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Posted by on Wednesday, October 5, 2011 at 12:38am.

A particle is moving along the curve y=5sqrt(3x+1). As the particle passes through the point (5,20) its x-coordinate increases at a rate of 3 units per second. Find the rate of change of the distance from the particle to the origin at this instant.

  • calculus - , Wednesday, October 5, 2011 at 1:01am

    dx/dt = 3 units/s at the time of interest
    dy/dt = dy/dx * dx/dt = 3 dy/dx

    dy/dx = (5/2)/sqrt(3x+1) = (5/2)/4 = 5/8

    Compute dy/dt and then use
    dR/dt = sqrt[(dy/dt)^2 + (dx/dt)^2] ]
    sqrt[3^2 + (15/8)^2] = 3.538

  • calculus - , Wednesday, October 5, 2011 at 1:38am

    I just plugged it in and it says it isn't right. =/

  • calculus - , Wednesday, October 5, 2011 at 8:23am

    drwls left out the factor of 3 in the derivative, it should have been

    dy/dx = (5/2)(3) / √(3x+1)
    so when x=5
    dy/dx = (15/2) / √16 = 15/8

    take it from there

  • calculus - , Wednesday, October 5, 2011 at 10:35am

    Thanks for the correction, Reiny :-)
    I am getting sloppier in my old age

  • calculus - , Monday, October 17, 2011 at 10:53pm

    at the point .

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