Calculus + Differentiation?

Question

A particle moves along the curve: y = sq.rt.(1 + x^3). As it reaches the point (2,3), the y-coordinate is increasing at a rate of 4 cm/s. How fast is the x-corrdinate of the point chaning at that instant?

Can someone show me how to figure this out?

Answer 1

At the designated point, dx/dt = 2 cm/s.

This is found as follows:

y = (1 + x^3)^(1/2).

Note first that (x, y) = (2, 3) is INDEED on this curve, since
y(2) = (1 + 2^3)^(1/2) = 9^(1/2) = 3.

[There is a convention that "sq.rt." means the POSITIVE square root, so that the SIGN to use is, for example, thrown onto the +/- sign that occurs in the standard solution to a quadratic equation.]

Thus we may safely proceed, first deducing the general result that

dy/dx = 1/2 * 3 x^2 / (1 + x^3)^(1/2).

Then dx/dt = (1 + x^3)^(1/2) * 2 / [3 x^2] * dy/dt

= (2y / [3 x^2]) * dy/dt

So if dy/dt = 4 cm/s, dx/dt = (6 / 12) * 4 cm/s = 2 cm/s.

QED

Live long and prosper.

Answer 2

y = sq.rt.(1 + x^3)
x = (y^2-1)^(1/3)
dx/dt = (dx/dy)(dy/dt)
dx/dt = (2y/3)(y^2-1)^-(2/3)(dy/dt)
dx/dt = (4)(2(3)/3)(3^2-1)^-(2/3)
dx/dt = 8/8^(2/3) = 8^(1/3) = 2 cm/s