Related Rates problems. Please Read.

  • Thread starter Thread starter jasonlr82794
  • Start date Start date
  • Tags Tags
    Related rates
jasonlr82794
Messages
34
Reaction score
0
I have an oil spill that spills out at a rate of 100 ft per second. I need to find the rate of change of the radius at 800ft. The problem I thought would have been set up like this...
A=∏rsquared

100ft per second= 2∏r, but this isn't correct, it is actually...

100ft per second - 2∏r (dr/dt), this is how it was shown in the problem.

Why is the derivative of the cirlces area multiplied by the radiuses derivative with respect to time? why is it set up this way? It mathematically makes sense to say this without the dr/dt, so why is it there? please help.
 
Physics news on Phys.org
jasonlr82794 said:
I have an oil spill that spills out at a rate of 100 ft per second. I need to find the rate of change of the radius at 800ft. The problem I thought would have been set up like this...
A=∏rsquared

100ft per second= 2∏r, but this isn't correct, it is actually...
Have you quoted the problem correctly? Is the area of the spill growing at 100 square feet per second? (100 ft2/sec)
jasonlr82794 said:
100ft per second - 2∏r (dr/dt), this is how it was shown in the problem.
There's a typo here and the units on the left side are wrong.
The problem probably had it as
100 ft2[/color]/sec = 2##\pi##r dr/dt.

The units on the left are ft2/sec. The units on the right have to agree, with r in ft and dr/dt in ft/sec.
jasonlr82794 said:
Why is the derivative of the cirlces area multiplied by the radiuses derivative with respect to time? why is it set up this way? It mathematically makes sense to say this without the dr/dt, so why is it there? please help.
 
Back
Top