10.1 curves defined by parametric equations

10.1 Curves Defined by Parametric Equations Ex: Consider the unit circle from Trigonometry. What is the equation of that circle?

There are 2 ways to describe it: x2 + y2 = 1 and x = cos ! y = sin ! When x and y are given as functions of a third variable, called a parameter, they describe a parametric curve. Each of the equations for x and y are called parametric equations. Ex: What curve is represented by the given parametric equations? y = cos 2! x = sin 2!

#× 2+-12=1

of8**•*:¥9

Fact : Swot cos 'o= 1

y2 = cos2 2! x2 = sin2 2! so, again we have x2 + y2 = 1, but the graph looks a little different when we look carefully at !: ! 2! y = cos 2! x = sin 2! 0 0 1 0

Notice that, although the curve x2 + y2 = 1 describes the same unit circle as in the previous example, the parametric curve actually traces the circle twice:

Yo,1)

<•

I¥±EE8÷'Effie

IT ZT 1 0 •

5¥ 5¥ 0 1 ÷*

Ex: x = 1 – 2t y = 0.5 t – 1 -2 < t < 4 a) Sketch the curve by using the parametric equations to plot points. Indicate with an arrow the direction in which the curve is traced as t increases.

t x = 1-2t y = 0.5t - 1 -2 5 -2 -1 3 -1.5 0 1 -1 1 2 3 4

Note – Desmos can help you graph parametric equations. Simply look to the upper right where the menu is, next to “Untitled Graph.” Scroll down to “Parametric: Introduction.”

x

EH¥

¥o¥tf÷f+- 1 - 0.5

- 3 0

5595

c.fr#t←

Y only

^← graph

th tipFeth> x

a#EEs•c•_ DEHJm=÷ R :[ -2 , D

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b) Eliminate the parameter to find a Cartesian equation of the curve. Hint: Either look for a relationship between the two (as in cos t and sin t) or solve for t and plug it into the other equation. x = 1 – 2t so 2t = 1 – x which means that t = 0.5(1 – x) and then Cycloid What is the curve described by following one point on a bicycle wheel as the bicycle moves forward?

* 1 -2 t

y = 0.5T.

1

= = ⇒ t= tzltx )#t in terms of ×

y = 0.5T - l

y + 1 = 0.5T

¥¥¥±.ie#efIEoEEEfIEm= - it( o

,

- ⇒

.

What are the parametric equations that describe this?

First, look at OT. Why is its length = r ! ? Does the rest of the image make sense? We want to find parametric equations for x and y: x = |OT| - |PQ| y = |QT| = |CT| - |CQ| From the picture, |OT| = r ! What is |PQ|? sin ! = |PQ|/r => |PQ| = r sin ! so x = r $ - r sin $ = r ($ - sin $) CT is a radius, so |CT| = r cos ! = |CQ|/r => |CQ| = r cos! so y = r - r cos$ = r ( 1 - cos$ ) (Now look at Desmos to see it in action.)

to fy=KH-KQI

*y=r - 1cal Circum .= ZIRl←0t→ '

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10.2 Calculus with Parametric Curves If x = f(t) and y = g(t) and both f and g are differentiable functions of x, then It’s possible to find the tangent line even if you don’t have a way to write y as a function of x directly: By The Chain Rule: %&%' =

%&%) ∙

%)%'

but we want to find dy/dx, so we divide to get

%&%) =

+,+-+.+-

provided that dx/dt is not 0.

Can we also calculate %/&%)/ ?

%/&%)/ = %%)

%&%) =

++-

+,+.+.+-

(This is just the first equation, but replace y with dy/dx.) Ex: A curve C is defined by the parametric equations x = t2 y = t3 – 3t a) Show that C has two tangents at the point (3, 0) and find their equations. If x = 3, then t = 3 or - 3 When t = 3, y =

=

⇐ ⇐±⇒¥

}wwenhaearkpni.lt/Y#g$x

T

( x, y )

(A)3

. 3 B= 3 B - 355 = 0

.

¥ . ¥÷Ft

If = Feedt

""

¥4 's =EE¥=⇐EE#at

* I - Zt from 10.1

y=÷t - I

y=. tx - E -→¥x= - ¥

I¥= -2

¥÷E¥÷⇐ I

When t = - 3, y = So, both values for t work. Next, we need to find dy/dx at (3, 0)

%&%) =

+,+-+.+-

%&%' = 3t2 - 3 %)%' = 2t %&%) =

1'/213'

When t = 3, %&%) =

so the equation of the tangent line is y – 0 = m (x – 3) y = When t = − 3,

%&%) =

y – 0 = m (x – 3) y = b) Find the points on C where the tangent is horizontal or vertical. In other words, we want the numerator or denominator of dy/dx to be 0.

ftp.3tr )= - 3 Bt3B= 0

y=t?3t* E

3t2 – 3 = 0 3t2 = 3 t2 = 1 t = -1, 1 The corresponding points are: x = t2 y = t3 – 3t and x = t2 y = t3 – 3t 2t = 0 t = 0 The corresponding points are: x = t2 y = t3 – 3t c) Determine the concavity of the curve. We want to find

%/&%)/ =

++-

+,+.+.+-

= ++-

5-/65/-3'

= 7-∗/-6/(5-/65)

;-/3'

= <3'/2='/>=?'5 = ='

/>=?'5 = 1'

/>1@'5

%/&%)/ > 0 when t %/&%)/ < 0 when t d) Sketch the curve. Let’s summarize what we know so far: Horizontal tangents at (1, 2) and (1, -2) Vertical tangent at (0, 0) Two tangents at the point (3, 0) Add a couple more points: t = -2: t = 2: Now graph

Areas Ex: Find the area under one arch of the cycloid x = r(< − =>?<) y = r(< − @A=<)

Remember that to find the area, we need to find A = ℎC>DℎE ∗ F>GEℎL

M = JGLLM = JGL3PQ

R y = r(< − @A=<) %)%S = %%S Q(< − =>?<) = so dx = which gives us A = Q 1 − @A=< Q 1 − @A=< G<3P

R = Q3 1 − 2@A=< + @A=3< G<3P

R = Q3 1 − 2@A=< + <

3 (1 + @A=2<) G<3PR

= Q3 < − 2=>?< + <

3 < +<@ =>?2<

2U0

=

10.3 Polar Coordinates When we graph using x and y, we use rectangles to describe and draw our graphs: There are other systems of graphing that don’t use rectangles. In 10.3, we’ll study a system that uses circles – or angles and radii – to graph.

When we use (x, y) to graph, x represents the horizontal distance from the origin y represents the vertical distance from the origin When we use (r, <) to graph, r represents the linear distance from O to P < represents the counter-clockwise angle from the polar axis (the positive part of the x-axis)

Plot the following points whose polar coordinates are given: a) (1, YP@ ) b) (2, 3U ) c) (2, 23P1 ) d) (-3, 1P@ ) a) b)

c) d)

Connecting Cartesian and Polar Coordinates

x = r cos < y = r sin < r2 = x2 + y2 tan < = &)

Ex: Convert the point (2, P1) from Polar to Cartesian coordinates. (2, P1) = (r, <) r = 2 < = P1 x = r cos < = 2 cos P1 = 2 (1/2) = 1 y = r sin < = 2 sin P1 = 2 ( 1

3 ) = 3 (1, 3 ) Ex: Convert the point (1, -1) from Cartesian to Polar coordinates. r2 = x2 + y2 = 1 + 1 = 2 r = tan < = &) = 2<< = -1 < = 1P@ , ZP@ , 2P@ , …. which one should we choose? (1, -1) lies in Quadrant IV so we should pick < = ZP@ or 2P@

Polar Curves Sketch the polar curve < = 1 Note: < is in radians What do we know about r? It can be anything, so the sketch looks like this:

Ex: (P 657) Sketch the curve with Polar equation r = 2 cos < and then find a Cartesian equation for it. Let’s set up a table of values first:

And then graph: Hmmm…that looks like a circle… Converting to Cartesian coordinates: r = 2 cos < r/2 = cos < But we also know that x = r cos < = r (r/2) so 2x = r2 But we also know that r2 = x2 + y2 so 2x = x2 + y2 or x2 + y2 – 2x = 0 Complete the square:

Ex: Sketch the curve r = cos 2 < First, let’s look at the graph in Cartesian coordinates:

And then graph it in Polar coordinates:

This is a four-leaved rose.

Areas in Polar Coordinates How do we find the area under a curve in Cartesian Coordinates? We estimate it with rectangles. So, how can we do it in Polar Coordinates? We should estimate it with sections of circles, called sectors. The formula for the area of a sector of a circle is given by: A = <3 Q

3< where r = the radius of the circle, and < is the angle that defines the sector

We will attempt to divide the region into many sectors and then add their areas. A = YQCZA[\Z@ℎ]C@EAQ`

ab< = <3 Qa

3∆<a`ab<

= <3 [(<a)3∆<a`

ab< = <

3 [(<)3G<L

M

Ex: Find the area enclosed by one loop of the four-leaved rose r = cos 2<

A = <3 [(<)

3G<LM

= <

3 @A=3(2<)G<

d;6d;

= <

@ (1 + @A= 4< )G<d;6d;

=