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Homework Help: 2 Problems. 1 regarding implicit differentiation

  1. Oct 21, 2008 #1

    A particle is moving along the curve y=3sqrt{3x+3}. As the particle passes through the point (2, 9), its x-coordinate increases at a rate of 2 units per second. Find the rate of change of the distance from the particle to the origin at this instant.


    Use implicit differentiation to find the slope of the tangent line to the curve
    at the point ( 1,10/-39 ).

    3. The attempt at a solution

    For the first problem I did it several times and I keep coming up with the wrong answer.

    This is my attempt:
    dy/dt = 3(1/2)(3x+3)^(-1/2)(3)(dx/dt)
    dy/dt = 3/(sqrt(3x+3)) (2)
    dy/dt = 3/(sqrt(3(2)+3))
    dy/dt = (3/3) * (2/1) = 2
    dy/dt = 2

    x^2 + y^2 = r^2
    2^2 +9^2 = r^2
    r = sqrt(85)
    2x(dx/dt) + 2y(dy/dt) = 2r(dr/dt)

    2(2)(2) + 2(9)(2) = 2(sqrt(85))dr/dt
    8+36 = 2(sqrt(85))(dr/dt)
    (44/2sqrt(85)) = dr/dt
    That is the wrong answer and I have no idea what I am doing wrong.

    For the second problem

    I did implicit differentiation and came up with

    dy/dx = (3x^2(x+4y)^2+y)/x

    I plugged in my coordinate values and it was still marked wrong.

    Please help me because I really want to understand what I am doing wrong here.

  2. jcsd
  3. Oct 22, 2008 #2
    This might help

    dr/dt=dr/dx * dx/dt, where dx/dt=2

    You are on the right track with x^2 + Y^2= r^2

    For part 2 I would do it a little different. I would cross multiply and find dy/dx at point
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