Find rate of change with implicate derivative

[tttcomrader]

Q: The altitude of a triangle is increasing at a rate of 3.000 centimeters/minute while the area of the triangle is increasing at a rate of 1.000 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 7.000 centimeters and the area is 89.000 square centimeters? (Note: the altitude of a triangle is the "h" in the formula "A=(1/2)bh".

Here is what I got so far:

Given - dh/dt = 3 cm/min; dA/dt = 1 cm^2/min, also A = (1/2)bh

Solve for db/dt when h=7 and A=89

I first find the relation between b and A in respect to h, therefore I have h = 2A/b, but I cannot plot it into the orginal A = (1/2)bh as the two b will cancel each other out. I have already learned to find the db/dt by taking dA/dt after plotting in b in relation to h. So what is the proper steps I need to take to solve this problem?

Thank you very much!

Kan

 

[galactus]

You need to find \(\displaystyle \frac{db}{dt}\) given \(\displaystyle \frac{dh}{dt}=3\) and \(\displaystyle \frac{dA}{dt}=1\), when h=7 and A=89


Solve the area formula for b: \(\displaystyle b=\frac{2A}{h}\)

Now, differentiate with respect to t:

\(\displaystyle \L\\\frac{db}{dt}=\frac{2h\frac{dA}{dt}-2A\frac{dh}{dt}}{h^{2}\)

Enter in your info and solve.

 

[skeeter]

A = (1/2)bh ... all three variables are functions of time

d/dt[A = (1/2)bh] ... use the product rule

dA/dt = (1/2)b(dh/dt) + (1/2)h(db/dt)

solve for db/dt ...

[2(dA/dt) - b(dh/dt)]/h = db/dt

now substitute in your given values ... you'll have to find the length of the base given that the area is 89 cm² and h = 7 cm.