## mathcalculus Group Title hello, i have a related rate problem with area. (attached below) 11 months ago 11 months ago

1. mathcalculus Group Title

2. mathcalculus Group Title

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

3. mathcalculus Group Title

the derivative of 1/2bh?

4. mathcalculus Group Title

yes i know the product rule

5. Psymon Group Title

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.

6. phi Group Title

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

7. mathcalculus Group Title

so not respect to b?

8. phi Group Title

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

9. mathcalculus Group Title

I'm confused with finding the derivative.

10. mathcalculus Group Title

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

11. mathcalculus Group Title

and also a=1/2*b*h

12. Psymon Group Title

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.

13. mathcalculus Group Title

right

14. Psymon Group Title

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.

15. phi Group Title

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

16. mathcalculus Group Title

dA/dt

17. phi Group Title

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

18. mathcalculus Group Title

oh okay so we ignore the 0.5 for now?

19. phi Group Title

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

20. mathcalculus Group Title

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

21. phi Group Title

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

22. phi Group Title

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

23. phi Group Title

?

24. mathcalculus Group Title

@phi sorry connection went off

25. mathcalculus Group Title

i understand what you're saying up there.

26. mathcalculus Group Title

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

27. mathcalculus Group Title

@Psymon

28. mathcalculus Group Title

@phi

29. phi Group Title

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