Results 1 to 15 of 15

Math Help - Related Rates Problem...don't know what to do, help.

  1. #1
    Newbie
    Joined
    Mar 2009
    Posts
    11

    Related Rates Problem...don't know what to do, help.

    I don't know what to do at all...I had an idea, but I got stuck, BIG TIME. Please show me what to do step by step, Thanks.

    Sand falls from a hopper at a rate of 0.1 cubic meters per hour and forms a conical pile beneath. If the side of the cone makes an angle Pi/6 radians with the vertical, find the rate at which the height of the cone increases. Give your answer in terms of "h," the height of the pile in meters.

    Thanks for any help.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by ilsj6 View Post
    I don't know what to do at all...I had an idea, but I got stuck, BIG TIME. Please show me what to do step by step, Thanks.

    Sand falls from a hopper at a rate of 0.1 cubic meters per hour and forms a conical pile beneath. If the side of the cone makes an angle Pi/6 radians with the vertical, find the rate at which the height of the cone increases. Give your answer in terms of "h," the height of the pile in meters.

    Thanks for any help.
    sketch a diagram ... half the cone forms a 30-60-90 triangle

    ratio between r and h is ...

    \frac{r}{h} = \frac{1}{\sqrt{3}}

    r = \frac{h}{\sqrt{3}}

    volume of a cone ...

    V = \frac{\pi}{3} r^2 h

    V = \frac{\pi}{3} \left(\frac{h}{\sqrt{3}}\right)^2 h = \frac{\pi}{9}h^3

    take the time derivative of the last equation, then determine \frac{dh}{dt} in terms of h.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2009
    Posts
    11
    So basically, my answer would be:

    h' = sqrt{v'9/3Pi}

    ?

    and since v' = 0.1, I would just compute a numerical answer, which would be approximately 0.3090?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by ilsj6 View Post
    So basically, my answer would be:

    h = SQRT(v'9/3Pi)

    ?

    and since v' = 0.1, I would just compute a numerical answer, which would be approximately 0.3090?
    no ... you're supposed to find \frac{dh}{dt} in terms of h.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2009
    Posts
    11
    Quote Originally Posted by skeeter View Post
    no ... you're supposed to find \frac{dh}{dt} in terms of h.

    I don't understand. Please elaborate...my Calculus Professor never explained this.

    I know that the derivative of the last equation is:

    V' = (3Pih^2)/9 = (Pih^2)/3
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member Pinkk's Avatar
    Joined
    Mar 2009
    From
    Uptown Manhattan, NY, USA
    Posts
    419
    We know the volume of the cone is \frac{1}{3}\pi r^{2}h. Now, we know that there must be a constant relationship between the sides of the right triangle formed by any height h and any radius r, since the angle between the vertical (the height) of the cone and the side of the cone is always \frac{\pi}{6}=30^{\circ}. It's a 30-60-90 triangle! Here is a diagram of the triangle created:



    h=r\sqrt{3} or r=\frac{h}{\sqrt{3}}

    Let's plug that into the cone volume equation:

    V=\frac{1}{9}\pi h^{3}. Now differentiate with respect to t:

    \frac{dV}{dt}=\frac{1}{3}\pi h^{2}\frac{dh}{dt}

    \frac{dV}{dt}=0.1, so plug that in and just solve for \frac{dh}{dt}, which will be in terms of h.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Mar 2009
    Posts
    11
    Quote Originally Posted by Pinkk View Post
    We know the volume of the cone is \frac{1}{3}\pi r^{2}h. Now, we know that there must be a constant relationship between the sides of the right triangle formed by any height h and any radius r, since the angle between the vertical (the height) of the cone and the side of the cone is always \frac{\pi}{6}=30^{\circ}. It's a 30-60-90 triangle! Here is a diagram of the triangle created:



    h=r\sqrt{3} or r=\frac{h}{\sqrt{3}}

    Let's plug that into the cone volume equation:

    V=\frac{1}{9}\pi h^{3}. Now differentiate with respect to t:

    \frac{dV}{dt}=\frac{1}{3}\pi h^{2}\frac{dh}{dt}

    \frac{dV}{dt}=0.1, so plug that in and just solve for \frac{dh}{dt}, which will be in terms of h.

    So I would have this as my answer then?:

    (0.1)/(Pih^2) = dh/dt

    Is this correct?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member Pinkk's Avatar
    Joined
    Mar 2009
    From
    Uptown Manhattan, NY, USA
    Posts
    419
    It would be \frac{.3}{\pi h^{2}}=\frac{dh}{dt}. That is for any height h. If the question did not give to a specific height or radius at which to measure \frac{dh}{dt}, then your answer is just the general formula that I wrote above. Now, if you were told a specific h, you would plug that into the equation. If you were given a specific r, you would convert it into an h value and then plug that into the equation. Remember, h and r are constantly changing; they are variables, not constants.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Mar 2009
    Posts
    11
    Quote Originally Posted by Pinkk View Post
    It would be \frac{.3}{\pi h^{2}}=\frac{dh}{dt}. That is for any height h. If the question did not give to a specific height or radius at which to measure \frac{dh}{dt}, then your answer is just the general formula that I wrote above. Now, if you were told a specific h, you would plug that into the equation. If you were given a specific r, you would convert it into an h value and then plug that into the equation. Remember, h and r are constantly changing; they are variables, not constants.

    Thanks, I re-worked the dh/dt part and got what you wrote. Now if I wanted to find the rate at which the radius increases and write my answer in terms of "h", would I just do the same work, but plug in h=rsqrt3 for the volume equation?

    Just trying to learn every type of question, so that I am prepared for any quiz.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Senior Member Pinkk's Avatar
    Joined
    Mar 2009
    From
    Uptown Manhattan, NY, USA
    Posts
    419
    Yes, you would substitute h=r\sqrt 3, differentiate with respect to t and solve for \frac{dr}{dt}.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie
    Joined
    Mar 2009
    Posts
    11
    I am going to work out the radius version by myself, let me know if this is correct.

    h = rSqrt3

    V = 1/3Pir^2(rSqrt3)

    = (Pi*r^3*Sqrt3)/3

    V' = (9*Pi*Sqrt3*r^2)/9

    = (Pi*Sqrt3*r^2)*(dr/dt)

    Solving for dr/dt:

    dr/dt = 0.1/(Pi*Sqrt3*r^2)

    That is my answer for dr/dt in terms of "h."

    Is this correct? I hope so.

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Senior Member Pinkk's Avatar
    Joined
    Mar 2009
    From
    Uptown Manhattan, NY, USA
    Posts
    419
    I don't know why you suddenly switched to 9, but your work seems to check out.

    V=\frac{\sqrt{3}}{3}\pi r^{3}

    \frac{dV}{dt}= \frac{3\sqrt{3}}{3}\pi r^{2}\frac{dr}{dt}

    \frac{dV}{dt}=\sqrt{3}\pi r^{2}\frac{dr}{dt}

    \frac{0.1}{\sqrt{3}\pi r^{2}} = \frac{dr}{dt}

    And this equation is in terms of r; you eliminated the h term with the substitution h=r\sqrt{3}.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Newbie
    Joined
    Mar 2009
    Posts
    11
    For the 9, I had this as the derivative:

    (3*(3*Pi*Sqrt3*r^2))/(3^2)

    so I just multiplied out the 3s to get the 9s, which canceled out.

    Thank you SO MUCH for everything. I understand these problems much better now.

    One last question, since my answer is in terms of "r," how would I put it in terms of "h?"

    I'm done after this.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Senior Member Pinkk's Avatar
    Joined
    Mar 2009
    From
    Uptown Manhattan, NY, USA
    Posts
    419
    Just rewrite your r term as \frac{h}{\sqrt{3}}, which would eventually simplify to:

    \frac{dr}{dt}=\frac{.3}{\sqrt{3}\pi h^{2}}

    You may notice that \frac{dr}{dt}=\frac{1}{\sqrt{3}}\frac{dh}{dt} or \sqrt{3}\frac{dr}{dt}=\frac{dh}{dt}. Remember how h=r\sqrt{3}. The same ratio applies with their respective derivatives.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Newbie
    Joined
    Mar 2009
    Posts
    11
    Quote Originally Posted by Pinkk View Post
    Just rewrite your r term as \frac{h}{\sqrt{3}}, which would eventually simplify to:

    \frac{dr}{dt}=\frac{.3}{\sqrt{3}\pi h^{2}}

    You may notice that \frac{dr}{dt}=\frac{1}{\sqrt{3}}\frac{dh}{dt} or \sqrt{3}\frac{dr}{dt}=\frac{dh}{dt}. Remember how h=r\sqrt{3}. The same ratio applies with their respective derivatives.
    I figured that.

    Thanks again!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Related rates problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 19th 2010, 07:12 AM
  2. related rates problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 25th 2010, 01:34 PM
  3. Related rates problem -- Am I right??
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 4th 2009, 12:12 AM
  4. related rates problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 22nd 2009, 03:11 PM
  5. Related rates problem..see pic
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 20th 2009, 09:05 AM

Search Tags


/mathhelpforum @mathhelpforum