Question

Suppose the height of a triangle decreases at a rate of 5 feet per minute while the area of the triangle increases at a rate of 8 square feet per minute. At what rate is the base of the triangle changing when the height is 12 feet and the area is 144 square feet? Do not assume you have a right triangle. Include a sketch with variables labeled for each problem.

Answer 1

\[\large A = \cfrac12 b\cdot h\]We're taking everything w.r.t the base, since we'retrying to find the rate of change of the base.. \[\large \frac{dA}{dt}(A)= \frac12 \frac{d}{dt} ( b\cdot h)\]\[\large A' = \frac12 \left( h\cdot \frac{db}{dt}+b \cdot \frac{dh}{dt}\right)\]Now we need to find b.\[\large A=144 \ ft^2 \ , \ h= 12 \ ft\]\[\large 144 = 12b\]\[\large b= 12\]\[\large A' = 8 \frac{ft^2}{min} \ , \ \frac{dh}{dt}= 5\frac{ft}{min} \ , \ h = 12 \ , \ \frac{db}{dt} = \ ?\] plug in all your variables, you shall find your answer :)

Answer 2

taking everything w.r.t time*

Answer 3

Oh, I see.
So is it easier to find each piece separately and then multiply the different derivatives to get the final answer w.r.t.?

Answer 4

Awesome, this is the 3000th question I'm answering in math :D

Answer 5

Yes you are correct.

Answer 6

Not sure why I even needed to finish this off. Looks pretty much done to me.

Answer 7

Thank you so much for showing the steps and the work. It really helped me understand how to approach the problem.

Answer 8

I've got to head out though, and I trust bahrom will aid you in finding the correct answer :)

Answer 9

Okay, thank you.

Answer 10

Good luck!

Answer 11

I am getting 19/3 as an answer.

Answer 12

Can you post your solution for me to take a quick look at?

Answer 13

\[\frac{ 8+\frac{ 1 }{ 2 }(12*5) }{ 6 }=\frac{ 19 }{ 3 }\]

Answer 14

Ok let's see:
\[\large A' = \frac12 \left( h\cdot \frac{db}{dt}+b \cdot \frac{dh}{dt}\right)\]\[8=\frac{1}{2}(12*\frac{db}{dt}+12*5)\]\[8=6\frac{db}{dt}+30\]Something's wrong in Jhan's answer

Answer 15

Oh I see it: \[A=144ft^2=\frac{1}{2}b*h=\frac{1}{2}b*12=6b\]\[144=6b;\space b=24\]

Answer 16

Hey, can you give me a moment to write this out on paper. I'm really not sure how you got positive 19/3 when I keep getting a negative answer when I plug in the numbers. Going to verify the solution.

Answer 17

Okay. That's fine. :)

Answer 18

Nvm I think I found the other mistake:
The height of a triangle decreases at a rate of 5 feet per minute. Since it's decreasing,
\[\frac{dh}{dt}\ne5\frac{ft}{min}\]\[\frac{dh}{dt}=-5\frac{ft}{min}\]

Answer 19

Oh, I see. That explains why my answer is coming out positive.

Answer 20

So finally, we get:
\[\large A' = \frac12 \left( h\cdot \frac{db}{dt}+b \cdot \frac{dh}{dt}\right)\]
\[A'=8\frac{ft^2}{min}\]\[h=12ft\]\[b=24ft\]\[\frac{dh}{dt}=-5\frac{ft}{min}\]

Answer 21

\[8=\frac{1}{2}(12*\frac{db}{dt}+24*(-5))\]\[16=12*\frac{db}{dt}-120\]\[4=3*\frac{db}{dt}-30\]\[34=3*\frac{db}{dt}\]\[\frac{db}{dt}=\frac{34}{3}(\frac{ft}{min})\]

Answer 22

Thank you so much! I never would have noticed those issues.

Answer 23

Now let me solve it out on paper anyway, just to make sure the answer's correct.

Answer 24

Okay.

Answer 25

Yup \[\frac{db}{dt}=\frac{34}{3}(\frac{ft}{min})\] is the final answer. Though I'm not sure how you were getting a positive result with the earlier values.