Rate of the change question

[Roark]

Hi, I've spent a week on this problem, and I think the answer in the book is wrong. Here it is: A rectangle is inscribed in a circle of radius 5 inches. If the length of the rectangle is decreasing at the rate of 2 inches per second, how fast is the area changing at the instant when the length is 6 inches. HINT: A diagonal of the rectangle is the diameter of the circle. The answer given is -7 sq in/sec.

 

It is not clear whether you are to find the rate of change of the area of the circle or the rectangle. First consider the rectangle: The width is 8 making the area 48. After 1.5 seconds the length is 3, width is 4 and area is 12. So the answer given of -7 in/sec can't be right since 48 - 21/2 does not equal 12.

 

Consider the circle. Initially the area is 25pi. After 1.5 seconds the radius is 5/2 so the area is 25/4pi. Again the answer given can't be.

 

For the rectangle I get an answer of -32 sq. in/sec. This bears out with my post 1.5 second test. For the circle I found dr/dt to be -5/3 and dA/dt = -50/3. The dr/dt bears out as well, but the dA/dt can't be right since after 1.5 seconds -50/3 * 1.5 = -25 and 25pi - 25 does not equal 25/4pi. I'm stumped. Thanks in advance for any help.

[Dave]

Hmm.

 

I did this question and got their answer. I drew the rectangle inscribed in the circle and called the width w. If you do a bit of trig etc (which I can help you with if you want - it's not all that hard) then you get the area of the rectangle (which I called A) to be:

 

A = 2w*sqrt(25-(w^2/4)) = w * sqrt(100 - w^2)

 

Now by differentiating, dA/dw = sqrt(100 - w^2) - w^2/(sqrt(100 - w^2))

 

But dA/dw = dA/dt * dt/dw (by the chain rule), and we know at this particular instant in time that dw/dt = -2 and w = 6, so by substituting in, we have:

 

dA/dt = -2 * (8 - 9/2) = -7 sq. inches per sec

 

I'm not entirely sure what you've done tbh, but it may have something to do with assuming starting values and whatnot. You definately need to differentiate - perhaps you're getting confused with the unit length and the rate of change of the unit length. I don't really know

[Roark]

Hi, you mixed up w and l but I get it. Express w in terms of l and differentiate since you know dl/dt. Here's what I don't get:

 

A rectangle inscribed in a circle looks like this: ([]). We know that when l=6, w=8 and A=48. If l=6 and is decreasing at a rate of -2/sec then after 1.5 seconds: l=6-(1.5*2) = 3. If A is decreasing at -7/sec then after 1.5 secs A=48-(1.5*7) = 37.5. This means w = 37.5/3 = 12.5 which means the area of the circle is getting bigger.

 

What in this problem tells you what is going on? Just because the length is decreasing at a certain rate, the width could be increasing or decreasing at any rate. I assumed that the rectangle was shrinking in porportion to itself. I drew both diagonals and kept the inner angles constant. It looks like you kept the diagonal constant yet after 1.5 secs it has grown to 12.855.

 

What's up with that? ...and thank you.

[Dave]

Originally posted by Roark

Hi, you mixed up w and l but I get it. Express w in terms of l and differentiate since you know dl/dt. Here's what I don't get:

 

A rectangle inscribed in a circle looks like this: ([]). We know that when l=6, w=8 and A=48. If l=6 and is decreasing at a rate of -2/sec then after 1.5 seconds: l=6-(1.5*2) = 3. If A is decreasing at -7/sec then after 1.5 secs A=48-(1.5*7) = 37.5. This means w = 37.5/3 = 12.5 which means the area of the circle is getting bigger.

 

What in this problem tells you what is going on? Just because the length is decreasing at a certain rate, the width could be increasing or decreasing at any rate. I assumed that the rectangle was shrinking in porportion to itself. I drew both diagonals and kept the inner angles constant. It looks like you kept the diagonal constant yet after 1.5 secs it has grown to 12.855.

 

What's up with that? ...and thank you.

 

I think I can see where you're coming from. In the question it tells you that "A diagonal of the rectangle is the diameter of the circle." and also that the rectangle is inscribed in the circle (i.e. always in it). So basically, when the width of the rectangle decreases, it's going to mean that the height will increase to ensure that it is still inscribed.

 

As for the other question about the rate of change of height - it doesn't matter. We basically have one important formula, which is the area of the rectangle, A = w*h. We don't want the h in there, so we're going to have to find a formula to get it. We know that the rectangle is always going to be touching the circle so we have one length of 5 inches from the circle's circumference to its centre, O. From O to the upper line of the rectangle the height is going to be (h/2). From the upper line to the circle's circumference we have a width (w/2). So by using Pythagoras's Theorem, we can find a formula of h in terms of w and substitute it into the formula for A, differentiate and do all the other stuff.

 

If you need more help, just ask.

[Roark]

Interesting. I assumed that the circle would have to be shrinking as the length of the rectangle decreased. You're assuming that the circle size is staying fixed and, after re-reading the problem, I think you're right. The radius of the circle should stay at 5. Indeed, in you're solution you fixed the radius at 5 instead of leaving it variable.

 

But there's a problem. The answer doesn't check out. Consider:

 

t0:l=6, w=8, A(rect)=48, r=5, d=10, dl/dt=-2

 

t1(one second later):l=4, A(rect)=48-7=41, w=41/4=10.25, d=11, r=5.5

 

For the area of the rectangle to be decreasing at -7, the circle would have to be getting bigger.

 

What is the correct answer if the circle doesn't change?

[Dave]

Originally posted by Roark

t0:l=6, w=8, A(rect)=48, r=5, d=10, dl/dt=-2

 

t1(one second later):l=4, A(rect)=48-7=41, w=41/4=10.25, d=11, r=5.5

 

For the area of the rectangle to be decreasing at -7, the circle would have to be getting bigger.

 

What is the correct answer if the circle doesn't change?

 

The rate of change of area isn't constant - it depends on the width of the rectangle. The answer quoted is only for a particular instant in time.

[Roark]

Got it! The rate of change of the area of the rectangle is -7 only when L=6. Incidently, the rate of change of the W is 1.5 only when L=6.

 

How would you go about testing it?