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mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0i 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

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0the derivative of 1/2bh?

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0yes i know the product rule

Psymon
 one year ago
Best ResponseYou've already chosen the best response.0Yep. 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
 one year ago
Best ResponseYou've already chosen the best response.0if 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

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0so not respect to b?

phi
 one year ago
Best ResponseYou've already chosen the best response.0the change in Area with respect to time is also known as the rate you want dA/dt , and so on..

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0I'm confused with finding the derivative.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0alli know is this formula: a=pi^2 r but im guessing it's not relevant for this problem.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0and also a=1/2*b*h

Psymon
 one year ago
Best ResponseYou've already chosen the best response.0This 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.

Psymon
 one year ago
Best ResponseYou've already chosen the best response.0Now 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
 one year ago
Best ResponseYou've already chosen the best response.0to take the derivative of A with respect to time you write a little d in front, and divide by dt

phi
 one year ago
Best ResponseYou've already chosen the best response.0yes that is the left side you have dA/dt = 0.5 * d/dt ( B * H) use the product rule on B*H

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0oh okay so we ignore the 0.5 for now?

phi
 one year ago
Best ResponseYou've already chosen the best response.0You don't ignore constants, but you can keep them on the outside of the derivative just like d/dt (2x) = 2 *dx/dt

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0so here: dA/dt=1/2(dh/dt)+(db/dt)

phi
 one year ago
Best ResponseYou've already chosen the best response.0except that is not the product rule http://www.1728.org/chainrul.htm

phi
 one year ago
Best ResponseYou've already chosen the best response.0the product rule of d/dt (x * y) = x * dy/dt + y * dx/dt

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0@phi sorry connection went off

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0i understand what you're saying up there.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0is this right? dA/dt= 1/2 (b*dh/dt)+(h*db/dt)

phi
 one year ago
Best ResponseYou've already chosen the best response.0is 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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