Question

A tanker truck has overturned and is spilling oil onto the highway. Amazingly, the region covered by the oil spill maintains a perfectly circular shape. The area covered by the spill is growing at a rate of 40π square feet per minute. How quickly is the radius of the oil spill growing when it covers an area of 100π square feet?

Answer 1

Answer:

The radius of the oil spill growing at 2 feet per minute

Step-by-step explanation:

Area of a circle is

$A=\pi r^2$

Differentiating A with respect to r,

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

When $A=100\pi$,

$100\pi = \pi r^2$

$r = 10$

The rate of growth of the spill is

$\dfrac{dA}{dt} = 40\pi$

This is equivalent to

$\dfrac{dA}{dt} = \dfrac{dA}{dr} \times \dfrac{dr}{dt}$

$\dfrac{dr}{dt} = \dfrac{dA}{dt}/\dfrac{dA}{dr} = \dfrac{40\pi}{2\pi r} = \dfrac{20}{r}$

$\dfrac{dr}{dt} = \dfrac{20}{10} = 2$

Answer 2

Answer:

$\frac{dr}{dt} = 2\,\frac{ft}{min}$

Step-by-step explanation:

The area of a circle is:

$A = \pi\cdot r^{2}$

The rate of change of the oil spill is:

$\frac{dA}{dt} = 2\pi\cdot r \cdot \frac{dr}{dt}$

The radio of the oil spill is:

$r =\sqrt { \frac{100\pi\,ft^{2}}{\pi}}$

$r = 10\,ft$

The rate of change of the oil spill is:

$40\pi\,\frac{ft^{2}}{min} = \left(20\pi\,ft\right)\cdot \frac{dr}{dt}$

$\frac{dr}{dt} = 2\,\frac{ft}{min}$