math 54 lecture 8 - parametric equations

Parametric Equations Calculus of Parametric Equations

Parametric Curves

Institute of Mathematics, University of the Philippines Diliman

Mathematics 54–Elementary Analysis 2

Parametric Curves 1/ 18


Introduction

Consider a particle moving along the curve below



Introduction

The curve cannot be expressed as the graph of an equation of the form y = f (x).

Sometimes, it’s impossible to find an equation relating the x− and y−coordinatesof the particle.

But we can express the x− and the y−component of the particle’s position asfunctions of time t, i.e.

x = x(t)

y = y(t)

In this process, we say that we are parametrizing the curve above.

The equations defining x and y as functions of t are called parametric equations,and t is called a parameter.



Parametric Curves

Examples.

Sketch the curve defined by C : x = t, y = t2, where t ∈R.

t → (x,y)−2 → (−2,4)−1 → (−1,1)0 → (0,0)1 → (1,1)2 → (2,4)3 → (3,9)

Plotting these points, we have

Indeed, if we relate x and y directly, eliminating t, we have y = x2.

We have just obtained a cartesian equivalent of the parametric curve.



Examples

Example.

Sketch the curve defined by x = cos t and y = sin t, where 0 É t É 2π.

t → (x,y)0 → (1,0)π4 →

(p2

2 ,p

22

)π2 → (0,1)π → (−1,0)3π2 → (0,−1)

2π → (1,0)

Indeed, if we relate x and y directly, eliminating t, we have x2 +y2 = 1.

The direction of trace is known as the direction of increasing parameter. In thiscase, it’s counterclockwise.

The point corresponding to t = 0 is the initial point.The point corresponding to t = 2π is the terminal point.



Examples

Parametrization of circles...

x = sin t

y = cos t

0 É t É 2π

’clockwise’ circle

x = 2cos t

y = 2sin t

0 É t É 2π

’counterclockwise’ circle of radius2 units



Examples

x = h+ r cos t

y = k+ r sin t

0 É t É 2π

circle of radius r centered at (h,k)

x = cos2t

y = sin2t

0 É t Éπ

’faster’ circle...



Interesting Examples

x = sin2t

y = sin3t

Lissajous curves

x = cos t +0.5cos7t +(

1

3

)sin17t

y = sin t +0.5sin7t +(

1

3

)cos17t

0 É t É 2π

(?)



Examples

Consider the curvedefined by

x = sec t

y = tan t

−π2É t É π

2.

Note that

x2 = sec2 t

y2 = tan2 t

Thus,

x2 −y2 = 1



Exercises

1 Describe the curve given by the parametric equations1 x = t2 and y = t4.2 x = sin t and y = sin2 t.

2 What can be said about the parametric equations1 x = cos t, y =−sin t2 x = cos2t, y =−sin2t

3 Describe the curve given by the parametric equations x = 2cos t andy = 3sin t.

4 Eliminate the parameter t, i.e., transform the following into theircorresponding cartesian form

1 x = 3− t, y = 2t −3

2 x = t2, y = 1−2t

3 x =pt +1, y = 6− t

4 x = t2, y = t3

5 x = et , y = e−t

6 x = cos t, y = cos2t



Slope of Parametric Curves

Smooth Curve

A parametric curve C : x = f (t),y = g(t) is said to be smooth if f and g aredifferentiable and no value of t satisfies f ′(t) = g′(t) = 0.

If C is a smooth curve, then its slope dydx is given by

dy

dx=

dy

dtdx

dt

Also, the concavity d2ydx2 is given by

d2y

dx2=

d(

dydx

)dtdx

dt

6=d2y

dt2

d2x

dt2



Slope of Parametric Curves

Example.

Find the slope of the astroid having the equation x = cos3 t, y = sin3 t, where t = 3π4 .

Solution:dy

dx= dy/dt

dx/dt= 3sin2 t cos t

−3cos2 t sin t=− tan t

So,dy

dx

∣∣∣t= 3π

4

=− tan3π

4= 1



Tangent Lines to Parametric Curves

Example.

Find the equation of the tangent line to the right branch of the hyperbolax = sec t,y = tan t, where −π

2 < t < π2 , at the point (

p2,1).

Solution: Note that the point (p

2,1) corresponds to t = π4 . Also,

dy

dx= dy/dt

dx/dt= sec2 t

sec t tan t= csc t

Thus,

dy

dx

∣∣∣(p

2,1)= csc t

∣∣∣t=π/4

=p2

Hence, the equation is

y−1 =p2(x−p

2)

y =p2x−1



Concavity

Example.

Find d2ydx2 if x = t − t2,y = t − t3.

Solution: y ′ = dy

dx= 1−3t2

1−2tThus,

d2y

dx2=

d(

1−3t21−2t

)dtdxdt

=(1−2t)(−6t)−(1−3t2)(−2)

(1−2t)2

1−2t

= 6t2 −6t +2

(1−2t)3

t < 12 t > 1

2

Caution :d2y

dt2

d2xdt2

= −6t1 =−6t 6= d2y

dx2 .



Length of a Parametric Curve

Recall that the length of arc of the smooth curve y = F(x), from t = a to t = b, isgiven by

L =∫ b

a

√1+

(dy

dx

)2dx.

Such a curve can be parametrized as x = t, y = F(t). Hence,

√1+

(dy

dx

)2dx =

√√√√√1+ dy

dtdxdt

2dx

dt·dt =

√(dx

dt

)2+

(dy

dt

)2dt.

For a general parametric curve, we prove later that the formula still works.

Length of Arc

Let C : x = x(t),y = y(t),a É t É b, be a smooth curve traced exactly once. Then thelength of C is given by

L =∫ b

a

√(dx

dt

)2+

(dy

dt

)2dt



Arc Length

Example.

Find the length of thecycloid

x = 4(t − sin t)

y = 4(1−cos t)

0 É t É 2π

dx

dt= 4(1−cos t) and

dy

dt= 4sin t

L =∫ 2π

0

√16(1−cos t)2 +16sin2 t dt

=∫ 2π

0

√16−32cos t +16cos2 t +16sin2 t dt

=∫ 2π

0

p32−32cos t dt =

∫ 2π

04√

2(1−cos t)dt

=∫ 2π

04

√4sin2

(t

2

)dt

=∫ 2π

08sin

(t

2

)dt

= −16cos

(t

2

)∣∣∣2π

0

= 32



Arc Length

Example.

Find the length of theastroid

x = cos3 t

y = sin3 t

0 É t É 2π

L = 8∫ π/4

0

√(−3cos2 t sin t)2 + (3sin2 t cos t)2 dt

= 12∫ π/4

02cos t sin t dt

= 12∫ π/4

0sin2t dt

= 6[−cos2t]π/40

= 6



Exercises

1 Findd2y

dx2of the curve with parametric equation x = t2 and y = t3 −3t.

2 Find the length of the curve given by x = et cos t and y = et sin t where0 ≤ t ≤π.

3 Set-up the iterated integral that will give the distance travalled by a particle

with position (x,y) as time t varies where x = t

1+ tand y = ln(1+ t), with

t ∈ [0,2].


math 54 lecture 8 - parametric equations

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