How fast is the area of the oil spill increasing when the radius is 30 m?

In summary, the conversation discusses finding the rate of change of the area of a circle as its radius increases. The first part of the conversation involves using the equation A=(pi)r^2 to find the derivative of the area with respect to time, dA/dt. The second part involves a real-life scenario of an oil spill that is spreading in a circular pattern, with a constant rate of increase for the radius. The conversation also mentions the use of the chain rule for finding the derivative of the area with respect to time.
  • #1
synergix
178
0

Homework Statement


a) if A is the area of circle with radius r and the circle expands as time passes, find dV/dt in terms of dr/dt
b)Suppose oil spills from a ruptured ranker and spreads in a circular pattern. If the radius of the oil spill increases at a constant rate of 1 m/s, how fast is the area of the spill increasing when the radius is 30 m?


Homework Equations



A=(pi)r2

The Attempt at a Solution



dA/dt= ((pi)r2)'=(r2+2(pi)r)

dA/dt= ((pi)2(30) + 302)

is this correct? it seems like an awfully big number what have I done wrong?
 
Physics news on Phys.org
  • #2
oh wait I am treating pi like a variable not a constant. durh
 
  • #3
so dA/dt = 2*30*pi
 
  • #4
You've got it dA/dt=pi*2*r*dr/dt.
 
  • #5
OK so I would need to have multiplied 2*pi*r*1m/s but since its one it doesn't matter but that is what is happening correct?

I am multiplying the derivative of the expression by the derivative of r?
 
  • #6
It matters if you are keeping track of units. That's where the m/s came from.
 
  • #7
synergix said:
OK so I would need to have multiplied 2*pi*r*1m/s but since its one it doesn't matter but that is what is happening correct?

so i am multiplying the derivative of the expression by the derivative of r?

You are using the chain rule. d/dt(f(r))=d/dr(f(r))*dr/dt. You knew that, right?
 
  • #8
Dick said:
You are using the chain rule. d/dt(f(r))=d/dr(f(r))*dr/dt. You knew that, right?

I know that now. I missed a good couple classes (long story short) because I had no other choice. Now I am trying to catchup. I guess I better do some reading..thanks
 

1. What is DA/dt of expanding circle?

DA/dt of expanding circle refers to the rate of change of the area of a circle as it expands. It is a measure of how fast the circle is increasing in size.

2. How is DA/dt of expanding circle calculated?

DA/dt of expanding circle can be calculated using the formula DA/dt = 2πr(dr/dt), where r is the radius of the circle and dr/dt is the rate of change of the radius.

3. What factors affect the DA/dt of expanding circle?

The DA/dt of expanding circle is affected by the rate of change of the radius, the initial size of the circle, and the type of expansion (linear or exponential).

4. Why is DA/dt of expanding circle important?

DA/dt of expanding circle is important in various fields of science, such as physics and biology, as it helps in understanding the growth and expansion of objects and organisms. It also has practical applications, such as in calculating the growth rate of tumors in cancer research.

5. How is DA/dt of expanding circle related to other mathematical concepts?

DA/dt of expanding circle is related to other mathematical concepts such as derivatives, area of a circle, and rates of change. It is also connected to the concept of instantaneous rate of change, which is the rate of change at a specific moment in time.

Similar threads

Replies
8
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
Replies
5
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
2K
  • Calculus and Beyond Homework Help
Replies
15
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
Replies
2
Views
1K
Back
Top