A point P is moving along the curve whose equation is
y=((x^3)+17)^(1/2)
When P is at (2,5), y is increasing at the rate of 2units/s. How fast is x changing?

Question

A point P is moving along the curve whose equation is 
y=((x^3)+17)^(1/2)
When P is at (2,5), y is increasing at the rate of 2units/s.  How fast is x changing?

amistre64 · Answer

i would simply take the derivative in an implicit manner

$$y=(x^3+17)^{1/2}$$
$$y'=\frac{2x^2x'}{2}(x^3+17)^{-1/2}$$
we know y' so lets solve for x'
$$y'=x^2x'(x^3+17)^{-1/2}$$
$$y'x^{-2}(x^3+17)^{1/2}=x'$$
$$2(2)^{-2}(2^3+17)^{1/2}=x'$$
$$2^{-1}(2^3+17)^{1/2}=x'$$