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Section 10.3 – Parametric Equations and Calculus
Derivatives of Parametric Equations
To analyze a parametric curve analytically, it is useful to rewrite the equations in the form . But what if the parametric equations are difficult to convert to a single Cartesian equation? Consider:
3
2 3
sin 15
arcsin5
x t t t t
ty t t
t
We must find a way to analyze the curves without having to convert them.
Derivatives of Parametric Equations
Let a differentiable parametric curve be defined as and . Consider :
'g t dy
dt
dy dx
dx dt
By the Chain Rule.
'dy
f tdx
So... ' 'dy
f t g tdx
OR...
'
'
g tdy
dx f t
Derivatives of Parametric Equations
If and are differentiable functions of and , then
'
'
y tdy dy dt
dx dx dt x t
If is also a differentiable function of , then
2
2
ddt dy dxd y d dy
dx dx dx dx dt
Nothing is new. All results about derivatives from earlier chapters still apply.
Example 1Consider the curve defined parametrically by and for .
(a) Find the highest point on the curve. Justify your answer.
dy dy dt
dx dx dtFind
dy/dx:
2cos
2
t
t
cos0
t
t
cos t
t
2t
Find the critical points. Test the critical points
and the endpoints to
find the maximum y.
t x y
0 -5 0
2
0
cos tundefined
t
0t
Example 1 (continued)Consider the curve defined parametrically by and for .
(b) Find all points of inflection on the curve. Justify your answer.
2
2
ddt dy dxd y
dx dx dtFind
d2y/dx2:
2
sin 1 cos
2
t t t
t
t
3
sin cos0
2
t t t
t
3
sin cos
2
t t t
t
2.798t
Find the critical
points of the first
derivative.
Check to see if there is a sign change in the second derivative.
2.798
''f x0 𝜋
Find the x and y value:22.798 5 2.831
2sin 2.798 0.673
x
y
2.831,0.673
is also undefined at the endpoint
White Board Challenge
Let and . Find the equation of the tangent line at .
2315 10
6y x
Example 2Let and . Find:
(a) The coordinate(s) where the tangent line is vertical.
(b) The coordinate(s) where the tangent line is horizontal.
dy dy dt
dx dx dtFind
dy/dx:
2
2
4
3 6
t
t t
2 2
3 2
t t
t t
' 0
0
dy dt
dx dt
This occurs when:
Although t=2 makes the denominator 0, t=0 is the only
value that satisfies both conditions.
3 20 3 0x
313 0 4 0y
0
0,0
' 0
0
dy dt
dx dt
This occurs when:
Although t=2 makes the numerator 0, t=-2 is the only value that
satisfies both conditions.
3 22 3 2x
313 2 4 2y
20163
16320,
0
Example 2Let and .(c) Prove the relation is differentiable at .
163
2& lim
ty t
The one-sided derivatives are equal and non-infinite.
2
lim 4t
x t
163
2& lim
ty t
Prove that it is Continuous
2
lim 4t
x t
23
2lim t
tt
2 2
3 2
Since the limits equal the values of the coordinate, the relation is continuous at t=2.
2 2
3 22lim t t
t tt
The
lim
it ex
ists
The
poi
nt (
x,y)
fo
r t=
2 ex
ists
16
3
2 4
2
x
y
16
32 2lim 4 & & limt tx t y t
Prove the Right Hand Derivative is the same as the Left Hand Derivative (and non-infinite)
46
23
2lim t
tt
2 2
3 2
2 2
3 22lim t t
t tt
46
Thus the derivative exists, at t=2.
Arc Length of Parametric Curves
Let and be continuous functions of . Consider:
21
b dydxa
L dx Regular Arc Length Formula.
2
1
21
t dy dtdx dtt
dx 2
1
22 2t dydt dxdx dt dtt
dx 2
1
22t dydxdt dtt
dtdx
dx
2
1
22t dydxdt dtt
dt
Arc Length of Parametric Curves
Let be the length of a parametric curve that is traversed exactly once as increases from to .
If and are continuous functions of , then:
2
1
22t dydxdt dtt
L dt
Example 1Calculate the perimeter of the ellipse generated by and .
We already graphed this curve. If the curve starts at , it will complete a cycle at .
We must find the limits for the integral.
2 2 2
03cos 2sind d
dt dtL t t dt
2 2 2
03sin 2cost t dt
2 2 2
09sin 4cost tdt
For most arc length problems, the calculator needs to evaluate the definite integral.
15.865
Example 2A particle travels along the path and .
Find the following:
(a) The distance traveled during the interval .
(b) The displacement during the interval .
Use
arc
leng
th.
24 2 3 2
02 1d d
dt dtL t t dt 4 22 1 2320
2 t dt 4
940
4 tdt 11.52
Use
the
Dis
tanc
e F
orm
ula
Coordinate at t=0: 3 2
2 0 0
1 0 1
x
y
Coordinate
at t=4: 3 2
2 4 8
1 4 9
x
y
2 28 0 9 1D 128 8 2 11.31
