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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 year 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
- Anonymous1 year 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.
