(adsbygoogle = window.adsbygoogle || []).push({}); Consider the graph of y=x^{2}. A point is moving along the x-axis in such a way, that its speed is proportional to its distance from the origin. At the same time, a point is moving along the curve of the graph, which always has the same x-value as the point moving along the x-axis. At what rate is the point on the graph moving away from the origin?

At any time, the horizontal distance of the point is 'x' units, and the vertical distance is 'x^{2}' units. By using pythagoras' theorem, i can show that the distance from the origin is thus given by:

[tex]

D = \sqrt {x^2 + x^4 }

[/tex]

Now, since the point moving along the x-axis has a speed proportional to its horizontal distance, i can say that:

[tex]

\frac{{dx}}{{dt}} \propto x\therefore\frac{{dx}}{{dt}} = kx

[/tex]

Where 'k' is some constant.

Since both 'D' and 'x' are functions of time, i differentiate implicitly with respect to time, to get:

[tex]

\frac{{dD}}{{dt}} = \frac{{kx^2 + 2kx^4 }}{{\sqrt {x^2 + x^4 } }}

[/tex]

Therefore, for any given distance of the first point from the origin, the second point, moving along the curve, has a speed of [itex]

\frac{{kx^2 + 2kx^4 }}{{\sqrt {x^2 + x^4 } }}

[/itex] relative to the origin.

I wasnt 100% sure if i did it correctly. If anybody would be able to double check my working, that would be great.

Thanks,

Dan.

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# Homework Help: Related Rates Question

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