# Homework Help: Point on a Curve

1. Jan 20, 2009

### jgens

1. The problem statement, all variables and given/known data

A particle moves along a path described by y = 4 - x^2. At what point along the curve are x and y changing at the same rate

2. Relevant equations

Simple equations regarding derivatives.

3. The attempt at a solution

It's been a while before I've done any related rates problems, could someone please let me know if this is correct:

Since, x and y must be changing at the same rate (presumably with respect to time) x' = y' and y' = -2xx'. Therefore, -2x = 1 and x = -1/2. Placing my x value into the original equation yields 15/4. Hence, the point is (-1/2, 15/4).

Thanks.

2. Jan 20, 2009

### mrdoe

SEEMS correct...

3. Jan 20, 2009

### Dick

Of course, it's right. What could go wrong?

4. Jan 20, 2009

### jgens

Plenty, I could have made an incorrect assumption ultimately leading to false conclusions.

5. Jan 20, 2009

### quasar987

Good answer but you do not need to assume that x and y are varying wrt an external parameter. The derivative y'(x) = dy/dx of y wrt x expresses the instantaneous rate of change of y wrt a change in x.

The points where y and x are changing at the same rate are those where y'(x)=1.