# Homework Help: X- and y-coordinate's rate of change

1. Nov 25, 2006

### gunch

Hi, I have a problem with an exercise in the book "A First Course in Calculus, fifth edition". The problem is stated as follows:
"A particle moves differentiably on the parabola $$y=x^2$$. At what point on the curve are its x- and y-coordinate moving at the same rate? (You may assume dx/dt and dy/dt is not equal to 0 for all t)"
According to the book the answers should be 1/2 and 1/4 however I only get 1/2.

I solve it, as follows:
Their rate is the same when
$$\frac{dx}{dy}=\frac{dy}{dx}$$
so I differentiate the equation with respect to x.
$$\frac{dy}{dx}=2x$$
And then with respect to y.
$$1=2x\frac{dx}{dy}$$
$$\frac{dx}{dy}=\frac{1}{2x}$$

Then to find when their rate of change is the same I write:
$$\frac{1}{2x}=2x$$

$$4x^2=1$$

$$x=\sqrt{1/4}$$

$$x=\frac{1}{2}$$

So, what have I done wrong? I'm not all that familiar with this stuff yet so I may have misunderstood the exercise.

Last edited: Nov 25, 2006
2. Nov 25, 2006

### dextercioby

Maybe that 1/4 is the "y" value for the 1/2 "x" value ?

Daniel.

3. Nov 25, 2006

### gunch

Thanks, that must be it. I can't believe I failed to notice that.