MATH 1110 Sections 9 & 10
Timothy Goldberg: 2009-09-12
TANGENT LINES TO A PARAMETRIZED CURVE
Question. Consider the curve C parametrized by
x = t3 − 3t,
y = t2 ,
−2 ≤ t ≤ 2.
(a) What values of t correspond to the point (0, 3) on the curve C? What does it mean if
more than one value of t gives you the same point?
(b) Find equations for all tangent lines to C at the point (0, 3).
(c) Sketch the curve C along with the tangent lines you determined in part (b). Use
arrows to indicate the direction in which this parametrization traces out the curve.
d2 y
(d) Compute the second derivative
of the curve C at the values of t you found in
dx2
part (a) above.
Solution.
(a) We set t3 − 3t = 0 and t2 = 3. Factoring the first equation, we obtain
√
√
0 = t3 − 3t = t(t2 − 3) = t(t − 3)(t + 3),
√
√
so t = 0, 3, − 3. Only the second two values are also solutions to t2 = 3, so
√
√
the values of t which yield the point (0, 3) are t = 3, − 3 . This means that as we
trace out the curve C, we go through the point (0, 3) twice, so the curve crosses itself.
(b) Since
dy/dt
2t
dy
=
= 2
,
dx
dx/dt
3t − 3
we have
and
√
√
√
dy 2 3
2 3
3
1
= √
=
=
=√
√
2
dx t= 3 3( 3) − 3
6
3
3
√
√
√
dy −2 3
−2 3
3
1
√
=
=
=−
= −√
√
dx t=− 3 3(− 3)2 − 3
6
3
3
1
Since both tangent lines contain the point (0, 3), they have equations y − 3 = √ x
3
1
and y − 3 = − √ x .
3
1
MATH 1110 Sections 9 & 10
Timothy Goldberg: 2009-09-12
(c) To sketch the curve C, we will evaluate the parametrization at some different values
of t.
t
−2
√
− 3
−1
0
√1
3
2
x = t3 − 3t y = t2
−2
0
2
0
−2
0
2
4
3
1
0
1
3
4
(x, y)
(−2, 4)
(0, 3)
(2, 1)
(0, 0)
(−2, 1)
(0, 3)
(2, 4)
If we plot these points (labeled with the corresponding values of t), connect the dots,
and draw arrows to indicate the direction of increasing values of t, we obtain the
picture shown in Figure 1.
5
t=-2
t=2
2.5
t=-1
t=1
-5
-2.5
0
2.5
5
t=0
Figure 1: The curve C, with information about how it was parametrized.
Oops! I forgot to put in the tangent lines. Sorry! I think you can see where they
should go.
2
MATH 1110 Sections 9 & 10
Timothy Goldberg: 2009-09-12
(d) First we compute
d
2t
(2)(3t2 − 3) − (2t)(6t)
6t2 − 6 − 12t2
−6t2 − 6
=
=
=
.
dt 3t2 − 3
(3t2 − 3)2
(3t2 − 3)2
(3t2 − 3)2
Then since
d2 y
=
dx2
d
(dy/dx)
dt
dx/dt
(−6t2 − 6)/(3t2 − 3)2
=
3t2 − 3
2
6t + 6
,
=− 2
(3t − 3)3
we have
√
6(± 3)2 + 6
24
1
d2 y 6(3) + 6
√
=−
=− 3 = − .
=−
2
3
√
2
3
dx t=± 3
(3(3) − 3)
(6)
9
(3(± 3) − 3)
3