dr}{dt} &= \dfrac{30}{72} = \dfrac{5}{12} \end{aligned} $$ Since $h=2r$, differentiating gives $\dfrac{dh}{dt} = 2\dfrac{dr}{dt}$. Thus: $$ \dfrac{dh}{dt} = 2\left(\dfrac{5}{12}\right) = \mathbf{\dfrac{5}{6}} $$ (b) Here $h$ is NOT always double $r$. We differentiate the original formula $A = \dfrac{1}{3} r^2 h + 2r^3$ implicitly using the product rule for the first term: $$ \dfrac{dA}{dt} = \dfrac{1}{3}\left(2r \dfrac{dr}{dt} h + r^2 \dfrac{dh}{dt}\right) + 6r^2 \dfrac{dr}{dt} $$ We substitute the known values at the instant $r=3$, $h=6$, $\dfrac{dA}{dt} = 30$, and $\dfrac{dr}{dt} = \dfrac{5}{12}$: $$ \begin{aligned} 30 &= \dfrac{1}{3}\left[ 2(3)\left(\dfrac{5}{12}\right)(6) + (3^2) \dfrac{dh}{dt} \right] + 6(3^2)\left(\dfrac{5}{12}\right) \\ 30 &= \dfrac{1}{3}\left[ 36\left(\dfrac{5}{12}\right) + 9 \dfrac{dh}{dt} \right] + 54\left(\dfrac{5}{12}\right) \\ 30 &= \dfrac{1}{3}\left[ 15 + 9 \dfrac{dh}{dt} \right] + \dfrac{45}{2} \\ 30 &= 5 + 3 \dfrac{dh}{dt} + 22.5 \\ 30 &= 27.5 + 3 \dfrac{dh}{dt} \\ 2.5 &= 3 \dfrac{dh}{dt} \\ \dfrac{dh}{dt} &= \dfrac{2.5}{3} = \mathbf{\dfrac{5}{6}} \end{aligned} $$