calculus

A particle is moving along the curve y= 3sqrt3x+1. As the particle passes through the point (5,12), 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.

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  1. Let P(x,y) be any point on the curve
    then we could write P as (x,3√(3x+1))
    and if D is the distance to the origin,
    D^2 = x^2 + (3√(3x+1))^2
    = x^2 + 27x + 9
    2D(dD/dt) = 2x(dx/dt) + 27(dx/dt)

    at (5,12) D^2 = 25+135+9 = 169
    so D = 13

    so dD/dt = (2(5)(3) + 27(3))/(2(13))
    = 4.26

    check my work please

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  2. your answer is good

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