Related Rates problem involving triangle

[biochem850]

Homework Statement

"At a given instant the legs of a right triangle are 8in. and 6in., respectively. The first leg decreases at 1in/min and the second increases at 2in/min. At what rate is the area increasing after 2 minutes?"

Homework Equations

A=\frac{1}{2}bh

\frac{db}{dt}=-1

\frac{dh}{dt}=2

The Attempt at a Solution

A=\frac{1}{2}bh

\frac{dA}{dt}=\frac{1}{2}(\frac{db}{dt}*h+b*\frac{dh}{dt})

\frac{dA}{dt}=\frac{1}{2}(-1*10+6*2)

\frac{dA}{dt}=\frac{1}{2}(2)=1

Therefore the area is increasing at a rate of \frac{1in^{2}}{min} after 2 minutes. Is my reasoning sound (I'm pretty sure my answer is correct but I want to be sure that my work is logical)?

 

[scurty]

I see no problem with this, looks nice!

 

[Mark44]

Homework Statement

"At a given instant the legs of a right triangle are 8in. and 6in., respectively. The first leg decreases at 1in/min and the second increases at 2in/min. At what rate is the area increasing after 2 minutes?"

Homework Equations

A=\frac{1}{2}bh

\frac{db}{dt}=-1

\frac{dh}{dt}=2

The Attempt at a Solution

A=\frac{1}{2}bh

\frac{dA}{dt}=\frac{1}{2}(\frac{db}{dt}*h+b*\frac{dh}{dt})

\frac{dA}{dt}=\frac{1}{2}(-1*10+6*2)

\frac{dA}{dt}=\frac{1}{2}(2)=1

Therefore the area is increasing at a rate of \frac{1in^{2}}{min} after 2 minutes. Is my reasoning sound (I'm pretty sure my answer is correct but I want to be sure that my work is logical)?

Your work is logical and mostly correct, but you have a small mistake. The two legs are 8" and 6", not 10" and 6" as you show in your work. The hypotenuse is 10", but it doesn't enter into this problem.

 

[biochem850]

I thought that you're supposed input the length of the two legs after 2 minutes (and according to the derivatives for both legs this would be 6 and 10 after a 2 minute interval)?

Your supposed to input the original lengths?

 

[Mark44]

Sorry, I missed that "after 2 minutes" part in the first post. Your work is fine.