## mathcalculus 2 years ago hello, i have a related rate problem with area. (attached below)

1. mathcalculus

2. mathcalculus

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

the derivative of 1/2bh?

4. mathcalculus

yes i know the product rule

5. 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.

6. 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

7. mathcalculus

so not respect to b?

8. phi

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

9. mathcalculus

I'm confused with finding the derivative.

10. mathcalculus

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

11. mathcalculus

and also a=1/2*b*h

12. 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.

13. mathcalculus

right

14. 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.

15. phi

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

16. mathcalculus

dA/dt

17. phi

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

oh okay so we ignore the 0.5 for now?

19. 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

20. mathcalculus

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

21. phi

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

22. phi

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

23. phi

?

24. mathcalculus

@phi sorry connection went off

25. mathcalculus

i understand what you're saying up there.

26. mathcalculus

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

27. mathcalculus

@Psymon

28. mathcalculus

@phi

29. 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