The radius of a circle is increasing at a rate of 2 cm/s. How fast is the area increasing when the radius is 5 cm?

Let $r$ = radius, $A$ = area, $t$ = time.

Given: $\frac{dr}{dt} = 2$ cm/s. Find: $\frac{dA}{dt}$ when $r = 5$.

Relationship: $A = \pi r^2$

Differentiate with respect to time: $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$

Substitute $r = 5$ and $\frac{dr}{dt} = 2$: $\frac{dA}{dt} = 2\pi(5)(2) = 20\pi \text{ cm}^2/\text{s}$

A 10-meter ladder rests against a vertical wall. If the bottom of the ladder slides away from the wall at 0.5 m/s, how fast is the top of the ladder sliding down the wall when the bottom is 6 meters from the wall?

Let $x$ = horizontal distance from wall, $y$ = vertical height on wall.

Given: $\frac{dx}{dt} = 0.5$ m/s. Find: $\frac{dy}{dt}$ when $x = 6$.

By the Pythagorean theorem: $x^2 + y^2 = 100$

When $x = 6$: $36 + y^2 = 100$, so $y = 8$.

Differentiate with respect to time: $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$

Substitute known values: $2(6)(0.5) + 2(8)\frac{dy}{dt} = 0$ $6 + 16\frac{dy}{dt} = 0$ $\frac{dy}{dt} = -\frac{3}{8} \text{ m/s}$

The top is sliding down at $\frac{3}{8}$ m/s (negative indicates downward motion).

Water is draining from a conical tank at 2 cubic meters per minute. The tank has height 10 m and radius 4 m at the top. How fast is the water level dropping when the water is 5 m deep?

Let $h$ = water height, $r$ = radius at water surface, $V$ = volume.

Given: $\frac{dV}{dt} = -2$ m³/min (negative for draining). Find: $\frac{dh}{dt}$ when $h = 5$.

By similar triangles: $\frac{r}{h} = \frac{4}{10}$, so $r = \frac{2h}{5}$.

Volume of cone: $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi\left(\frac{2h}{5}\right)^2 h = \frac{4\pi h^3}{75}$

Differentiate: $\frac{dV}{dt} = \frac{4\pi}{75} \cdot 3h^2 \frac{dh}{dt} = \frac{4\pi h^2}{25}\frac{dh}{dt}$

When $h = 5$: $-2 = \frac{4\pi(25)}{25}\frac{dh}{dt} = 4\pi\frac{dh}{dt}$ $\frac{dh}{dt} = -\frac{1}{2\pi} \approx -0.159 \text{ m/min}$