Tangent Lines and Arc Length Parametric Equations Objective: Use the formulas required to find slopes, tangent lines, and arc lengths of parametric and

Tangent Lines and Arc LengthParametric Equations

Objective: Use the formulas required to find slopes, tangent lines, and arc

lengths of parametric and polar curves.

Parametric Equations

• Suppose that a particle moves along a curve C in the xy-plane in such a way that its x- and y-coordinates, as functions of time, are and . We call these the parametric equations of motion for the particle and refer to C as the trajectory of the particle or the graph of the equations. The variable t is called the parameter for the equation.

)(tgy )(tfx

Example 1

• Sketch the trajectory over the time interval 0 < t < 10 of the particle whose parametric equations of motion are ty cos34 ttx sin3

Example 1

• Sketch the trajectory over the time interval 0 < t < 10 of the particle whose parametric equations of motion are

• One way to sketch the trajectory is to choose a representative succession of times, plot the (x, y) coordinates of points on the trajectory at those times, and connect the points with a smooth curve.

ty cos34 ttx sin3

Example 1

• Sketch the trajectory over the time interval 0 < t < 10 of the particle whose parametric equations of motion are

• Observe that there is no t-axis; the values of t appear only as labels on the plotted points, and even these are usually omitted unless it is important

to emphasize the location of the particle at specific times.

ty cos34 ttx sin3

Example 1

• Although parametric equations commonly arise in problems of motion with time as the parameter, they arise in other contexts as well. Thus, unless the problem dictates that the parameter t in the equation represents time, it should be viewed simply as an independent variable that varies over some interval of real numbers. If no restrictions on the parameter are stated explicitly or implied by the equations, then it is understood that it varies from

)(tgy )(tfx

to

Example 2

• Find the graph of the parametric equationsty sintx cos )20( t

Example 2

• Find the graph of the parametric equations

• One way to find the graph is to eliminate the parameter t by noting that Thus the graph is contained in the unit circle. Geometrically, the parameter t can be interpreted as the angle swept out by the radial line from the origin to the point (x, y) = (cost, sint) on the unit circle. As t increases from 0 to 2, the point

traces the circle counterclockwise.

ty sintx cos )20( t

1sincos 2222 ttyx

Orientation

• The direction in which the graph of a pair of parametric equations is traced as the parameter increases is called the direction of increasing power or sometimes the orientation imposed on the curve by the equations. Thus, we make a distinction between a curve, which is a set of points, and a parametric curve, which is a curve with an orientation imposed on it by a set of parametric equations.

Orientation

• For example, we saw in example 2 that the circle represented parametrically is traced counterclockwise as t increases and hence has counterclockwise orientation.

• To obtain parametric equations for the unit circle with clockwise orientation,

we can replace t by –t.

Example 3

• Graph the parametric curve by eliminating the parameter and indicate the

orientation on the graph.

76 ty32 tx

Example 3


orientation on the graph.• To eliminate the parameter we will solve the first

equation for t as a function of x, and then substitute this expression for t into the second equation.

76 ty32 tx

2

3x

t 2376 23 xy x

Example 3


orientation on the graph.• The graph is a line of slope 3 and y-intercept 2. To

find the orientation we must look at the original equations; the direction of increasing t can be deduced by observing that x

increases as t increases or that y increases as t increases.

76 ty32 tx

23 xy

Tangent Lines to Parametric Curves

• We will be concerned with curves that are given by parametric equations x = f(t) and y = g(t) in which f(t) and g(t) have continuous first derivatives with respect to t. If can be proved that if dx/dt is not zero, then y is a differentiable function of x, in which case the chain rule implies that

• This formula makes it possible to find dy/dx directly from the parametric equations without eliminating the parameter.

dtdx

dtdy

dx

dy

/

/

Example 1

• Find the slope of the tangent line to the unit circle

at the point where tytx sin,cos 20 t

6t

Example 1

• Find the slope of the tangent line to the unit circle

at the point where• The slope at a general point on the circle is

• The slope at is

tytx sin,cos 20 t

6t

tt

t

dtdx

dtdy

dx

dycot

sin

cos

/

/

6t

3cot 66/

tdx

dy

Tangent Lines

• It follows from the formula that the tangent line to a parametric curve will be horizontal at those points where dy/dt = 0 and dx/dt does not (0/#).

Tangent Lines


• Two different situations occur when dx/dt = 0. At points where dx/dt =0 and dy/dt does not (#/0), the tangent line has infinite slope and a vertical tangent line at such points.

Tangent Lines


• Two different situations occur when dx/dt = 0. At points where dx/dt =0 and dy/dt does not (#/0), the tangent line has infinite slope and a vertical tangent line at such points.

• When dx/dt and dy/dt =0, we call such points singular points. No general statement can be made about singular points; they must be analyzed case by case.

Example 2

• In a disastrous first flight, an experimental paper airplane follows the trajectory

but crashes into a wall at time t = 10. (a) At what times was the airplane flying horizontally? (b) At what times was it flying vertically?

tyttx cos34,sin3 )0( t

Example 2


but crashes into a wall at time t = 10. (a) At what times was it flying horizontally? (a) The airplane was flying horizontally at those times

when dy/dt = 0 and dx/dt does not.


3,2,,0

sin3

t

tdt

dy

Example 2


but crashes into a wall at time t = 10. (b) At what times was it flying vertically? (b) The airplane was flying vertically at those times

when dx/dt = 0 and dy/dt does not.


tdt

dxcos31

3/1cos

cos310

t

t 51.7,05.5,23.1

)3/1(cos 1

Example 3

• The curve represented by the parametric equations

is called a semicubical parabola. The parameter t can be eliminated by cubing x and squaring y, from which it follows the y2 = x3. The graph of this equation consists of two branches;

an upper branch obtained by graphing y = x3/2 and a lower branch obtained by graphing y = -x3/2.

32 , tytx )( t

Example 4

• Without eliminating the parameter, find and at (1, 1) and (1, -1) on the semicubical parabola given in example 3.

dxdy

2

2

dxyd

Example 4


dxdy

2

2

dxyd

tt

t

dtdx

dtdy

dx

dy

2

3

2

3

/

/ 2

Example 4


dxdy

2

2

dxyd

tt

t

dtdx

dtdy

dx

dy

2

3

2

3

/

/ 2

ttdtdx

dtdy

dx

yd

4

3

2

2/3

/

//

2

2

Example 4


• Since the point (1, 1) on the curves corresponds to t = 1 in the parametric equations, it follows that

dxdy

2

2

dxyd

tt

t

dtdx

dtdy

dx

dy

2

3

2

3

/

/ 2

ttdtdx

dtdy

dx

yd

4

3

2

2/3

/

//

2

2

4

3;

2

3

1

2

2

1

tt dx

yd

dx

dy

Example 4


• Since the point (1, -1) on the curves corresponds to t = -1 in the parametric equations, it follows that

dxdy

2

2

dxyd

tt

t

dtdx

dtdy

dx

dy

2

3

2

3

/

/ 2

ttdtdx

dxdy

dx

yd

4

3

2

2/3

/

//

2

2

4

3;

2

3

1

2

2

1

tt dx

yd

dx

dy

Tangent Lines to Polar Curves

• Our next objective is to find a method for obtaining slopes of tangent lines to polar curves of the form r = f() in which r is a differentiable function of . A curve of this form can be expressed parametrically in terms of the parameter by substituting f() for r in the equation x = rcos and y = rsin. This yields

sin)(,cos)( fyfx


• From this we obtain

sin)(,cos)( fyfx

cossincos)(sin)( /

d

drrff

d

dx

sincossin)(cos)( /

d

drrff

d

dy


• Thus, if and are continuous and if then y is a differentiable function of x, and with in place of t yields

ddx /

ddr

ddr

ddx

ddy

r

r

dx

dy

cossin

sincos

ddy / 0ddx

cossincos)(sin)( /

d

drrff

d

dx

sincossin)(cos)( /

d

drrff

d

dy

Example 5

• Find the slope of the tangent line to the circle r =4cos at the point where .4

Example 5

• Find the slope of the tangent line to the circle r =4cos at the point where .

• Substituting into the formula gives

sincos4sincos4

sin4cos4

cossin

sincos 22

ddr

ddr

ddx

ddy

r

r

dx

dy

4

Example 5


• Substituting into the formula gives

2cot2sin4

2cos4

sincos8

)sin(cos4 22

sincos4sincos4

sin4cos4

cossin

sincos 22

ddr

ddr

ddx

ddy

r

r

dx

dy

4

0cot 24/

dx

dym

Example 5


• Another way we can do this is if we express the cardioid parametrically by substituting r = 4cos into the conversion formulas x = rcos and y = rsin. This yields

2sin2sincos4

cos4 2

y

x

4

2cos4d

dy

2sin4sincos8 d

dx

Example 5


• Another way we can do this is to solve for dr/d, cos, and the sin and substitute.

22

22

cos

sin

22

r

4

224

sin4

22

ddr

ddr

ddr

ddx

ddy

r

r

dx

dy

cossin

sincos

Example 5


• Another way we can do this is to solve for dr/d, cos, and the sin and substitute.

22

22

cos

sin

22

r

4

224

sin4

22

ddr

ddr

ddr

ddx

ddy

r

r

dx

dy

cossin

sincos

016

22

)22(222222

)22(22 22

22

ddx

ddy

dx

dy

Example 6

• Find the points on the cardioid r = 1 – cos at which there is a horizontal tangent line, a vertical tangent line, or a singular point.

Example 6


sincossin)cos1(

sinsincos)cos1(

cossin

sincos

ddr

ddr

ddx

ddy

r

r

dx

dy

Example 6


sincossin)cos1(

sinsincos)cos1(

cossin

sincos

ddr

ddr

ddx

ddy

r

r

dx

dy

)1cos2(sin

)cos1)(1cos2(

sincossin2

)cos(sincos 22

ddx

ddy

dx

dy

2cos1 1coscos2 2

Example 6


• This is easiest if we express the cardioid parametrically by substituting r = 1 – cos into the conversion formulas x = rcos and y = rsin. This yields

)1cos2(sin ddx

sin)cos1( y cos)cos1( x

)cos21)(cos1(

d

dy

Example 6


• A horizontal tangent occurs when

35

3 ,,2,,0

)1cos2(sin

d

dx

2,,,0

)cos21)(cos1(

34

32

d

dy

0,0 ddx

ddy

34

32 ,

Example 6


• A vertical tangent occurs when

35

3 ,,2,,0

)1cos2(sin

d

dx

2,,,0

)cos21)(cos1(

34

32

d

dy

0,0 ddx

ddy

35

3 ,,

Example 6


• A singular point occurs when

35

3 ,,2,,0

)1cos2(sin

d

dx

2,,,0

)cos21)(cos1(

34

32

d

dy

0,0 ddx

ddy

2,0

Tangent Lines to Polar Curves at the Origin

• The following theorem could prove useful.

Tangent Lines to Polar Curves at the Origin

• The following theorem could prove useful.

• This theorem tells us that equations of the tangent lines at the origin to the curve r = f() can be obtained by solving the equation f() = 0. It is important to keep in mind that r = f() may be zero for more than one value of , so there may be more than one tangent line at the origin.

Example 7

• The three-petal rose r = sin3 has three tangent lines at the origin, which can be found by solving the equation sin3 = 0. The solutions are 3

23 ,,0

Arc Length of a Parametric Curve

• Here is the formula for the arc length of a parametric curve:

b

a dt

dy

dt

dxL

22

Arc Length of a Polar Curve

• A formula for the arc length of a polar curve r = f() can be derived by expressing the curve in parametric form and applying the formula for the are length of a parametric curve.

Arc Length of a Polar Curve

• A formula for the arc length of a polar curve r = f() can be derived by expressing the curve in parametric form and applying the formula for the are length of a parametric curve.

Example 8

• Find the arc length of the spiral r = e between = 0 and = .

Example 8

• Find the arc length of the spiral r = e between = 0 and = .

)1(222 0

0

eedeL

0

222

2 )()( deedd

drrL

Example 9

• Find the total arc length of the cardioid r = 1 + cos.

Example 9


2

0

222

2 )sin()cos1( ddd

drrL

Example 9


Aappendix 45Identity

|cos|2)(cos2cos122

02

2

0212

2

0

dd

2

0

222

2 )sin()cos1( ddd

drrL

Example 9


2

02

2

0212

2

0

|cos|2)(cos2cos12 dd

2

0

222

2 )sin()cos1( ddd

drrL

8sin8cos4|cos|202

02

2

02

d

Other Important Ideas

• Here are some formulas that you will need to know for the AP Exam. These are not in the book. These formulas are for equations defined parametrically.

2/2/ ))(())(( tytxspeed

dttytxb

a

2/2/ ))(())((Travled Distance


• A particle moves in the xy-plane so that its velocity vector at time t is and the particle’s position vector at time t = 0 is (1,0). What is the position vector of the particle when t = 3?

a) d)

b) e)

c)

))sin(,()( 2 tttv

),9( 1

),10( 2

)2,6(

)2,10(

),10(


• A particle moves in the xy-plane so that its velocity vector at time t is and the particle’s position vector at time t = 0 is (1,0). What is the position vector of the particle when t = 3?

a) d)

b) e)

c)

))sin(,()( 2 tttv

),9( 1

),10( 2

)2,6(

)2,10(

),10( 10)(,3;1

01

3

32

3

3

tst

C

Cdtt

t

t

211

1

1

)(,3;)cos(

0

)cos()sin(

tstt

C

Ctdtt


• A particle moves in the xy-plane so that at any time t its coordinates are and . At t = 1, its acceleration vector is

a) d)

b) e)

c)

2tx

)3,2(

)6,2(

)6,1(

)6,2(

)2,1(

34 ty


• A particle moves in the xy-plane so that at any time t its coordinates are and . At t = 1, its acceleration vector is

a) d)

b) e)

c)

2tx

)3,2(

)6,2(

)6,1(

)6,2(

)2,1( 2

2//

/

2

x

tx

tx

34 ty

ty

ty

ty

6

3

4

//

2/

3


• Which of the following is an equation of the line tangent to the curve with parametric equations

at the point where t = 0?

a) d) b) e)

c)

tt eyex 6,3

0122 yx

0122 yx

092 yx

02 yx

0152 yx


• Which of the following is an equation of the line tangent to the curve with parametric equations

at the point where t = 0?

a) d) b) e)

c)

tt eyex 6,3

0122 yx

0122 yx

092 yx

02 yx

0152 yx

6,3

0

yx

t

23

6

t

t

e

e

dx

dy

0122

)3(26

yx

xy

Homework

• Page 701

• 1-11 odd

• 41-49 odd

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Tangent Lines and Arc Length Parametric Equations Objective: Use the formulas required to find slopes, tangent lines, and arc lengths of parametric and