AP Calculus BC Monday, 16 November 2015 OBJECTIVE TSW (1) find the slope of a tangent line to a parametric curve, and (2) find the concavity of a parametric

AP Calculus BCMonday, 16 November 2015

• OBJECTIVE TSW (1) find the slope of a tangent line to a parametric curve, and (2) find the concavity of a parametric curve.

• TODAY’S ASSIGNMENT (due tomorrow)

– Sec. 11.2: Problems given on the next slide.

• TEST: Parametric Equations, Polar Equations, and Vectors is this Friday, 20 November 2015.

Sec. 11.2: Calculus with Parametric Equations

Find (a) dy/dx, (b) d 2y/dx2, (c) the slope, (d) the equation of the tangent line, and (e) concavity (if possible) at the given value of the parameter.

, 3 11 , 1) x t y t t

2 3 2, 2 ,2 0) x t t y t t

cos , 3sin3 , 0) x y

, ,4 1 2) x t y t t

2 , 2 1 cos , ,) 17 : 2 4x y x

Find all points (if any) of horizontal and vertical tangency to the curve defined by the given parametric equations.

Due tomorrow, Tuesday, 17 November 2015.

2 15 , 3) x t y t t co6 s , i) 2s n2x y



How do you find the derivative of a set of parametric equations?


11.1


Ex: Find dy / dx for the curve given bysin and cos .x t y t

dy

dx

dy dt

dx dt sin

cos

t

t

tan t


For higher order derivatives, use Theorem 11.1 repeatedly.

Notice that the denominator for each higher-order derivative is always dx/dt.

2

2

d dyd y d dy dt dx

dx dx dx dtdxSECOND DERIVATIVE

2

23 2

3 2

d d ydt dxd y d d y

dx dx dtdx dxTHIRD DERIVATIVE


Ex: For the curve given by

find the slope and concavity at the point

(2, 3).

21and 4 ,

4x t y t

dy

dx

dy dt

dx dt

121

2

t

t

t t 3 2t


The second derivative is

1 232

1

2

t

t

2

2

d y

dx

ddy dx

dtdx dt

3 2dt

dtdx dt

1 23t t

3t

2

2

, 2, 3

3 4x y

d y

dx


We’re given the point (2, 3) & 21and 4 .

4x t y t

Since x = 2, that means that or t = 4.

The slope at (2, 3) is:

And the concavity at (2, 3) is:

2 ,t3 2dyt

dx

8

2

23

d yt

dx

12

0

3 2

, 2, 3

4x y

dy

dx

∴ concave up


Ex: The prolate cycloid given by

2 sin and 2 cosx t t y t crosses itself at the point (0, 2). Find the equations of both tangent lines at this point.








A point is given; you need only determine the slope, dy/dx.

dy

dx

sin

2 cos

t

t

Now you need to determine t.

Use the original parametric equations to determine t.


0 2 sin and 2 2 cost t t

Solve one of these equations for t.

The second equation would be the easiest.

0 cos t 0 cos t

1cos 0t

2t

2 2 cos t


When t = /2,

dy

dx

sin

2 cos

t

t

2

sin / 2

2 cos / 2t

dy

dx

/ 2

and the equation is

2 02

y x


When t = –/2,

2

sin / 2

2 cos / 2t

dy

dx

/ 2

and the equation is

2 02

y x


0 and 0dy dx

dt dt

Horizontal Tangents

If

when t = t0, then the curve represented by

andx f t y g t

has a horizontal tangent at 0 0 .,f t g t


0 and 0dx dy

dt dt

Vertical Tangents

If

when t = t0, then the curve represented by

andx f t y g t

has a vertical tangent at 0 0 .,f t g t


Ex: Find all points (if any) of horizontal and vertical tangency to the curve defined by

x = t + 1 and y = t 2 + 3t.

2 3dy

tdt



x = t + 1 and y = t 2 + 3t.

2 3dy

tdt

0dx

dt NEVER 0

dy

dt

3

2t


3

2t

Horizontal tangency:

0dx

dt NEVE

R0

dy

dt

3

2t

0 & 0dy dx

dt dt

1 9,

2 4

Vertical tangency: 0 & 0dx dy

dt dt

NEVER NONE



x = cosθ and y = 2sin2θ.


Find (a) dy/dx, (b) d 2y/dx2, (c) the slope, (d) the equation of the tangent line, and (e) concavity (if possible) at the given value of the parameter.

, 3 11 , 1) x t y t t

2 3 2, 2 ,2 0) x t t y t t

cos , 3sin3 , 0) x y

, ,4 1 2) x t y t t

2 , 2 1 cos , ,) 17 : 2 4x y x

Find all points (if any) of horizontal and vertical tangency to the curve defined by the given parametric equations.

Due tomorrow, Tuesday, 17 November 2015.

2 15 , 3) x t y t t co6 s , i) 2s n2x y

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AP Calculus BC Monday, 16 November 2015 OBJECTIVE TSW (1) find the slope of a tangent line to a parametric curve, and (2) find the concavity of a parametric