Question

When a circular plate of metal is heated in an oven, its radius increases at rate of 0.01 cm per minute. At what rate is the plate's area increasing when the radius is 50 cm?

A conical water tank with vertex down has a radius of 10 ft at the top and is 24 ft high. If water flows to the tank at a rate of 20 cube ft per minute, how fast is the depth of the water increasing when the water is 16 ft deep?

Answer 1

Area of a circle of radius r

$$S =\pi r^2=\pi r^2(t),\ where \ t\ is \ time$$

rate is the plate's area increasing

$$\frac{dS}{dt}=2\pi r(t)r'(t)=2\pi\cdot 50\cdot 0.01=\\\pi\cdot100\cdot 0.01=\pi cm^2/min$$

2. Radius of the tank at height h is

$$r(h)=h\cdot\frac{10}{24}$$

volume of water in the tank, when depth is h is

$$V=\frac{\pi}{3}r^2h=\frac{\pi}{3}r(h(t))^2h(t)=\frac{\pi}{3}(h(t)\cdot\frac{10}{24})^2h(t)=\\ \frac{\pi}{3}(\frac{5}{12})^2h^3(t)$$

water volume increase rate is

$$\frac{dV}{dt}=(\frac{\pi}{3}(\frac{5}{12})^2h^3(t))'= \frac{\pi}{3}(\frac{5}{12})^2\cdot 3 h^2(t)h'(t)=\\ \pi(\frac{5}{12})^2h^2(t)h'(t)$$

we have

$$\pi(\frac{5}{12})^2 16^2h'(t)=20$$

depth increase rate is

$$h'(t)=\frac{20\cdot 12^2}{\pi5^2 16^2}=\frac{9}{20\pi}ft/min$$

Answer:

1)

$$\pi cm^2/min$$

2)

$$\frac{9}{20\pi}ft/min$$