Calculus: Analytic Geometry and Calculus, with Vectors

414 Curves, lengths, and curvatures

terminology, the semicubical parabola is a graph (not

necessarily a curve) which is the point set shown in Figure

7.195. The graph has two branches, one being the upper

branch consisting of points for which y = x34-, and the

other being the lower branch consisting of points for

which y = -x' x. Each of these branches contains the

origin.

When the points on the upper branch are ordered in

such a way that P(xi,yi) precedes P(x2,y2) when x, < x2,

-i} the branch becomes a curve Cl in the first quadrant

having an end (or end point) at the origin. Similarly,

when the points on the lower branch are ordered in such

Figure 7.195 a way that P(x,,y,) precedes P(x2,y2) when x, < x2. this
branch becomes another curve C2 having an end at the
origin. With these orderings, the graph consists of the points on the two curves C,
and C2, the orderings being indicated by the arrows of Figure 7.195 which lie between
the curves and the positive x axis. When x > 0, the curve C, has a tangent at the
point (x,x34) which has slope T634, and the curve C2 has a tangent at the point
(x,-x'6) which has slope -W4. We can observe that Ci has a forward tangent
at the origin whose slope is 0 and that C2 has a forward tangent at the origin whose
slope is 0. Although we have definitions involving tangents to graphs of equa-
tions of the form y = f(x) and have definitions involving tangents to curves, we
have no definitions which we can apply to decide whether the x axis is tangent
to the semicubical parabola at the origin. In accordance with terminology
other applications of which are not explained here, we say that the semicubical
parabola has a cusp at the origin.
11 Prove that the curve C consisting of points P for which

(1) OP = t2i + t3j

coincides with the semicubical parabola having the equation y2 = x3. Sketch

a new graph of the semicubical parabola and insert arrows which show how the

points on the semicubical parabola are ordered to produce the single curve C.

Put (1) in the equivalent form

(2) r = t21 + t3j

and, supposing that a particle moves along C in such a way that its displacement

vector at time t is r, find its velocity and acceleration at time t. See what hap-

pens when you try to write the equation of the line tangent to C at the point

occupied by the particle at time t = 0.

12 The folium of Descartes is the graph of the parametric equations

(1) X 3at 3at2

1+t3, y=+a

The vector r running from the origin to the point P(x,y) on the folium is

(2) r = 3a^112

1 + t3i +1 .+ toj ],