How to solve a related rates problem with an expanding square?

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To solve the related rates problem involving an expanding square, start by differentiating the area equation A = s^2 with respect to time, yielding dA/dt = 2s(ds/dt). Given that the area is increasing three times faster than the side, set dA/dt equal to 3(ds/dt). This leads to the equation 2s(ds/dt) = 3(ds/dt), simplifying to 2s = 3, resulting in s = 3/2. The correct answer to the problem is therefore 3/2.
carlodelmundo
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Homework Statement



When the area of an expanding square, in square units, is increasing three times as fast as its side is increasing, in linear units, the side is

a.) 2/3
b.) 3/2
c) 3
d) 2
e) 1

Homework Equations



A = s^2
dA/dt = 3s^2


The Attempt at a Solution



Can anyone give me hints on how to start this problem?
 
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Start by differentiating A=s^2 with respect to time correctly. Use implicit differentiation.
 
Okay, Dick.

I get dA/dt = 2s dS/Dt. Since it's saying that the area is increasing three times as fast as its side is increasing... 2s must equal to 3. or s = 3/2

is this correct?
 
carlodelmundo said:
Okay, Dick.

I get dA/dt = 2s dS/Dt. Since it's saying that the area is increasing three times as fast as its side is increasing... 2s must equal to 3. or s = 3/2

is this correct?

You betcha.
 
Thank you, Sir.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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