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A 10-m ladder leans against a vertical wall. The foot of the ladder slides away from the wall at 2 m/s. How fast is the top of the ladder sliding down the wall when the foot is 6 m from the wall?

Let $x$ = distance of foot from wall, $y$ = height of top on wall. By Pythagorean theorem:

$x^2 + y^2 = 100$

When $x = 6$: $y = \sqrt{100 - 36} = 8$ m.

Differentiate with respect to time:

$2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt} = 0$
$\dfrac{dy}{dt} = -\dfrac{x}{y}\cdot\dfrac{dx}{dt} = -\dfrac{6}{8}(2) = -\dfrac{3}{2}$

The top slides down at $\boxed{1.5\ \text{m/s}}$. (Negative sign indicates downward.)

An inverted conical tank has a base radius of 4 m and a height of 8 m. Water drains out at $2\ \text{m}^3/\text{min}$. Find the rate at which the water level is dropping when the water depth is 4 m.

By similar triangles: $\dfrac{r}{h} = \dfrac{4}{8} = \dfrac{1}{2}$, so $r = \dfrac{h}{2}$.

Volume of a cone:

$V = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3}\pi\!\left(\dfrac{h}{2}\right)^2\!h = \dfrac{\pi h^3}{12}$

Differentiate with respect to time:

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

Solve for $\dfrac{dh}{dt}$ (with $dV/dt = -2$, $h = 4$):

$\dfrac{dh}{dt} = \dfrac{4}{\pi h^2}\cdot\dfrac{dV}{dt} = \dfrac{4}{\pi(16)}(-2) = -\dfrac{1}{2\pi}$

The water level drops at $\boxed{\dfrac{1}{2\pi} \approx 0.159\ \text{m/min}}$.

The radius of a circle is increasing at a rate of 3 cm/s. How fast is the area increasing when the radius is 5 cm?

$A = \pi r^2$. Differentiate with respect to time:

$\dfrac{dA}{dt} = 2\pi r\dfrac{dr}{dt}$

Substitute $r = 5$ cm and $\dfrac{dr}{dt} = 3$ cm/s:

$\dfrac{dA}{dt} = 2\pi(5)(3) = \boxed{30\pi \approx 94.25\ \text{cm}^2/\text{s}}$

Car A travels north at 60 km/h and Car B travels east at 80 km/h. Both depart from the same intersection at the same time. How fast is the distance between them increasing after 1 hour?

After 1 hour: $y = 60$ km (north), $x = 80$ km (east).

$d = \sqrt{x^2 + y^2} = \sqrt{80^2 + 60^2} = \sqrt{10000} = 100\ \text{km}$

Differentiate $d^2 = x^2 + y^2$:

$2d\dfrac{dd}{dt} = 2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt}$
$\dfrac{dd}{dt} = \dfrac{x\,\dot{x} + y\,\dot{y}}{d} = \dfrac{80(80) + 60(60)}{100} = \dfrac{6400 + 3600}{100} = \boxed{100\ \text{km/h}}$

A street lamp is mounted 10 m above ground. A person 2 m tall walks away from the lamp at 1.5 m/s. How fast is the tip of the person's shadow moving?

Let $x$ = distance from lamp to person, $s$ = length of shadow. By similar triangles:

$\dfrac{10}{x + s} = \dfrac{2}{s} \implies 10s = 2(x+s) \implies s = \dfrac{x}{4}$

The tip of the shadow is at distance $T = x + s = x + \dfrac{x}{4} = \dfrac{5x}{4}$ from the lamp.

$\dfrac{dT}{dt} = \dfrac{5}{4}\dfrac{dx}{dt} = \dfrac{5}{4}(1.5) = \boxed{1.875\ \text{m/s}}$

The shadow length itself grows at $ds/dt = \dfrac{1}{4}(1.5) = 0.375$ m/s.

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