In calm waters, the oil spilling from the ruptured hull of a grounded tanker spr

Question

In calm waters, the oil spilling from the ruptured hull of a grounded tanker spreads in all directions. Assuming that the are polluted is circular, determine how fast the radius of the circle is changing when the area of the circle is 1600(pi) and increading at a rate of 80(pi) ft^2/sec.

I know the formula is A=(pi)r^2 and then if you solve for the radius then you get 40. But then when I solved for da/dt=(pi)2r (dr/dt) I got my answer as 1ft^2/sec and my book says the answer is 188.5ft^2/sec. Please help very confused.

Explanation / Answer

We know that

A=(pi)r^2

or, 1600(pi) =(pi)r^2

or, 1600 =r^2

or, r = 40 m

To determine how fast the radius of the circle is changing ,we have to differentiate this equation with respect to t (time)

So, dA/dt = (pi) x 2r x dr/dt

or,80(pi) = (pi) x 2 x 40 x dr/dt

or, 80(pi) = (pi) x 80 x dr/dt

or, dr/dt = 80(pi)/(pi) x 80 = 1

or, dr/dt = 1

Rate of change of Radius = 1 ft/sec

I think your answer is right and it might be some printing mistake in the book. Please note that the unit will be "ft/sec" and not "ft^2/sec".

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