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differentiating relative rates: a stone dropped into a still pond sends out a circular...
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- At t = 0 the radius of the circular ripple is zero or negligible
- The radius of the pond is larger than 80ft.
Let us say the radius of the circular ripple at any point is rft. It is given that the rate of increasing of r is 5ft/s.
`(dr)/(dt) = 5`
If the area of the circular ripple is A then;
`A = pir^2`
`(dA)/(dt) = pixx2rxx(dr)/(dt)`
When the time is 16 seconds;
`r = txx(dr)/(dt)`
`r = 16xx5 = 80ft`
`((dA)/(dt))_(t = 16) = pixx2rxx(dr)/(dt)`
`((dA)/(dt))_(t = 16) = pixx2xx80xx5`
`((dA)/(dt))_(t = 16) = 800pi`
So the area is increasing at a rate of `800pift/s` when the time is 16 seconds.
Posted by jeew-m on July 11, 2013 at 3:55 AM (Answer #1)
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