Expert: Paul Klarreich - 12/15/2006

Question
A square is inscribed in a circle.  As the square expands, the circle expands to maintain the four parts of intersection.  The perimeter of the square is increasing at the rate of 8 inches per second.
a) Find the rate at which the circumference of the circle is increasing.
b) At the instant when the area of the square is 16 square inches, find the rate at which the area enclosed between the square and the circle is increasing.

Answer
Questioner:  Stephanie
Category:  Calculus
 
Subject:  Related rates
Question:  A square is inscribed in a circle.  As the square expands, the circle expands to maintain the four parts of intersection.  The perimeter of the square is increasing at the rate of 8 inches per second.

a) Find the rate at which the circumference of the circle is increasing.

b) At the instant when the area of the square is 16 square inches, find the rate at which the area enclosed between the square and the circle is increasing.
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Hi, Stephanie,

I reworded your Question: Compare it with the original and make sure you understand why it is the same.  

Four points are marked on a circle to form a square.  As the circle expands, one side of that square increases at the rate of 2 inches per second.

a) Find the rate at which the circumference of the circle is increasing.

b) At the instant when one side of the square is 4 square inches, find the rate at which the area enclosed between the square and the circle is increasing.

...........................
Is this your first attempt at R-R problems?  If so, the scheme is something like this:

1. Identify the variables in the problem -- the things that change.  Give them names.

2. Write their rates of change as derivatives WITH RESPECT TO time.  Note which are known and which is to be found.

3. Determine a relationship (yes, it is called 'related rates' for a reason) between the variables.  Use a diagram, use your life experience, use your general knowledge and brilliance, do whatever you have to.  This is the key step.

4. Now differentiate implicitly, then substitute the known quantities and rates, and solve for the unknown rate.




In this case, (part a, that is) the variables are:

s = length of one side of the square.
r = radius of the circle.
C = circumference of the circle.

The rates are:

ds/dt = the rate of increase of a side, GIVEN as  2 in/sec.
dr/dt = the rate of increase of the radius.
dC/dt = the rate of increase of the circumference, TO BE FOUND.

Now, how about a relationship:

C = 2 pi r.

But we want that in terms of s, don't we?  Alas, I can't draw diagrams for

you, so you will have to go with my description.  

Draw the square and mark one side as s.
Mark the diagonal as  2r.  Yes, 2r, because the diagonal of the square is the

same as the diameter of the circle, or 2r.

Now s^2 + s^2 = (2r)^2
   2 s^2 = 4 r^2
     s^2 = 2 r^2
       r = s/sqrt(2)

OK, now:  C = 2 pi (s/sqrt(2))

C = sqrt(2) pi s

Ready to go.  Differentiate:

dC/dt = sqrt(2) pi ds/dt

Substitute numbers:

dC/dt = sqrt(2) pi (2) = 2 sqrt(2) pi.
............... DONE WITH PART a ..............

For part b, we have these variables:

s = length of one side of the square.
r = radius of the circle.
A = area between the square and circle.

The rates are:

ds/dt = the rate of increase of a side, GIVEN as  2 in/sec.
dr/dt = the rate of increase of the radius.
dA/dt = the rate of increase of that area, TO BE FOUND.

Relationships:

A = circle area - square area
A = pi r^2 - s^2

Get rid of one of those:

s^2 = 2 r^2, remember?

A = pi r^2 - 2r^2 = (pi - 2)r^2

Differentiate:

dA/dt = 2(pi - 2)r dr/dt    <<<< correction.

Substitute values.  We have  s = 4.

And, since  r = s/sqrt(2),  dr/dt = (ds/dt)/sqrt(2) = 2/sqrt/2     <<<< correction.

Small difficulty.  We don't have a value of r.  
Resolution:  Use one of the relations to find it.

r = s/sqrt(2) = 4/sqrt(2) = 2 sqrt(2)

Ready now:

dA/dt = 2(pi - 2)(2 sqrt(2))(2/sqrt(2))     <<<< correction.
dA/dt = 8(pi - 2)      <<<< correction.

I think that does it.