Related Rates

$x$ = distance from the wall to the foot, $y$ = height of the top. The ladder stays $10$ m.

Given $\dfrac{dx}{dt} = 0.4$ m/s (foot moves away, $x$ grows). Wanted $\dfrac{dy}{dt}$ (expect it negative).

$x^2 + y^2 = 10^2$.

$2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt} = 0$.

At $x = 6$: $y = \sqrt{100 - 36} = 8$. Then $2(6)(0.4) + 2(8)\dfrac{dy}{dt} = 0 \Rightarrow \dfrac{dy}{dt} = -\dfrac{4.8}{16} = -0.3$ m/s.

$y$ = A's distance north, $x$ = B's distance east, $z$ = distance between the cars. Right angle at the intersection.

$\dfrac{dy}{dt} = +50$ km/h (moving away), $\dfrac{dx}{dt} = -60$ km/h (moving toward). Wanted $\dfrac{dz}{dt}$.

$x^2 + y^2 = z^2$.

$2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt} = 2z\dfrac{dz}{dt}$, so $x\dfrac{dx}{dt} + y\dfrac{dy}{dt} = z\dfrac{dz}{dt}$.

$z = \sqrt{40^2 + 30^2} = 50$. Then $40(-60) + 30(50) = 50\dfrac{dz}{dt} \Rightarrow \dfrac{dz}{dt} = \dfrac{-900}{50} = -18$ km/h.

$x$ = horizontal distance along the ground, $z$ = straight-line distance to the drone. The height is fixed at $30$.

Given $\dfrac{dx}{dt} = 2$ m/s (moving away). Wanted $\dfrac{dz}{dt}$. The height's rate is $0$.

$x^2 + 30^2 = z^2$.

$2x\dfrac{dx}{dt} = 2z\dfrac{dz}{dt}$ (the $30^2$ is a constant, its derivative is $0$).

At $z = 50$: $x = \sqrt{2500 - 900} = 40$. Then $40(2) = 50\dfrac{dz}{dt} \Rightarrow \dfrac{dz}{dt} = 1.6$ m/s.

$y$ = climber's height (opposite), $8$ m to the cliff (adjacent, fixed), $\theta$ = angle of elevation.

Given $\dfrac{dy}{dt} = 0.4$ m/s (going up). Wanted $\dfrac{d\theta}{dt}$ in rad/s.

$\tan\theta = \dfrac{y}{8}$.

$\sec^2\theta\,\dfrac{d\theta}{dt} = \dfrac{1}{8}\dfrac{dy}{dt}$.

At $y = 6$ the hypotenuse is $10$, so $\sec\theta = \dfrac{10}{8}$ and $\sec^2\theta = \dfrac{25}{16}$. Then $\dfrac{d\theta}{dt} = \dfrac{0.4}{8} \cdot \dfrac{16}{25} = 0.032$ rad/s.

$x$ = distance from $A$ to the spot (opposite), $2$ km from lighthouse to $A$ (adjacent, fixed), $\theta$ = angle of the beam from the line to $A$.

$\dfrac{d\theta}{dt} = 3 \cdot 2\pi = 6\pi$ rad/min. Wanted $\dfrac{dx}{dt}$.

$\tan\theta = \dfrac{x}{2}$, i.e. $x = 2\tan\theta$.

$\dfrac{dx}{dt} = 2\sec^2\theta\,\dfrac{d\theta}{dt}$.

At $x = 1$: $\tan\theta = \dfrac{1}{2}$, so $\sec^2\theta = 1 + \tan^2\theta = \dfrac{5}{4}$. Then $\dfrac{dx}{dt} = 2 \cdot \dfrac{5}{4} \cdot 6\pi = 15\pi \approx 47.1$ km/min.

$h$ = kite's height (opposite), string $= 100$ m (hypotenuse, fixed), $\theta$ = angle with the ground.

$\dfrac{d\theta}{dt} = -0.05$ rad/s (decreasing). Wanted $\dfrac{dh}{dt}$.

The hypotenuse is fixed, so use sine: $\sin\theta = \dfrac{h}{100}$, i.e. $h = 100\sin\theta$.

$\dfrac{dh}{dt} = 100\cos\theta\,\dfrac{d\theta}{dt}$.

At $\theta = \dfrac{\pi}{6}$: $\cos\theta = \dfrac{\sqrt{3}}{2}$, so $\dfrac{dh}{dt} = 100 \cdot \dfrac{\sqrt{3}}{2} \cdot (-0.05) = -2.5\sqrt{3} \approx -4.33$ m/s.

$h$ = water depth. The radius is fixed at $2$ m.

$\dfrac{dV}{dt} = 500$ L/min $= 0.5$ m$^3$/min (filling, positive). Wanted $\dfrac{dh}{dt}$.

$V = \pi r^2 h$.

$r = 2$ is fixed: $V = 4\pi h$.

$\dfrac{dV}{dt} = 4\pi\dfrac{dh}{dt}$.

No instant needed — the rate does not depend on $h$: $0.5 = 4\pi\dfrac{dh}{dt} \Rightarrow \dfrac{dh}{dt} = \dfrac{1}{8\pi} \approx 0.040$ m/min.

$h$ = water depth, $r$ = radius of the water surface. Both change.

$\dfrac{dV}{dt} = -2$ m$^3$/min (leaking). Wanted $\dfrac{dh}{dt}$.

$V = \dfrac{1}{3}\pi r^2 h$.

Similar triangles: $\dfrac{r}{h} = \dfrac{4}{10}$, so $r = \dfrac{2h}{5}$ and $V = \dfrac{1}{3}\pi\left(\dfrac{2h}{5}\right)^2 h = \dfrac{4\pi h^3}{75}$.

$\dfrac{dV}{dt} = \dfrac{4\pi h^2}{25}\dfrac{dh}{dt}$.

At $h = 5$: $-2 = \dfrac{4\pi(25)}{25}\dfrac{dh}{dt} = 4\pi\dfrac{dh}{dt} \Rightarrow \dfrac{dh}{dt} = -\dfrac{1}{2\pi} \approx -0.159$ m/min.

$h_B$ = depth in the box, $h_C$ = depth in the cylinder. All other dimensions are fixed.

$\dfrac{dh_B}{dt} = 0.3$ m/min. Both tanks share the same $\dfrac{dV}{dt}$. Wanted $\dfrac{dh_C}{dt}$.

$V_B = 2 \cdot 5 \cdot h_B$ and $V_C = \pi (2)^2 h_C$.

Already one variable each: $V_B = 10h_B$, $V_C = 4\pi h_C$.

$\dfrac{dV}{dt} = 10\dfrac{dh_B}{dt}$ and $\dfrac{dV}{dt} = 4\pi\dfrac{dh_C}{dt}$.

$\dfrac{dV}{dt} = 10(0.3) = 3$ m$^3$/min, so $4\pi\dfrac{dh_C}{dt} = 3 \Rightarrow \dfrac{dh_C}{dt} = \dfrac{3}{4\pi} \approx 0.239$ m/min.

$x$ and $y$ change; $P$ is constant.

$\dfrac{dx}{dt} = 3$ workers/month, $\dfrac{dP}{dt} = 0$. Wanted $\dfrac{dy}{dt}$.

Product rule: $0 = \dfrac{4}{3}x^{-1/3}y^{1/3}\dfrac{dx}{dt} + \dfrac{2}{3}x^{2/3}y^{-2/3}\dfrac{dy}{dt}$.

$x^{-1/3} = \tfrac{1}{4}$, $y^{1/3} = 3$, $x^{2/3} = 16$, $y^{-2/3} = \tfrac{1}{9}$: $0 = \dfrac{4}{3}\cdot\dfrac{1}{4}\cdot 3 \cdot 3 + \dfrac{2}{3}\cdot 16 \cdot \dfrac{1}{9}\dfrac{dy}{dt} = 3 + \dfrac{32}{27}\dfrac{dy}{dt}$.

$\dfrac{dy}{dt} = -\dfrac{81}{32} \approx -2.53$ thousand dollars per month.

$P$ and $V$ change; $C$ is a constant.

$\dfrac{dP}{dt} = 20$ kPa/min. Wanted $\dfrac{dV}{dt}$.

Product rule: $\dfrac{dP}{dt}V + P\dfrac{dV}{dt} = 0$.

$20(600) + 150\dfrac{dV}{dt} = 0$.

$\dfrac{dV}{dt} = -\dfrac{12000}{150} = -80$ cm$^3$/min.

$E$ and $L$ change only because $a$ changes.

Wanted $\dfrac{dE}{dL}$; we need $\dfrac{dE}{da}$ and $\dfrac{dL}{da}$.

$\dfrac{dL}{da} = 2e^{1/a} + 2a e^{1/a}\left(-\dfrac{1}{a^2}\right) = 2e^{1/a}\left(1 - \dfrac{1}{a}\right)$ and $\dfrac{dE}{da} = 2a e^{2/a} + a^2 e^{2/a}\left(-\dfrac{2}{a^2}\right) = e^{2/a}(2a - 2)$.

At $a = 2$: $\dfrac{dL}{da} = 2e^{1/2}\cdot\dfrac{1}{2} = e^{1/2}$ and $\dfrac{dE}{da} = e \cdot 2 = 2e$.

$\dfrac{dE}{dL} = \dfrac{2e}{e^{1/2}} = 2\sqrt{e} \approx 3.30$ energy units per unit length.

$x$ = person's distance from the post, $s$ = shadow length. Lamp $6$ m, person $1.8$ m.

$\dfrac{dx}{dt} = 1.4$ m/s. Wanted $\dfrac{ds}{dt}$ and $\dfrac{d}{dt}(x + s)$.

$\dfrac{6}{x + s} = \dfrac{1.8}{s} \Rightarrow 6s = 1.8x + 1.8s \Rightarrow 4.2s = 1.8x \Rightarrow s = \dfrac{3}{7}x$.

$\dfrac{ds}{dt} = \dfrac{3}{7}\dfrac{dx}{dt}$.

$\dfrac{ds}{dt} = \dfrac{3}{7}(1.4) = 0.6$ m/s; tip: $1.4 + 0.6 = 2$ m/s. The $10$ m is not needed — both rates are constant.

$x$ = child's distance from the light, $s$ = shadow length. Light $5$ m, child $1$ m.

$\dfrac{dx}{dt} = -1.2$ m/s (toward). Wanted $\dfrac{ds}{dt}$.

$\dfrac{5}{x + s} = \dfrac{1}{s} \Rightarrow 5s = x + s \Rightarrow s = \dfrac{x}{4}$.

$\dfrac{ds}{dt} = \dfrac{1}{4}\dfrac{dx}{dt}$.

$\dfrac{ds}{dt} = \dfrac{1}{4}(-1.2) = -0.3$ m/s.

$x$ = man's distance from the spotlight, $H$ = shadow height on the wall. Spotlight to wall $12$ m, man $2$ m.

$\dfrac{dx}{dt} = 1.6$ m/s (away from the light). Wanted $\dfrac{dH}{dt}$.

$\dfrac{2}{x} = \dfrac{H}{12} \Rightarrow H = \dfrac{24}{x}$.

$\dfrac{dH}{dt} = -\dfrac{24}{x^2}\dfrac{dx}{dt}$.

$4$ m from the building means $x = 12 - 4 = 8$: $\dfrac{dH}{dt} = -\dfrac{24}{64}(1.6) = -0.6$ m/s.