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4.11. Tangents to Parametric Curves
Let a parametric curve be given by equations x = x(t), y = y(t).
Suppose both functions are differentiable and we want to find the
tangent line at a point on the curve where y is also a differentiable
function of x. Then the Chain Rule gives
dy
dt
=
dy
dx
dx
dt
.If
dx
dt
=0,wecansolvefor
dy
dx
:
dy
dx
=
dy
dt
:
dx
dt
if
dx
dt
=0.
Example 20
Consider a parametric circle (part of) given by x = a cos t, y =
a sin t; 0 <t<π. Find
dy
dx
: 1) at an arbitrary t; 2) at t =
π
4
.
1) y
x
=
(a sin t)
(a cos t)
=
a cos t
−a sin t
= −ctg t;
2) y
x
t=π/4
= −ctg
π
4
= −1.
Example 21
Find an equation of the tangent line to the parametric curve
x =2sin2t, y =2sint at the point (
√
3, 1). Where does this curve
have horizontal or vertical tangent?
At the point with parameter value t, the slope is
dy
dx
=
dy/dt
dx/dt
=
2cost
2(cos 2t)2
=
cos t
2cos2t
.
The point (
√
3, 1) corresponds to the parameter value t =
π
6
,so
the slope of the tangent at that point is
dy
dx
t=π/6
=
cos
π
6
2cos
π
3
=
√
3/2
2 ·
1
2
=
√
3
2
.
An equation of the tangent line is therefore
y − 1=
√
3
2
(x −
√
3) or y =
√
3
2
x −
1
2
.
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