Problem F*. Consider the "twisted cubic" curve r(t)=(t,t^2,t^3) described in the text's example 7 (p. 889). Is it possible for a plane to intersect this curve in more than three different points? Explain why or why not.

Solution to problem F:

Any plane in R^3 has an equation of the form

ax + by + cz = d

for some specific numbers a, b, c, d (and a, b, c not all zero).

If r is the twisted cubic, r(t)=, then the point r(t) is on the plane when (and only when)

at + bt^2 + ct^3 = d.

This is a degree-3 (cubic) polynomial in t, so it can have at most three roots in real numbers. (Or in complex numbers, for that matter.) So, there can be at most three values of t for which the last equation is true. So, there can be at most three points r(t) of the twisted cubic that are on the plane.

(A minor quibble: If c=0, the polynomial has degree 2, not 3. Or, if b also is 0, it has degree 1. Sometimes it's important to be careful about the degree of a polynomial. But not this time --- in all of these cases, the polynomial still can't have more than 3 roots.

If a, b, and c were all zero, we would have a zero-degree polynomial, and that can have infinitely many roots. But if a, b, c were all zero, we wouldn't have a plane in the first place.)