Expert: Ahmed Salami - 1/16/2006

Question
The length of a rectangle is changing at a rate of 2 cm/s, while the width is changing is such a way that the are remains constant at 20 cm^2.  What is the rate of change of the perimeter when the length is 5 cm?

Answer
Hi Aarti,
Take the width and length of the rectangle to be x and y
respectively and the formula for the area of a rectangle
A = xy
Since A is constant at 20cm^2, we have
20 = xy
y = 20/x
dy/dx = -20/x^2
But we know that dy/dx = (dy/dt)(dt/dx)
and dy/dt = 2cm/s
Therefore,
-20/x^2 = 2(dt/dx)
dt/dx = -10/x^2
dx/dt = -x^2/10
Or we could have simply by differentiating both sides
with respect to time, t had
dy/dt = (-20/x^2)dx/dt
2 = (-20/x^2)dx/dt
dx/dt = -x^2/10
Moving on, the formula for the perimeter is
P = 2(x+y)
dP/dt = 2(dx/dt + dy/dt)
     = 2(-x^2/10 + 2)
     = -x^2/5 + 4
At y = 5, x = 4 (since xy = 20)
Therefore, at y = 5
dP/dt = -4^2/5 + 4
     = -16/5 + 4
     = -3.2 + 4
     = 0.8cm/s
I hope i have helped. You can always get back to me.
Regards.