Related Rates triangle area

[NewsboysGurl91]

1. A rectangle has a constant area of 200 square meters and its length L is increasing at the rate of 4 meters per second. Find the width W at the instant the width is decreasing at the rate of 0.5 meters per second.



2. Know: dL/dt= 4 m/s, dW/dt=.5 m/s Want: W



3. LW=A. Then DL/dt * DW/dt = 200 m^2. How do you get W from this?

I have some other questions too. Related rates are hard.

 

[HallsofIvy]

1. A rectangle has a constant area of 200 square meters and its length L is increasing at the rate of 4 meters per second. Find the width W at the instant the width is decreasing at the rate of 0.5 meters per second.



2. Know: dL/dt= 4 m/s, dW/dt=.5 m/s Want: W



3. LW=A. Then DL/dt * DW/dt = 200 m^2. How do you get W from this?

I have some other questions too. Related rates are hard.

dL/dt*dw/dt= 200 can't possibly be correct, can it? 4*0.5 is not 200! Recheck your differentiation- particularly the "product rule"! Also, it is the area that is the constant 200, not the rate of change. What is the derivative of a constant?

 

[NewsboysGurl91]

Never mind, I totally forgot about the product rule.

 

[NewsboysGurl91]

Okay, I got stuck again.
DL/dt*W+ L*DW/dt = 0.
4*W + L* -.5 = 0. How do you find L? Once you find L, isn't it obvious what W will be? Then you wouldn't have to go through this whole rate problem.

 

[teken894]

use the two equations

solve the original area equation and differential equation simultaneously

That is,

A = wl
0 = 4w - .5l

solve the above for w.

 

[NewsboysGurl91]

Kay, thanks.