Related Rates

$30\pi$ cm²/s

$\tfrac{dA}{dt} = 2\pi r \tfrac{dr}{dt} = 2\pi(5)(3) = 30\pi$.

$24$ cm³/s

$V = s^3$, so $\tfrac{dV}{dt} = 3s^2\tfrac{ds}{dt} = 3(16)(0.5) = 24$.

$\tfrac{1}{\pi}$ cm/s, about $0.32$ cm/s

$V = \tfrac{4}{3}\pi r^3$, so $\tfrac{dV}{dt} = 4\pi r^2\tfrac{dr}{dt}$. Then $100 = 4\pi(25)\tfrac{dr}{dt}$, so $\tfrac{dr}{dt} = \tfrac{1}{\pi}$.

$\tfrac{3}{4}$ ft/s

$x^2 + y^2 = 100$; at $x = 6$, $y = 8$. $2x x' + 2y y' = 0$ gives $12 + 16y' = 0$, so $y' = -\tfrac{3}{4}$: the top slides down at $\tfrac{3}{4}$ ft/s.

The area is decreasing at $2$ m²/s.

$A = LW$, so $\tfrac{dA}{dt} = L'W + LW' = (2)(4) + (10)(-1) = 8 - 10 = -2$.

$\tfrac{25}{3}$ ft/s

Let $x$ be the person’s distance from the pole and $s$ the tip’s. Similar triangles: $\tfrac{s}{15} = \tfrac{s - x}{6}$, so $6s = 15s - 15x$ and $s = \tfrac{5}{3}x$. Then $s' = \tfrac{5}{3}(5) = \tfrac{25}{3}$.

About $0.25$ ft/min

$\tfrac{dV}{dt} = 0.16\pi h^2\tfrac{dh}{dt}$ with $\tfrac{dV}{dt} = -2$ and $h = 4$: $-2 = 2.56\pi\tfrac{dh}{dt}$, so $\tfrac{dh}{dt} = -\tfrac{2}{2.56\pi} \approx -0.249$.

$25$ km/h

After $3$ hours they are $45$ and $60$ km out, so $z = 75$. $z' = \tfrac{45(15) + 60(20)}{75} = \tfrac{675 + 1200}{75} = 25$.

$10\sqrt{3}$ cm²/min

$A = \tfrac{\sqrt{3}}{4}s^2$, so $\tfrac{dA}{dt} = \tfrac{\sqrt{3}}{2}s\tfrac{ds}{dt} = \tfrac{\sqrt{3}}{2}(10)(2) = 10\sqrt{3}$.

Is $x$ equal to $5$ at every moment?

$x$ was treated as a constant. The correct rate is $-\tfrac{5}{6}$ ft/s.

$x = 5$ only at one instant; the foot is moving, so $\tfrac{dx}{dt} = 2$.

Differentiating $x^2 + y^2 = 169$ first keeps the term $2x\tfrac{dx}{dt}$, which gives $\tfrac{dy}{dt} = -\tfrac{5}{6}$.