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# I need help with a math problem please!!!?

If snowball melts so that its surface area decreases at a rate of 1 cm^2/min, find the rate at which the diameter decreases when the diameter is 10cm.

### 3 Answers

- AshLv 71 month ago
Surface area of Sphere, SA = πD² ....where D = diameter

Find derivative with time

d(SA)/dt = π(2 D) dD/dt

d(SA)/dt = 2πD dD/dt

dD/dt = (d(SA)/dt)/(2πD)

Given d(SA)/dt = -1 cm²/min .... (negative sign shows decrease in rate)

D = 10 cm

dD/dt = (-1) /(2π*10)

dD/dt = -1/(20π)

dD/dt = -0.016 cm/min

The diameter of snowball decreases by 0.016 cm/min

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- Anonymous1 month ago
Area of a sphere = pi*diameter^2

Take the derivative

dA/dT = 2 * pi * d(diameter)dT

substituting

1 cm^2/sec = 2 * pi * d(diameter)dT

The rate at which the diameter decreases = 0.5 / pi cm^2/sec

I'm a little rusty on this so hopefully it help you to spot any mistakes I've done.

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