Calc prob: width of a rectnagle is decreasing at 4cm/s, height is increasing at 2cm/s

[jonathanmiller]

I have a homework problem that states that the width of a rectnagle is decreasing at 4cm/s and the height is increasing at 2cm/s. Find the rate of change of the area of the rectangle when the height is 5cm and the width is 50cm. Thanks for your help.

 

[stapel]

I have a homework problem that states:

The width of a rectangle is decreasing at 4cm/s and the height is increasing at 2cm/s. Find the rate of change of the area of the rectangle when the height is 5cm and the width is 50cm.

You started with the algebra: You drew the rectangle. You labelled the width "w" and the height "h".

Then you moved on to calculus: You labelled the width and height also with their respective growth (or shrinkage) rates. You plugged the variables into the formula for the area of a rectangle, and... then what?

Please be complete. Thank you!

 

[pka]

I have a homework problem that states that the width of a rectnagle is decreasing at 4cm/s and the height is increasing at 2cm/s. Find the rate of change of the area of the rectangle when the height is 5cm and the width is 50cm. Thanks for your help.

Think of each of \(\displaystyle A,~h,~\&~w\) as function of time.
Then \(\displaystyle A=h\cdot w\) so \(\displaystyle \dfrac{dA}{dt}=h\frac{dw}{dt}+w\frac{ds}{dt}\)
Can you finish?

 

[lookagain]

Think of each of \(\displaystyle A,~h,~\&~w\) as function of time.
Then \(\displaystyle A=h\cdot w\) so \(\displaystyle \dfrac{dA}{dt}=h\frac{dw}{dt}+w\frac{ds}{dt} \ \ \ \ \ \) <-----No, that last factor is incorrect.

Can you finish?

\(\displaystyle A=h\cdot w \ \) so \(\displaystyle \ \dfrac{dA}{dt}= \ h\frac{dw}{dt}+w\frac{dh}{dt} \)