Find how fast the radius of a circle get smaller as its area gets smaller

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SUMMARY

The discussion focuses on calculating the rate at which the radius of a circle decreases as its area diminishes from 75π cm² at a rate of 2π cm²/s. The initial approach involved using the area formula A = r²π and resulted in an incorrect rate of change for the radius. The correct method, as clarified by a peer, involves differentiating the area with respect to time, leading to the conclusion that dr/dt = -1/sqrt(75). This method aligns with the established relationship between area and radius, confirming the correct answer.

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  • Understanding of calculus, specifically differentiation
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  • Knowledge of related rates in physics and mathematics
  • Ability to solve equations involving square roots
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  • Learn how to differentiate functions involving multiple variables
  • Explore the implications of the chain rule in calculus
  • Practice solving problems involving the area and circumference of circles
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Students studying calculus, particularly those focusing on related rates, as well as educators looking to clarify concepts related to the geometry of circles and differentiation techniques.

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Homework Statement


(Translated, sorry my English) Find how fast does the radius of a circle get smaller when its area is 75*pi cm^2 and it (the area) gets smaller at a rate of 2*pi cm^2/s?

Homework Equations


Not really any given equations, but I used these:
A = r^2 * pi
s = v*t + s0 (If v is a constant)
t=A/Va (where t is the time when A=0)

The Attempt at a Solution


From now on I'll call area, A = 75*pi cm^2 and the rate v_a = 2*pi cm^2/s.
I gave myself that the speed (v) was constant, that shouldn't really change the answer.
So s = dv/dt = v*t + s0

By using A = r^2 * pi we know that:
r^2 = A/pi
r = sqrt(A/pi) and since at t = 0 then A = 75*pi so
r = sqrt(75) at t = 0, let's call r from now on s.

Let's find the time until A = 0, and therefore s=0
t = A/v_a = 75*pi/(2*pi) = 37,5 sec

So in 37,5 sec the circle area will be zero with is radius. Therefore we can see that:
s = v*t + s0
0 = v*37,5 + sqrt(75)
-sqrt(75) = v*37,5
v = -sqrt(75)/37,5
v = -2*sqrt(75)/75
v = -2/sqrt(75)

So my final answer was v = -2/sqrt(75) (or v = -2/(5*sqrt(3))But according to my teacher the correct answer is v = -1/sqrt(75).--
Here's how my teacher did it:

We know that F(t) = pi * r(t)^2 and
(1) F'(t0) = 2*pi*r(t0)*r'(t0) = -2*pi
(2) F(t0) = pi*r(t)^2 = 75*pi
for a specific t0.

According to (2) we know that r(t0) = sqrt(75) = 5*sqrt(3) and according to (1) we get r'(t0) = -1/r(t0) = -1/sqrt(75)

--
He called my solution "a waste of time" although he said it should work too, but couldn't explain to me why I didn't get the same answer.

Where did I go wrong (or even better, where did he go wrong?)
 
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The area of the circle is A(r) = pi r2. Assuming that both A and r are differentiable functions of t, and differentiating with respect to t, you get
dA/dt = 2pi r dr/dt

Solve this equation for dr/dt. That will give you dr/dt for any arbitrary t.
Now evaluate dr/dt at the time when A is 75 pi cm2 and when dA/dt is -2 pi cm2/sec. You will need to find the radius r for which the area A is 75 pi cm2 to do this.
 
Thank you for a quick reply!

dr/dt = 1/(2*pi*r) dA/dt and dA/dt = -2*pi so
dr/dt = -1/r

And like I showed before r = sqrt(75) so the final answer is indeed:
dr/dt = -1/sqrt(75)**SOLVED**
Thank you again. I understand your solution a lot better than my teacher's.
 

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