Expert: Scotto - 8/1/2010

Question
A particle is moving along the graph of cube root of x. When x=8, the y-component of its position is increasing at the rate of 1 cm per second.
a) How fast is the x-component changing at this moment?
b) How fast is the distance from the origin changing at this moment?
c) How fast is the angle of inclination (line between origin and point) changing at this moment?
I've looked at the graph and am not for sure where to use derivatives (or not use them) and am just generally confused about how to approach it.
Thanks for your help!

Answer
a) We are given that the point is moving along y = x^(1/3).
The rate of x and y motion are related to the tangent line at this point.
The tangent line is the derivative, and that is y = 1/(3x^(2/3)).
At 8, the value of x^(2/3) is (x^(1/3))^2.  Now the cuberoot of 8 is 2, and 2^2 is 4.
So at x = 8, y = 1/(3*4) = 1/12.  This means that the speed in the y direction is 1/12 as much as the speed in the x direction.  So if the y direction is moving at 1 cm per second, this is 1/12 of 12 cm per second, and that is the speed in the x direction.

b) To find the rate at which the distance from the origin is changing, take the squareroot of (x^2+y^2).

c) If the change in y is at 1 cm per second and the change in x is at 8 cm per second,
then the change in the angle Θ is given by tanΘ = y/x.  This means that sec^2Θ dΘ =
(x*dy' - y*dx)/x^2.