Calculus with Algebra and Trigonometry II Lecture 5 More ...frooney/Math_217_5_Polar.pdf · Calculus with Algebra and Trigonometry II Lecture 5 More Parametric equations, polar coordinates

Calculus with Algebra and Trigonometry IILecture 5

More Parametric equations, polar coordinates andL’Hopital’s rule

Feb 3, 2015

Calculus with Algebra and Trigonometry II Lecture 5More Parametric equations, polar coordinates and L’Hopital’s ruleFeb 3, 2015 1 / 18

Different parametric equations for the same curve

Consider the three different parametrizations

x = cos t y = sin t 0 ≤ t < 2π (a)

x = cos 2t y = sin 2t 0 ≤ t < π (b)

x = cos t2 y = sin t2 0 ≤ t <√

2π (c)

They are all parametrizations of the unit circle. The differences betweenthe three are in the interval for t and in how fast the point (x , y) moves ast changes.

For example at t = 1.5 the points on the circle are

(.071, .997) for (a) (−.99, .141) for (b) (−.628, .778) for (c)

The situation is illustrated below


: (a) : (b)

: (c)


Speed and arclength

To find the speed of the point, consider a small piece of the curve

∆s is approximately the hypotenuse of the triangle

(∆s)2 ≈ (∆x)2 + (∆y)2(

∆s

∆t

)2

=

(∆x

∆t

)2

+

(∆y

∆t

)2

Taking the limit as ∆t → 0 gives the speed dsdt(

ds

dt

)2

= x ′2 + y ′2 ⇒ ds

dt=√x ′2 + y ′2

The quantity s is called the arc length.


Curvature

If we denote the angle the tangent line makes with the x axis by φ

The curvature of the curve is determined from the rate of change of φ asyou move along the curve. Clearly if the point moves along the curve fasterthen φ will change more rapidly. The curvature is defined to be the rate ofchange of φ as the point moves along the curve with unit speed, in otherwords the curvature is the derivative of φ with respect to the arc length.

κ =dφ

dsκ > 0 means that the curve bends to the left as you move along it, κ < 0means it bends to the right.


Use of the chain rule gives the curvature for an arbitrary parametrization

dφ

dt=

dφ

ds

ds

dt⇒ κ =

dφdtdsdt

Since φ is the angle the tangent line makes with the x axis then

tanφ = slope =dy

dx=

y ′(t)

x ′(t)⇒ φ = tan−1

(y ′

x ′

)Using the expression for speed derived earlier

κ =1√

x ′2 + y ′2d

dt

(tan−1

(y ′

x ′

))=

1√x ′2 + y ′2

(1

1 + ( y′

x ′ )2

)(y ′′x ′ − y ′x ′′

x ′2

)


Simplifying gives the formula for the curvature

κ =y ′′x ′ − y ′x ′′

(x ′2 + y ′2)3/2

If the curve is given in the form y = f (x) then we can think of x as theparameter and in this case the curvature is given by

κ =f ′′(x)

(1 + (f ′(x))2)3/2

It is clear that the curvature is closely related to the convexity/concavitythe difference is that for curvature in general you are considering the shapeof the curve as you move along it at unit speed, convexity/concavity isrelated to the shape of the curve as you move in the x direction.


Geometrically there is a circle tangent to the curve at the point, called theosculating circle.

The radius of this circle is called the radius of curvature and is

Radius of curvature =1

κ

and the center of the circle is called the center of curvature.


An example

Determine the critical points and curvature for the curve

x(t) = 3t2 y(t) = t3 − 3t −∞ < t <∞

The horizontal tangents are determined from

y ′(t) = 0 ⇒ 3t2 − 3 = 0 ⇒ t = ±1

So the points are (3, 2) and (3,−2).

The vertical tangents are determined from

x ′(t) = 0 ⇒ 6t = 0 ⇒ t = 0

So the only point with a vertical tangent is (0, 0).


The curve looks like

x ′2 + y ′2 = (6t)2 + (3(t2 − 1))2 = 9(t2 + 1)2

and the curvature is

κ =(6t)(6t)− (3t2 − 3)(6)

27(t2 + 1)3=

18(t2 + 1)

27(t2 + 1)2=

2

3(t2 + 1)2

The curvature is positive everywhere so the curve is continuously turningto the left.


Polar coordinates

An alternative to using rectangular coordinates (x and y) to specify pointsin the plane is to specify how far the point is from the origin and thedirection it lies in.

Given x and y we can determine r and θ from

r =√x2 + y2 θ = tan−1

(yx

)Alternatively given r and θ, x and y can be found using

x = r cos θ y = r sin θ


The curves of constant r are circles and the curves of constant θ are raysfrom the origin, so poal graph paper resembles a spider web

An unusual feature is that each point has two different polarrepresentations.

(r , θ) (−r , θ + π)

represent the same point.


Polar graphs

Curves can be specified in polar form

r = f (θ)

Some examples


Polar graphs as parametric graphs

The curve r = f (θ) can be re-expressed in the parametric form

x(θ) = f (θ) cos θ y(θ) = f (θ) sin θ

with θ as the parameter.

For example find the critical points of the cardioid

r = 2(1 + cosθ)

The parametric equations are

x(θ) = 2(1 + cos) cos θ y(θ) = 2(1 + cos θ) sin θ

The points with horizontal tangents are determined from y ′(θ) = 0

2 cos θ + 2 cos2 θ − 2 sin2 θ = 0 2(cos θ + 2 cos2 θ − 1) = 0

this factors as(2 cos θ − 1)(cos θ + 1) = 0


Thus either

2 cos θ = 1 ⇒ cos θ =1

2⇒ θ =

π

3,

5π

3or

cos θ = −1 ⇒ θ = π

The points with vertical tangents are given by x ′(θ) = 0

−2 sin θ − 4 sin θ cos θ = 0 − 2 sin θ(1 + 2 cos θ) = 0

Then eithersin θ = 0 ⇒ θ = 0, π

or

2 cos θ + 1 = 0 ⇒ θ =2π

3,

4π

3Note that at θ = π. x ′(π) = y ′(π) = 0 so to determine the slope of thetangent we have to calculate

limθ→π

y ′(θ)

x ′(θ)


That is

− limθ→π

2 cos2 θ + cos θ − 1

sin θ + 2 sin θ cos θ

we will defer this until after we discuss L’Hopital’s rule. The graph of thecardioid with the tangents is


L’Hopital’s Rule

L’Hopital’s Rule is a extremely useful method to solve indeterminate limits.

L’Hopital’s Rule : Given two functions f (x) and g(x) withf (a) = g(a) = 0 then

limx→a

f (x)

g(x)= lim

x→a

f ′(x)

g ′(x)

Example 1:

limx→3

x4 − 81

x3 − 27= lim

x→3

4x3

3x2= 4

Example 2:

limt→0

sin t

1− cos t= lim

t→0

cos t

sin t→∞


Example 3:

limx→0

1− x2

2 − cos x

x4= lim

x→0

−x + sin x

4x3= lim

x→0

−1 + cos x

12x2

= limx→0

− sin x

24x= lim

x→0

− cos x

24= − 1

24

To finish the calculation of the slope of the tangent to the cardioid atθ = π

Slope = − limθ→π

2 cos2 θ + cos θ − 1

sin θ + 2 sin θ cos θ

It has the correct 00 form to use L’Hopital’s Rule so

Slope = − limθ→π

−4 sin θ cos θ − sin θ

cos θ + 2 cos2 θ − 2 sin2 θ= 0

So the tangent is horizontal.


Documents

Calculus with Algebra and Trigonometry II Lecture 5 More ...frooney/Math_217_5_Polar.pdf · Calculus with Algebra and Trigonometry II Lecture 5 More Parametric equations, polar coordinates