# Thread: Rates of Change involving a triangle.

1. ## Rates of Change involving a triangle.

The altitude of a triangle is increasing at the rate of 1cm/min while the area of the triangle is increasing at the rate of 2cm^2/min. At what rate is the base of the triangle changing when the altitude is 10cm and the area is 100cm^2?

I can't seem to get a single answer as I have got a few possible answers from -1.6cm/min, 2cm/min and 0.08cm/min. It depends on how you split up the db/dt notation to include the height and area variables.

I have trouble deciding which way to split the notation up. i.e. I get db/dt=db/dA*dA/dh*dh/dt which yeilds an answer of 2cm/min or db/dt=db/dh*dh/dA*dA/dt which yeilds an answer of 0.08cm/min or
dA/dt = [b×dh/dt + db/dt×h] which yeilds an answer of -1.6cm/min.

dA/db=5 and b=20 substituting in the known h and A values. db/dA then goes on to equal 2/5.

If anyone could see what they think it would be great. I think the answer may be 0.08cm/min as this seems to link the most variables. What does anyone think?

2. First, we take all the information given:

\begin{aligned}
\frac{dh}{dt} &= 1\,\frac{\mbox{\small cm}}{\mbox{\small min}} \\
\frac{dA}{dt} &= 2\,\frac{\mbox{\small cm}^2}{\mbox{\small min}} \\
h &= 10\,\mbox{\small cm} \\
A &= 100\,\mbox{\small cm}^2.
\end{aligned}

Then we remember the formula for the area of a triangle:

$A=\frac{1}{2}bh.$

Knowing the information above, we differentiate both sides using the Product Rule:

\begin{aligned}
\frac{dA}{dt}&=\frac{d}{dt}\left(\frac{1}{2}bh\rig ht)\\
2&=\frac{1}{2}\frac{db}{dt}h+\frac{1}{2}b\frac{dh} {dt}\\
2&=5\frac{db}{dt}+\frac{1}{2}\left(\frac{2A}{h}\ri ght)\\
2&=5\frac{db}{dt}+10\\
-\frac{8}{5}\,\frac{\mbox{\small cm}}{\mbox{\small min}}&=\frac{db}{dt}.
\end{aligned}