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[scarletfan]

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 when the length is 6 inches?

I am stuck at finding w and w'
A = lw
A'= l'w + w'l

Here is what I did.., which is probably wrong.
Since the radius is 5 inches, then the diagonal must be 10 inches.
so l^2 + w^2 = 100
w = 10/l

2ll' + 2ww" = 0
2(6)(-2) 10/3*w' = 0
w' = 24 *3 / 10 = 7.2

A' = -2(5/3) + 7.2(6)
A' = 42.27

The answer was supposed to be decreasing 7 in/s.

 

[Deleted member 4993]

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 when the length is 6 inches?

I am stuck at finding w and w'
A = lw
A'= l'w + w'l

Here is what I did.., which is probably wrong.
Since the radius is 5 inches, then the diagonal must be 10 inches.
so l^2 + w^2 = 100
w = 10/l <<<< It should be

\(\displaystyle w \ = \ \sqrt{100 \ - \ l^2}\)

and so on...

2ll' + 2ww" = 0
2(6)(-2) 10/3*w' = 0
w' = 24 *3 / 10 = 7.2

A' = -2(5/3) + 7.2(6)
A' = 42.27

The answer was supposed to be decreasing 7 in/s.

 

[galactus]

Call the width w instead of x, then the area of the rectangle can be described as:

\(\displaystyle A=2w\sqrt{5^{2}-(\frac{w}{2})^{2}}\)

To see this, draw a line from the center of the circle/rectangle up to the corner of the rectangle. This is also the radius of the circle.

By Pythagoras, the height is \(\displaystyle 2\sqrt{25-\frac{w^{2}}{4}}\)

Differentiate:

\(\displaystyle \frac{dA}{dw}=\sqrt{100-w^{2}}-\frac{w^{2}}{\sqrt{100-w^{2}}}\)

But, \(\displaystyle \frac{dA}{dt}=\frac{dA}{dw}\cdot \frac{dw}{dt}\)

You are given dw/dt=-2, and w=6. Find dA/dw from above using w=6, and finish.