1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Rate of change problem (differentiation)

  1. Oct 7, 2016 #1
    1. The problem statement, all variables and given/known data
    Refer to the photo, please verify my answer

    2. Relevant equations
    calculus

    3. The attempt at a solution
    For c, can I do it by assuming Ah=V.
    A(dh/dt) + h(dA/dt) = dV/dt then find dA/dt?
     

    Attached Files:

  2. jcsd
  3. Oct 7, 2016 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Don't you have an expression for A(h)?
     
  4. Oct 7, 2016 #3
    A(h) =2 (pi) root[1+(h+1)^2] .... then ?
     
  5. Oct 7, 2016 #4

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Do you have an expression for V(h) then?

    (BTW: very few people will trouble to read images.)
     
  6. Oct 7, 2016 #5
    I don't know what is V(h), I try to assume it as V(h)= A(h)[h] = 2 (pi) root[1+(h+1)^2] h for small interval of h, but I don't know if it is correct.
     
  7. Oct 7, 2016 #6

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Didn't you just do a "volume of a solid of revolution" calculation?
    Didn't you use the method of disks?

    Lessee ... the disk between y and y+dy will have area A(y) and width dy, so it's volume is dV = A(y).dy
    You need the volume between the bottom and y=h ... in the problem "h" is not a "small interval", it is a value of y.

    For the rest ... you are given the volume flow rate of water dV/dt, and you need dh/dt and dA/dt.
    If you know how the height changes with time already, and you know how the area depends on the height ...
     
    Last edited: Oct 7, 2016
  8. Oct 7, 2016 #7
    A(h) =2 (pi) root[1+(h+1)^2], so for c part, can I just differentiate both side, since I know dh/dt = 4/5 in b part.
    dA(h)/ dt = 2(pi)(1/2)(1+9)^(-1/2)(2)(2+1) dh/dt
    so dA(h)/dt = 24pi/5root(10)
     
  9. Oct 7, 2016 #8

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    That looks like the circumference of the circular surface, not the area.
     
  10. Oct 7, 2016 #9
    A(h) = pi(1+(h+1)^2)
    =pi (2+2h+h^2)
    dA(h) / dt = pi(2h+2)dh/dt
    =24/5
    thanks
     
  11. Oct 7, 2016 #10

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    That looks better.


    See what Simon said:
    Notice: It's pretty easy to display your image directly in a post:

    untitled-png.107082.png
     
  12. Oct 7, 2016 #11
    Thank you for the advice from you both. Can you also teach me how to show that "1-t^2/2 <=cost <=1 for 0<=t<=1 "
     
  13. Oct 7, 2016 #12

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    You should start a new thread for this.
     
  14. Oct 7, 2016 #13

    Mark44

    Staff: Mentor

    @Clara Chung, problems involving differentiation should be posted in the calculus subsection, not the precalculus section.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted