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differentiating relative rates: oil spilled from a ruptured tanker spreads in a circle...

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user7230927 | Salutatorian

Posted July 11, 2013 at 3:39 AM via web

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differentiating relative rates:

oil spilled from a ruptured tanker spreads in a circle whose area increases at a rate of 4 `mi^2` /hr.  how fast is the radius of the spill increasing when the area is 25 `mi^2`

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llltkl | College Teacher | Valedictorian

Posted July 11, 2013 at 4:32 AM (Answer #1)

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Area of a circle `A=pir^2`

Both A and r are functions of t, so differentiate implicitly with respect to t.

Thus, `dA/dt=pi*2r(dr)/dt`

It is given that the area of the circle increases at a rate of `4 (mi^2)/(hr).`

So, `(dA)/dt=4`

`rArr 4= pi*2r(dr)/dt`   ...............(i)

We have to find `(dr)/dt` when A=25.

So, `pir^2=25`

`rArr r^2=25/pi`

`rArr r=5/sqrtpi`

Substituting the value of r in eq(i) we get:

`4=pi*2*5/sqrtpi(dr)/dt`

`rArr 4=10sqrtpi(dr)/dt`

`rArr (dr)/dt=4/(10sqrtpi)=2/(5sqrtpi)`

Therefore, when the area is `25 mi^2` the radius of the spill is increasing at a rate of `2/(5sqrtpi)` mi/hr.

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