What is the radius of the oil slick after 10 minutes?

[Noah1]

Hi I am stuck on this integral question:
An oil tanker aground on a reef is losing oil and producing an oil slick that is radiating out at a rate approximated by the function (dr/dt)=20/√t, t is greater than or equal to 1 where r is the radius of the circular slick in metres after t minutes. If the radius of the slick is 21 metres, after 15 seconds, find, to the nearest metre, the radius of the slick after 10 minutes.

 

[HallsofIvy]

Re: Stuck on this question please help

Hi I am stuck on this integral question:
An oil tanker aground on a reef is losing oil and producing an oil slick that is radiating out at a rate approximated by the function (dr/dt)=20/√t, t is greater than or equal to 1 where r is the radius of the circular slick in metres after t minutes. If the radius of the slick is 21 metres, after 15 seconds, find, to the nearest metre, the radius of the slick after 10 minutes.

This is asking you to solve the first order differential equation \(\displaystyle \frac{dr}{dt}= 20t^{-1/2}\) with initial condition r(1/4)= 21. That is, specifically, a "separable" equation, $$dr= 20t^{-1/2}dt$$. Integrate both sides of that. Of course there will be a "constant of integration". Use the fact that r(1/4)= 21 to find that const

 

[Noah1]

Re: Stuck on this question please help

This is asking you to solve the first order differential equation \(\displaystyle \frac{dr}{dt}= 20t^{-1/2}\) with initial condition r(1/4)= 21. That is, specifically, a "separable" equation, $$dr= 20t^{-1/2}dt$$. Integrate both sides of that. Of course there will be a "constant of integration". Use the fact that r(1/4)= 21 to find that const

∫▒20/√t dt
20∫▒1/√t dt
20∫▒t^(-1/2) dt
40√t+c
Now subsitute in 10 to t and 21 into c
40√10+21
=147.4911064
So the slick will be 147 metres.
am I on the right track