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

arcsin 5

x t t t t

ty t tt

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 dydt

dy dxdx 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 dtdx dx dt

Find dy/dx:

2cos2t

t

cos 0tt

cos tt

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 t undefinedt

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 cos 02

t t tt

3

sin cos2

t t tt

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 0.673xy 2.831,0.673

is also undefined at the endpoint

White Board ChallengeLet and . Find the equation of the tangent line at .

2315 106

y 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 dtdx dx dt

Find dy/dx:

2

2

43 6tt t

2 2

3 2t tt t

' 00

dy dtdx 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

' 00

dy dtdx 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 .

1632

& limt

y t

The one-sided derivatives are equal and non-infinite.

2

lim 4t

x t

1632

& limt

y t

Prove that it is Continuous

2

lim 4t

x t

232

lim ttt

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

limit

exis

ts

The

poin

t (x,

y)

for t

=2 e

xist

s

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

232

lim ttt

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

dt dxdx

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 Fo

rmul

a 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