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lara [203]
3 years ago
6

Assume that the radius r of a sphere is expanding at a rate of 40 cm/min. The volume of a sphere is V = 4 3 πr3 and its surface

area is 4πr2. Determine the rate of change in surface area when r = 20 cm.
Mathematics
1 answer:
Scrat [10]3 years ago
6 0

Answer:

The rate of change in surface area when r = 20 cm is 20,106.19 cm²/min.

Step-by-step explanation:

The area of a sphere is given by the following formula:

A = 4\pi r^{2}

In which A is the area, measured in cm², and r is the radius, measured in cm.

Assume that the radius r of a sphere is expanding at a rate of 40 cm/min.

This means that \frac{dr}{dt} = 40

Determine the rate of change in surface area when r = 20 cm.

This is \frac{dA}{dt} when r = 20. So

A = 4\pi r^{2}

Applying implicit differentiation.

We have two variables, A and r, so:

\frac{dA}{dt} = 8r\pi \frac{dr}{dt}

\frac{dA}{dt} = 8*20\pi*40

\frac{dA}{dt} = 20106.19

The rate of change in surface area when r = 20 cm is 20,106.19 cm²/min.

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