I'm drawing a quarter circle with a pencil on a sheet of graph paper. The radius is 10cm. At a given moment in time the velocity at which I am moving the pencil is 5cm/s (centimeters per second). I need to calculate the velocity of the pencil on the x and y axis of the paper at that moment in time.

What is the mathematical equation that is used to solve this type of problem?

[EDIT] I realize that the answer depends on how far along the curve the pencil is. Let's say that it is 25% of the way.


The speed, not velocity of the pencil is constant at $5$ cm/sec. Velocity is a vector and has a direction, speed is the magnitude of velocity. In general, the slope of the curve is given by the derivative. Through a point $(x_0,y_0)$ on the curve, the slope is $m=\frac {dy}{dx}|_{(x_0,y_0)}$ and the tangent line is $y-y_0=m(x-x_0)$. Then a vector along the tangent is $(1,m)$ (of length $\sqrt{1+m^2}$)and we can make it of length $L$ by multiplying by $\frac L{\sqrt{1+m^2}}$ to get $(\frac L{\sqrt{1+m^2}},\frac {mL}{\sqrt{1+m^2}})$

In the case of a circle of radius $10$ centered at the origin, the equation of the top half is $y=\sqrt{100-x^2}$ so $\frac {dy}{dx}=\frac {-x}{\sqrt{100-x^2}}$. If we want the velocity at $(8,6)$ we have $\frac {dy}{dx}=\frac {-8}{\sqrt{100-64}}=-\frac 43$. The vector along the tangent is then $(-1,\frac 43)$ (note we have to resolve the direction, I assume you are drawing counterclockwise), the length of that vector is $\sqrt{1+\frac {16}9}=\frac 53$, so the velocity is $\dfrac 5{\frac 53}(-1,\frac 43)=(-3,4)$

  • $\begingroup$ Yes, the arc would be to the top left of the origin, drawing CCW. $\endgroup$ – user14993 Mar 6 '13 at 21:23
  • $\begingroup$ @kurtnelle: I followed mathematical practice of having $+x$ to the right and $+y$ up, but in fact all that matters is that you are following the circle from $+x$ to $+y$ in the $+y$ half-plane. $\endgroup$ – Ross Millikan Mar 6 '13 at 21:26
  • $\begingroup$ would this problem fall under derivatives? $\endgroup$ – user14993 Mar 6 '13 at 21:30
  • $\begingroup$ @kurtnelle: Specifically for the circle, you can use the fact that the tangent is perpendicular to the radius to avoid them. If you are at $(r \cos \theta, r \sin \theta)$ you are moving in direction $(-\sin \theta,\cos theta)$ so your velocity is $(-s\sin \theta,s\cos theta)$ where $s$ is the speed. You need the derivative unless you can find a nice geometric way to get the tangent. $\endgroup$ – Ross Millikan Mar 6 '13 at 21:55
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    $\begingroup$ @kurtnelle: Yes. When I divided by $\frac 53$ I normalized the vector, so it has no units at that point. It wouldn't change if you asked for the result in m/sec. Then we multiply by the desired speed, in your case 5 cm/sec. If you had asked for 0.05 m/sec I would have multiplied by that, getting $(-0.03 m/s, 0.04 m/sec)$ $\endgroup$ – Ross Millikan Mar 8 '13 at 14:24

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