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Section 3.11 : Related Rates

4. A thin sheet of ice is in the form of a circle. If the ice is melting in such a way that the area of the sheet is decreasing at a rate of 0.5 m2/sec at what rate is the radius decreasing when the area of the sheet is 12 m2?

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Start Solution

We’ll call the area of the sheet \(A\) and the radius \(r\) and we know that the area of a circle is given by,

\[A = \pi {r^2}\]

We know that \(A' = - 0.5\) and want to determine \(r'\) when \(A = 12\).

Show Step 2

Next step is to simply differentiate the equation with respect to \(t\).

\[A' = 2\pi r\,r'\] Show Step 3

Now, to finish this problem off we’ll first need to go back to the equation of the area and use the fact that we know the area at the point we are interested in and determine the radius at that time.

\[12 = \pi {r^2}\hspace{0.25in} \Rightarrow \hspace{0.25in}r = \sqrt {\frac{{12}}{\pi }} = 1.9544\]

The rate of change of the radius is then,

\[ - 0.5 = 2\pi \left( {1.9544} \right)\,r'\hspace{0.5in} \Rightarrow \hspace{0.5in}\require{bbox} \bbox[2pt,border:1px solid black]{{r' = - 0.040717}}\]