Related Rates problems. Please Read.

  • Context: Undergrad 
  • Thread starter Thread starter jasonlr82794
  • Start date Start date
  • Tags Tags
    Related rates
Click For Summary
SUMMARY

The discussion centers on solving a related rates problem involving an oil spill that spreads at a rate of 100 square feet per second. The correct formulation of the problem is established as 100 ft²/sec = 2πr (dr/dt), where the left side represents the rate of change of area and the right side incorporates both the radius and its rate of change. The inclusion of dr/dt is crucial as it accounts for the changing radius over time, ensuring dimensional consistency in the equation. Participants clarify the importance of units and the correct setup for related rates problems in calculus.

PREREQUISITES
  • Understanding of calculus concepts, specifically related rates
  • Familiarity with the formula for the area of a circle, A=πr²
  • Knowledge of derivatives and their application in real-world scenarios
  • Ability to manipulate units in mathematical equations
NEXT STEPS
  • Study the concept of related rates in calculus
  • Learn how to derive and apply the formula for the area of a circle
  • Explore examples of related rates problems in physics and engineering
  • Practice unit conversion and dimensional analysis in calculus problems
USEFUL FOR

Students and educators in mathematics, particularly those studying calculus, as well as professionals in fields requiring mathematical modeling of physical phenomena, such as environmental science and engineering.

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.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K