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gunch

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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 [tex]y=x^2[/tex]. 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

[tex]\frac{dx}{dy}=\frac{dy}{dx}[/tex]

so I differentiate the equation with respect to x.

[tex]\frac{dy}{dx}=2x[/tex]

And then with respect to y.

[tex]1=2x\frac{dx}{dy}[/tex]

[tex]\frac{dx}{dy}=\frac{1}{2x}[/tex]

Then to find when their rate of change is the same I write:

[tex]\frac{1}{2x}=2x[/tex]

[tex]4x^2=1[/tex]

[tex]x=\sqrt{1/4}[/tex]

[tex]x=\frac{1}{2}[/tex]

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

"A particle moves differentiably on the parabola [tex]y=x^2[/tex]. 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

[tex]\frac{dx}{dy}=\frac{dy}{dx}[/tex]

so I differentiate the equation with respect to x.

[tex]\frac{dy}{dx}=2x[/tex]

And then with respect to y.

[tex]1=2x\frac{dx}{dy}[/tex]

[tex]\frac{dx}{dy}=\frac{1}{2x}[/tex]

Then to find when their rate of change is the same I write:

[tex]\frac{1}{2x}=2x[/tex]

[tex]4x^2=1[/tex]

[tex]x=\sqrt{1/4}[/tex]

[tex]x=\frac{1}{2}[/tex]

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

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