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Problem If a snowball melts so that its surface area
decreases at a rate 1cm2/min, find the rate at which the
diameter decreases when the diameter is 10cm.
Solution: The quantities which are changing they are surface area, A, and diameter, D.
Given: A'(t)=­1cm2/min. Find D'(t) when D=10cm.
We know that D=2R(where R is radius). Thus A=4πR2=4π(D/2)2=πD2,
or A(t)=π[D(t)]2. Differentiating the last formula we get:
A'(t)=2πD(t)D'(t). Plug in the given numerical data and get
­1=2π10D'(t), or D'(t)=­1/(20π) cm/min.
Answer: The rate of decrease is 1/(20π) cm/min.
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