The hodograph is very useful for solving complicated problems in dynamics. By simple geometrical arguments students can directly obtain the answer to problems that would otherwise be complicated exercises in algebra. Although beyond the level of undergraduates, we also use the hodograph to calculate by variational geometrical techniques, the well-known brachistochrone curve, thus illustrating this approach.

## REFERENCES

1.

William Rowan

Hamilton

, “The Hodograph, or a new method of expressing in symbolical language the Newtonian law of attraction

,” Proc. R. Ir. Acad.

3

, 344

–353

(1847

).2.

James Clerk Maxwell,

*Matter and Motion*(Dover, New York, 1991).3.

David and Judith Goodstein,

*Feynman’s Lost Lecture*(W. W. Norton, New York, 1996).4.

If $sin\u200a\phi =sc,$ so that the two graphs just touch each other [see Fig. 5(b)], the slope of the ground at the intersection point is equal to the inverse of the slope of $f1,$ because $[dyground/dx]0=[df2/dy]0\u22121=[df1/dy]0\u22121=[(d/dy)sc2a1+y/a]0\u22121=(1+y0/a)/sc.$ If we call $\phi 1$ and $\phi 2$ the angles formed by the initial and final velocities, respectively, with the horizon we have that $\phi 1+\phi 2=\phi c,$ and $cos\u200a\phi 1=1+y0/a\u200acos\u200a\phi 2,$ where the last relation comes from the geometry of Fig. 4. With a little algebra we obtain $tan\u200a\phi 2=(1+y0/a)/sc\u2212cot\u200a\phi c.$ Therefore if $\phi c$ is the oblique angle that satisfies $sin\u200a\phi c=sc,$ the slope of the projectile trajectory is smaller than the slope of the ground at the landing point, which is impossible (because this would mean that the projectile arrives at this point from inside the ground). Thus, only the corresponding obtuse-angle solution can lead the projectile to this point. The same reasoning holds for values of φ such that $sin\u200a\phi $ is slightly less than $sc.$ Both intersection points then can be reached by the obtuse-angle solution, and the nearest point can be reached by the oblique-angle solution only if $dyground/dx\u2a7dtan\u200a\phi 2.$

5.

The stationary point we calculated here corresponds to a minimum, because an infinitely deep well would lead to infinite time travel.

6.

Because all expressions are written as functions of $u,$ we try to optimize the inverse function of $u(\theta );$ namely, $\theta (u).$

7.

The end points of θ are $\theta in=\u2212\pi /2,$ and $\theta fin=\pi /2,$ because these values lead to zero velocities when substituted in $u=(1/\lambda )cos\u200a\theta ,$ the hodographic curve.

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2003

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