hello, i have a related rate problem with area. (attached below)

- anonymous

hello, i have a related rate problem with area. (attached below)

- chestercat

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- anonymous

##### 1 Attachment

- anonymous

i first started the problem like this:
facts:
altitude: increasing 2.000 A=4.000
altitude: increasing 11.500 A=82.000
so Area= (1/2) * base* height
At what rate is the base of the triangle changing when the altitude is 11.500 centimeters and the area is 82.000 square centimeters?
82.000=(1/2)*base* 11.500
82.000=5.75b
14.2609=b
?? help

- anonymous

the derivative of 1/2bh?

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## More answers

- anonymous

yes i know the product rule

- Psymon

Yep. Thats what these related rate problems require. If you have a change in area then you definitely need an area equation. And since the area is changing, you know its the derivative of the area. And yeah, the derivative of 1/2bh.

- phi

if it helps, the area A, base B and height H are all functions of time (they are changing)
A(t) = 0.5 * H(t) * B(t)
take the derivative with respect to time

- anonymous

so not respect to b?

- phi

the change in Area with respect to time is also known as the rate
you want dA/dt , and so on..

- anonymous

I'm confused with finding the derivative.

- anonymous

alli know is this formula: a=pi^2 r but im guessing it's not relevant for this problem.

- anonymous

and also a=1/2*b*h

- Psymon

This is a triangle, though, so we don't even have pi. But yes, you would use 1/2bh since its the area of a triangle, which is what we have.

- anonymous

right

- Psymon

Now its just that every time you take the derivative of a variable, you also write dx/dt, dy/dt, etc along with it. These dx/dt things represent the changes given in your problem. So if you decided to make the altitude h, when you take the derivative of it, you would have dh/dt. This dh/dt is what gets replaced with 2.000cm/min value given. But if I'm confusing ya I'm sure phi can help.

- phi

to take the derivative of A with respect to time you write a little d in front, and divide by dt

- anonymous

dA/dt

- phi

yes that is the left side
you have
dA/dt = 0.5 * d/dt ( B * H)
use the product rule on B*H

- anonymous

oh okay so we ignore the 0.5 for now?

- phi

You don't ignore constants, but you can keep them on the outside of the derivative
just like d/dt (2x) = 2 *dx/dt

- anonymous

so here: dA/dt=1/2(dh/dt)+(db/dt)

- phi

except that is not the product rule
http://www.1728.org/chainrul.htm

- phi

the product rule of d/dt (x * y) = x * dy/dt + y * dx/dt

- phi

?

- anonymous

@phi
sorry connection went off

- anonymous

i understand what you're saying
up there.

- anonymous

is this right? dA/dt= 1/2 (b*dh/dt)+(h*db/dt)

- anonymous

- anonymous

- phi

is this right? dA/dt= 1/2 (b*dh/dt)+(h*db/dt)
yes, but the 1/2 multiplies both terms
\[ \frac{dA}{dt}= \frac{1}{2}\left( b\frac{dh}{dt} + h\frac{db}{dt} \right)\\ 2\frac{dA}{dt}= b\frac{dh}{dt} + h\frac{db}{dt} \]
now fill in the various quantities and solve for db/dt

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