differentiating relative rates:
a stone dropped into a still pond sends out a circular ripple whose radius increases at a constant rate of 5 ft/sec. how rapidly is the area enclosed by the ripple increasing at the end of 16 sec?
1 Answer | Add Yours
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.
- At t = 0 the radius of the circular ripple is zero or negligible
- The radius of the pond is larger than 80ft.
We’ve answered 319,863 questions. We can answer yours, too.Ask a question