Question

Ok I am now officially stressed. PLEASE HELP!!!! I can't get my pdf to attach but I am trying another. The second part of a related rates problem says:What is the rate of change of the surface area when the water is .5 meter deep? The solution to the first part: how fast is the water level changing when the water level is .5 meter from the bottom of the trough. They give you the radius: radius = 1meter, and the water is flowing out of a hole in the bottom at a rate of1.5m^3/hr

Answer 1

please help My test is tomorrow and the professor promissed there would be a problem like this on the test so I need help figuring the second part out.

Answer 2

if you got an answer for the first part the 2nd part should be easy just plug the value you have into the answer you got from the first part

Answer 3

no the problem is asking for the rate of change of the surface area. not the surface area itself. I actually know the answer but cannot legitametly arrive at that answer

Answer 4

if you have a problem with the "related rates" in general you can use this link http://www.youtube.com/user/patrickJMT/videos?query=related+rates if only in this specific problem wait for someone to help, because i am not good at these kind of problems i dont wwant to confuse you

Answer 5

I have about four hours to sleep before my test. I am going to bed but if anyone can help I would reall appreciate it.

Answer 6

just watch couple of videos before your exam, believe me they are incredibly good

Answer 7

i got 2 tests tomorrow, 2 presentations, a paper to turn in, ..... its a busy day tomorrow

Answer 8

Ok thanks but again my professor said this exact problem with different values would be on there. she said it is a very difficult one so doesn't mind giving us the heads up

Answer 9

I understand being busy. I have a son a disabled parent a mortgage and a business to take care of on top of full time school

Answer 10

:) so what about this problem is still giving you issues

Answer 11

I paid a tutor to explain it to me. But now the second part asks me to find the rate of change of the surface area when the water level is.5

Answer 12

the surface is just a rectangle right?

Answer 13

yes so length times width

Answer 14

Sa = 5*bc

Sa' = 5*bc'

Answer 15

if we can define bc in terms of volume or height or ....

Answer 16

That was what I thought. so I used the fact that BD is the square root of three over two and tried to differentiate using the formula A=sin(theta)times two times 5

Answer 17

so then we would need to determine the rate of change of theta after that ....

Answer 18

how can we pick this from the change in volume

Answer 19

If you look at the pdf I already have the fact that the rate of change of the angle OBC is -2/5. I am not sure how to relate that to the angle OBD

Answer 20

lets call bc, w for width

how can we define volume in terms of w; we know volume is changing so if we can relate that with the width we can solve for w'

Answer 21

i see it there in the middle

Answer 22

and you call theta the sweep of the central angle above the water right?

Answer 23

yes

Answer 24

this is what i would call theta but if we use yours then half the theta; sin it; and double the results .... right?