BigNewton2 From: US Posts: 45 Votes: 135 |
If a snowball melts so that its surface area decreases at arate of 1cm2/ min, find the rate at which the diameterdecreases when the diameter is 10 cm. |

Anonymous From: - Posts: - Votes: - |
First we know that dA/dt is -1 cm2/min and we know thesurface area(A) of a sphere is 4πr2 so now we get the derivative of A and we get (8πr dr/dt) andthat is equal to dA/Dt so since we need to find the rate of change of the diameter we setdr/dt= dA/dt*1/(8πr) and we get that the rate at which the radius is changing is -.008cm/min and since it asks for the rate at which the diameter changesyou multiply it by two and its -.016 cm/min or .1/(20π) |

Anonymous From: - Posts: - Votes: - |
....The area of the Sphere is: A=4πr^2=πD^2,....Now theproblem says that dA/dt=-1cm^2/min,and we search for dA/dt=2πDdD/dt,....and now ifD=10cm we will have: dD/dt=(1/20π)(cm)^(-1)(-1)cm^2/min...sothe result is: dD/dt=-(1/20π)cm/min. |

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