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Differentiation Q - C3 Jan 10 ocr

The question: Leaking oil is forming a circular patch on the surface of the sea. The area of the patch is increasing at a rate of 250 square metres per hour. Find the rate at which the radius of the patch is increasing at the instant when the area of the patch is 1900 square metres.

My working : dA/dt = 250 therefore: ∫dA = ∫250 dt => A = 250t

Find time to get 1900: 1900 = 250t => t= 7.6 hours

Find radius at 1900m^2 : πr^2 = 1900 => r = 24.592

Therefore dA/dt = 24.592/7.6 = 3.2 metres per hour

Just looking to see why this is wrong (answer is 1.6) a the mark scheme doesn't seem to help. Thanks :smile:
Original post by luke5675


Therefore dA/dt = 24.592/7.6 = 3.2 metres per hour

Just looking to see why this is wrong (answer is 1.6) a the mark scheme doesn't seem to help. Thanks :smile:



I'm sure you meant dr/dt there.

This, in bold, assume that the radius is increasing at a constant rate over the whole time period. However this is not the case, and the rate of increase in radius is decreasing as the radius increases.

Edit:

You need to use dA/dt = dA/dr x dr/dt

You know dA/dt, you can work out dA/dr when A=1900, hence dr/dt.
(edited 8 years ago)
Original post by luke5675
The question: Leaking oil is forming a circular patch on the surface of the sea. The area of the patch is increasing at a rate of 250 square metres per hour. Find the rate at which the radius of the patch is increasing at the instant when the area of the patch is 1900 square metres.My working : dA/dt = 250 therefore: ∫dA = ∫250 dt => A = 250tFind time to get 1900: 1900 = 250t => t= 7.6 hoursFind radius at 1900m^2 : πr^2 = 1900 => r = 24.592Therefore dA/dt = 24.592/7.6 = 3.2 metres per hourJust looking to see why this is wrong (answer is 1.6) a the mark scheme doesn't seem to help. Thanks :smile:


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luke5675
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Ah ok, that makes much more sense, thanks a lot for the responses :smile:
(edited 8 years ago)

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