Results 1 to 4 of 4

Math Help - More Related Rates

  1. #1
    Junior Member
    Joined
    Nov 2006
    Posts
    59

    More Related Rates

    A ruptured oil tanker causes a circular oil slick on the surface of the ocean. When its radius is 150 meters, the radius of the slick is expanding by 0.1 meter/minute and its thickness is 0.02 meter. At that moment:

    (a) How fast is the area of the slick expanding?
    (b) The circular slick has the same thickness everywhere, and the volume of oil spilled remains fixed. How fast is the thickness of the slick decreasing?

    (a) For this one, I'm not sure whether by "area" they mean surface area (since I think we'd be dealing with a cylinder), or just an area of the top circle (the top of the cylinder)?

    (b) I'll come back and work this one if you guys help me with the first one.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    I think, for part a, they're asking for change in area


    A={\pi}r^{2}

    \frac{dA}{dt}=2{\pi}r\frac{dr}{dt}

    You can use the volume of a cylinder formula to find part b.

    Remember, volume is fixed and the answer should be negative because it's decreasing.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2006
    Posts
    59
    all right, let's see..

    See, for part A, i had worked it out as if they were asking for the entire surface area (the area of the cylinder part and the top and bottom of the cylinder all added together)... but, if they're asking for the area of just the top circle, I'd do it like this:

    (a)
    A = \pi r^2
     \frac{dA}{dt} = 2 \pi r \frac{dr}{dt}
     \frac{dA}{dt} = 2* \pi * 150 * 0.1
     \frac{dA}{dt} = 30 \pi \approx 94.2478 m^2/min

    (b)
     V = \pi r^2 h
     h = \frac{V}{ \pi r^2 }
     \frac{dh}{dt} = \frac{-2V}{ \pi r^3} \frac{dr}{dt}
     \frac{dh}{dt} = \frac{-2*( \pi 150^2 0.02 )}{ \pi 150^3 } * 0.1
     \frac{dh}{dt} \approx -0.000026667
    decreasing at 0.000026667 m/min.

    Correct?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Good. That's what I got. Except maybe write your answer as \frac{dh}{dt}=\frac{-1}{37500}. Looks better.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Related Rates
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 26th 2009, 08:54 PM
  2. Rates and Related Rates!!
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 2nd 2008, 10:53 AM
  3. Another Related Rates
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 13th 2008, 06:32 AM
  4. Related Rates
    Posted in the Calculus Forum
    Replies: 6
    Last Post: March 12th 2008, 05:49 PM
  5. rates and related rates
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 29th 2007, 09:51 PM

Search Tags


/mathhelpforum @mathhelpforum