If a snowball melts so that its surface area decreases at a rate of , find the rate at which the diameter decreases when the diameter is

carra im on it

what are the rates. Are any numbers given?

sorry...if a snowball melts so that its surface area decreases at a rate of 0.3cm^2/min. find the rate at which the diameter decreases when the diameter is 15cm.

i thought it would be dA/dt=0.3cm^2...so would it be dD/dt=2*pi*(7.5)*0.3

sorry will respond in a few mins, i have to do something

Okay so you have: \[A = 4\pi r ^{2}\] <=Surface area of a ball

Now you know that d = 2r (diameter is twice the radius), therefore r = d/2 (radius is half of the diameter) Rewrite the equation in terms of diamater

\[A = 4\pi (d/2)^{2} = 4\pi * (d ^{2}/2^{2}) = 4\pi * (d^{2}/4)\]

Sorry meant diameter in the previous reply. Now you know that: A = 4 pi * (d^2/4), cancel the 4s to get: \[A = \pi * d ^{2}\]

Now differentiate with respect to t, or time: \[dA/dt = (2 *\pi * d) * dd/dt\]

You know the change in surface area; and the diameter at which you are required to find the change in radius: dA/dt = -0.3 cm^2/min (THE "-" SIGN HAS TO BE THERE BECAUSE THE SURFACE AREA IS DECREASING) and the diameter = 15 cm so plug those into your equation

\[-0.3 = 2\pi*15*(dd/dt)\]

-0.3 = 30pi * (dd/dt) [ -0.3 / (30pi) ] = dd/dt \[dd/dt = -0.3 / (30\pi)\]

woops forgot units, just add cm/min at the end, check with the answer

sorry carra wrong answer, lucky someone caught me

i was right

then y am i getting dd/dt=2*pi*15*0.3

Okay let me try this again; I will write it out on a piece of paper and upload it, i hate typing

http://i55.tinypic.com/8wdsp1.jpg anything unclear? there? ask me, i can explain steps

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