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Math Help - Related Rate problem

  1. #1
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    Related Rate problem

    At noon, ship A is 30 nautical miles due west of ship B. Ship A is sailing west at 16 knots and ship B is sailing north at 24 knots. How fast (in knots) is the distance between the ships changing at 3 PM?

    I get that the initial distance between them is 30 on a horizontal line, and then A moves west at x'(t) and B moves north at y'(t) but I think we would use Pythagorean theorem to solve it but I just cant get the set up for it.
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  2. #2
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    Hello, Nikhiln25!

    Did you make a sketch?


    At noon, ship A is 30 nautical miles due west of ship B.
    Ship A is sailing west at 16 knots and ship B is sailing north at 24 knots.
    How fast (in knots) is the distance between the ships changing at 3 PM?
    Code:
                                        * B    - 
                                     *  |      :
                                  *     | 24t  :
                               *        |      :
                      D     *           * S    :
                         *              |   24t+72
                      *                 |      :
                   *                    | 72   :
                *                       |      :
             *                          |      :
          * - - - - * - - - * - - - - - *      -
          A   16t   R  48   P    30     Q
    
          : - - - - - - 16t+78  - - - - :

    At noon, ship A is at P and ship B is at Q\!:\;PQ\,=\,30

    By 3 PM, ship A has sailed 48 miles west to R,
    . . and ship B has sailed 72 miles north to S.

    In the next t hours, ship A sails 16t miles west to A
    . . and ship B sails 24t miles north to B.

    Their distance is: D = AB, the hyptenuse of right triangle ABQ
    . . with sides: . 16t + 78 and 24t + 72.

    Hence: . D \;=\;\sqrt{(16t+78)^2 + (24t + 72)^2} \;=\;\sqrt{832t^2 + 5952t + 11,\!268}

    Then: . \dfrac{dD}{dt} \;=\;\dfrac{1}{2}\,\dfrac{1664t + 5952}{\sqrt{832t^2 + 5952t + 11,\!268}}  \;=\;\dfrac{416t + 1488}{\sqrt{208t^2 + 1488t + 2817}}


    When t=0\!:\;\;\dfrac{dD}{dt} \;=\;\dfrac{1488}{\sqrt{2817}} \;=\;\dfrac{496}{\sqrt{313}} \;=\;28.03557758


    Their distance is increasing at about 28\text{ knots.}

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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, Nikhiln25!

    Did you make a sketch?


    Code:
                                        * B    - 
                                     *  |      :
                                  *     | 24t  :
                               *        |      :
                      D     *           * S    :
                         *              |   24t+72
                      *                 |      :
                   *                    | 72   :
                *                       |      :
             *                          |      :
          * - - - - * - - - * - - - - - *      -
          A   16t   R  48   P    30     Q
    
          : - - - - - - 16t+78  - - - - :

    At noon, ship A is at P and ship B is at Q\!:\;PQ\,=\,30

    By 3 PM, ship A has sailed 48 miles west to R,
    . . and ship B has sailed 72 miles north to S.

    In the next t hours, ship A sails 16t miles west to A
    . . and ship B sails 24t miles north to B.

    Their distance is: D = AB, the hyptenuse of right triangle ABQ
    . . with sides: . 16t + 78 and 24t + 72.

    Hence: . D \;=\;\sqrt{(16t+78)^2 + (24t + 72)^2} \;=\;\sqrt{832t^2 + 5952t + 11,\!268}

    Then: . \dfrac{dD}{dt} \;=\;\dfrac{1}{2}\,\dfrac{1664t + 5952}{\sqrt{832t^2 + 5952t + 11,\!268}}  \;=\;\dfrac{416t + 1488}{\sqrt{208t^2 + 1488t + 2817}}


    When t=0\!:\;\;\dfrac{dD}{dt} \;=\;\dfrac{1488}{\sqrt{2817}} \;=\;\dfrac{496}{\sqrt{313}} \;=\;28.03557758


    Their distance is increasing at about 28\text{ knots.}


    Thank you, that helped a lot. I did draw the diagram and I got to 48 and 72, but I didn't know after that I also had to add 16t and 24 t to the triangle, and that was what was causing me not to get the answer.



    Another one I have a question with is the following;

    Gravel is being dumped from a conveyor belt at a rate of 40 cubic feet per minute. It forms a pile in the shape of a right circular cone whose base diameter and height are always equal to each other. How fast is the height of the pile increasing when the pile is 24 feet high?
    Hint: The volume of a right circular cone with height h and radius of the base r is given by V = \frac{1}{3} \pi r^2 h

    I understand we have to get dh/dt and radius would be 12 in this instance but im messing up on my steps and can't seem to solve for dh/dt
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  4. #4
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    Quote Originally Posted by Nikhiln25 View Post
    Thank you, that helped a lot. I did draw the diagram and I got to 48 and 72, but I didn't know after that I also had to add 16t and 24 t to the triangle, and that was what was causing me not to get the answer.



    Another one I have a question with is the following;

    Gravel is being dumped from a conveyor belt at a rate of 40 cubic feet per minute. It forms a pile in the shape of a right circular cone whose base diameter and height are always equal to each other. How fast is the height of the pile increasing when the pile is 24 feet high?
    Hint: The volume of a right circular cone with height h and radius of the base r is given by V = \frac{1}{3} \pi r^2 h

    I understand we have to get dh/dt and radius would be 12 in this instance but im messing up on my steps and can't seem to solve for dh/dt
    You probably should have started another thread. However, if you draw the cone with the center of the base at the origin, then you can reason that:

    h=y

    The diamter d=2r=y therefore r=\frac{1}{2}y. So now you can write the volume as:

    V=\frac{1}{3}\pi(\frac{1}{2}y)^2y=\frac{1}{12}\pi y^3

    \frac{dV}{dt}=\frac{1}{12}\pi (3y^2)\frac{dy}{dt}

    Now just rearrange and plug in your value of y

    \frac{dy}{dt}=\frac{24}{\frac{3\pi}{12}y^2}

    That should be it!
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