# Thread: Changing dimension in a rectangle

1. ## Changing dimension in a rectangle

so where do i start for this problem?

the length l of a rectangle is decreasing at the rate of 2 cm/sec while the width w is increasing at the rate of 2 cm/sec. when l=12cm and w= 5cm, find the rates of change of area, perimeter, and the lengths of the diagonals of the rectangle.

2. Originally Posted by cyberdx16
so where do i start for this problem?

the length l of a rectangle is decreasing at the rate of 2 cm/sec while the width w is increasing at the rate of 2 cm/sec. when l=12cm and w= 5cm, find the rates of change of area, perimeter, and the lengths of the diagonals of the rectangle.
First we know that,
A=w*l

Now A,w,l are function of t (time)
Thus, (differenciating and using product rule),
dA/dt=w*dl/dt+l*dw/dt
We are given that,
dl/dt=-2 cm/sec
dw/dt=2 cm/sec
And, w=5 cm with l=12 cm
Thus,
dA/dt=(5)(-2)+12(2)=10 cm^2/sec

For perimeter consider,
P=2w+2l
Thus,
dP/dl=2(dw/dt)+2(dl/dt)
Thus,
dP/dl=2(2)+2(-2)=0 cm/sec
Which makes sense since if you are increasing one and decreasing one at the same rate there is no change in rate.

For the diagnol, (by Pythagorean theorem)
d^2=l^2+w^2
Thus, (implicit differenciation)
2d*(dd/dt)=2l(dl/dt)+2w(dw/dt)
When (l,w)=(5,12) we have d=13
2(13)(dd/dt)=2(5)(-2)+2(12)(2)
Thus,
26(dd/dt)=28
Thus,
dd/dt=28/26 cm/sec