A particle is moving along the curve below.
y = sqrt(x)
As the particle passes through the point (4,2), its x-coordinate increases at a rate of 4 cm/s. How fast is the distance from the particle to the origin changing at this instant?

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  1. let P(x,y) be any point on the curve
    y = x^(1/2)

    dy/dt = (1/2)x^(-1/2) dx/dt

    when x-4 and y=2 and dx/dt=4

    dy/dt = (-1/2)(4^(-1/2)(4)
    = ...

    you finish it

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  2. A particle is moving along the curve y = 2 √{3 x + 7}. As the particle passes through the point (3, 8), its x-coordinate increases at a rate of 4 units per second. Find the rate of change of the distance from the particle to the origin at this instant.

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  3. Well, we know x=3 and y=8 and dx/dt=4. We need to find z (the distance from the particle to the origin) dy/dt, and finally our answer dz/dt.
    First solve for dy/dt
    Now we can draw a right triangle, with x=3 and y=8, and using the pythagorean theorem, z=8.544
    So we have z^2=x^2+y^2

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