A particle moves along the curve below.
y = sqrt(1 + x^3)
As it reaches the point (2, 3), the y-coordinate is increasing at a rate of 12 cm/s. How fast is the x-coordinate of the point changing at that instant?

Question

A particle moves along the curve below.
y = sqrt(1 + x^3)
As it reaches the point (2, 3), the y-coordinate is increasing at a rate of 12 cm/s. How fast is the x-coordinate of the point changing at that instant?

zepdrix · Answer

This is calculus question, yes?
Have you learned about derivative shortcuts yet?

anonymous · Answer

well yeah, however word problems are not my strong suit.

zepdrix · Answer

So our derivative gives us something like this.
$$\Large\rm y=\sqrt{1+x^3},\qquad y'=\frac{3x^2}{2\sqrt{1+x^3}}$$
Evaluate the derivative at x=2, you're get a numerical value.
This represents the slope of the function at the point (2,3).

If the instantaneous rate of change of y, $\Large\rm dy$ is 12 at that point,
then what is the instantaneous rate of change of x?$$\Large\rm \frac{dy}{dx}=y'(2)$$
$$\Large\rm \frac{12}{dx}=y'(2)$$Hopefully I'm interpreting that correctly...