The altitude of a triangle is increasing at a rate of 2.500 centimeters/minute while the area of the triangle is increasing at a rate of 2.000 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 7.000 centimeters and the area is 100.000 square centimeters?

Question

anonymous · Answer

related rate

anonymous · Answer

Note: The "altitude" is the "height" of the triangle in the formula "Area=(1/2)*base*height". Draw yourself a general "representative" triangle and label the base one variable and the altitude (height) another variable. Note that to solve this problem you don't need to know how big nor what shape the triangle really is.

anonymous · Answer

Yes it's related rates but I can't seem to get the right answer for some reason.

anonymous · Answer

|dw:1322612140318:dw|

anonymous · Answer

Derivaitve in respect to time

1/2 hw = A

slaaibak · Answer

$$A= {1\over2} * b * h$$

$$dA/dT = {1\over2}db/dt * h + {1\over2}b * dh/dt$$

I'm not completely sure, but isn't it something like this?

anonymous · Answer

question is looking for db/dt

and you know dA/dt = 2000
dh/dt = 2500
h = 7000
b =  100.000/ 7000 = 100/7

substitute these variables and then find the value of db/dt

slaaibak · Answer

The altitude of a triangle is increasing at a rate of 2.500 centimeters/minute while the area of the triangle is increasing at a rate of 2.000 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 7.000 centimeters and the area is 100.000 square centimeters?




2.000 = 0.5(db/dt)* 7.000 + 0.5 * b * 2.5000

Now you can get B by using:

A = .5 b h

anonymous · Answer

I will try it now. Thank you both a lot!

anonymous · Answer

For some reason I did not get the right answer... Can any of you guys please help?