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Related Rates Question

  1. Jul 15, 2007 #1

    danago

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    Consider the graph of y=x2. A point is moving along the x-axis in such a way, that its speed is proportional to its distance from the origin. At the same time, a point is moving along the curve of the graph, which always has the same x-value as the point moving along the x-axis. At what rate is the point on the graph moving away from the origin?

    At any time, the horizontal distance of the point is 'x' units, and the vertical distance is 'x2' units. By using pythagoras' theorem, i can show that the distance from the origin is thus given by:

    [tex]
    D = \sqrt {x^2 + x^4 }
    [/tex]

    Now, since the point moving along the x-axis has a speed proportional to its horizontal distance, i can say that:

    [tex]
    \frac{{dx}}{{dt}} \propto x\therefore\frac{{dx}}{{dt}} = kx
    [/tex]

    Where 'k' is some constant.

    Since both 'D' and 'x' are functions of time, i differentiate implicitly with respect to time, to get:

    [tex]
    \frac{{dD}}{{dt}} = \frac{{kx^2 + 2kx^4 }}{{\sqrt {x^2 + x^4 } }}
    [/tex]

    Therefore, for any given distance of the first point from the origin, the second point, moving along the curve, has a speed of [itex]
    \frac{{kx^2 + 2kx^4 }}{{\sqrt {x^2 + x^4 } }}
    [/itex] relative to the origin.


    I wasnt 100% sure if i did it correctly. If anybody would be able to double check my working, that would be great.

    Thanks,
    Dan.
     
  2. jcsd
  3. Jul 15, 2007 #2

    Dick

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    Ok, so dx/dt=kx. Then dx/x=k*dt. Solve that ode and then rethink the problem. You are currently a ways off from a correct solution.
     
  4. Jul 15, 2007 #3

    danago

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    Alright. Solving that i get:

    [tex]
    x = e^{kt + c} = x_0 e^{kt}
    [/tex]

    Where x0=ec which is equal to the initial horizontal displacement?

    Do i then just substitute that value of x into the equation i came up with in my first post?
     
  5. Jul 15, 2007 #4
    Calc 1 or 2? I did related rates in Calc 1 but I didn't do a problem anything close to that ...
     
  6. Jul 16, 2007 #5

    danago

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    Well its highscool calc. Im from West Australia, and all we can study is "calculus". Theres no calc 1 or calc 2.
     
  7. Jul 16, 2007 #6

    HallsofIvy

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    I am confused by Dick's response. You are given dx/dt and asked to find dD/dt. You can do that using the chain rule. I see no reason to solve for x as a function of t and then D as a function of t.
    What you did looks perfectly good to me.

    Even simpler, though, would be to use D2= x2+ x4 so that 2D dD/dt= (2x +4x3)(kx) Then
    [tex]\frac{dD}{dt}= \frac{kx^2+ 2kx^4}{D}[/tex]
    which is exactly what you have.

    Perhaps Dick is thinking you must find the rate of change of D in terms of t only. In that case, you can take the x(t)= x0ekt and substitute that into the answer you originally got. That would be exactly the same as substituting it into the distance function and differentiating that.
     
  8. Jul 16, 2007 #7

    Dick

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    Halls is quite correct, finding an explicit function of t is completely unnecessary. Sorry not to recognize an already correct solution.
     
  9. Jul 16, 2007 #8

    danago

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    Ahhh i was wondering where you were going with that. No harm done though, all with good intentions :smile: Thanks for the help guys!
     
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