Multivariate Calculus - Bruce E. Shapiro

LECTURE 4. LINES AND CURVES IN 3D 35
Tangent Line to a Parametrized Curve in 3-Space
Suppose we know the parameterization of some curve
r(t) = (f(t), g(t), h(t)) (4.14)
and would like to find a line that is tangent to this curve at t. The derivative of our
position is
r ′ (t) = dr(t)
dt
= lim
ɛ→0
r(t + ɛ) − r(t)
ɛ
(f(t + ɛ), g(t + ɛ), h(t + ɛ)) − (f(t), g(t), h(t))
= lim
ɛ→0
(
ɛ
f(t + ɛ) − f(t) g(t + ɛ) − g(t)
= lim
, ,
ɛ→0 ɛ
ɛ
=
(
lim
ɛ→0
f(t + ɛ) − f(t)
, lim
ɛ
g(t + ɛ) − g(t)
ɛ→0 ɛ
)
h(t + ɛ) − h(t)
ɛ
, lim
ɛ→0
h(t + ɛ) − h(t)
ɛ
(4.15)
(4.16)
(4.17)
)
(4.18)
= ( f ′ (t), g ′ (t), h ′ (t) ) (4.19)
In other words, you can differentiate term by term.
The derivative r ′ (t) gives a vector that is tangent to the curve r(t),
(4.20)
Example 4.6 Find the symmetric equations of the tangent line to the curve parameterized
by
r(t) =
(3t, −4t 2 , 1 )
π sin πt
at t = 1.
Solution. A tangent vector to the curve at any time t is given by the derivative,
At t = 1 we have the tangent vector
and a point on the curve
r ′ (t) = (3, −8t, cos πt)
r ′ (1) = (3, −8, −1)
r(1) = (3, −4, 0)
The symmetric equations of the line are then
x − 3
3
= − y + 4
8
= −z
Math 250, Fall 2006 Revised December 6, 2006.