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Parametric Equations. Eliminating the Parameter. 1). 2). 11.2 Slope and Concavity. For the curve given by Find the slope and concavity at the point (2,3). At (2, 3) t = 4 and the slope is 8. The second derivative is positive so graph is concave up. Horizontal and Vertical tangents.

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11 2 slope and concavity
11.2 Slope and Concavity

For the curve given by

Find the slope and concavity at the point (2,3)

At (2, 3) t = 4 and the slope is 8. The second derivative is positive so graph is concave up

horizontal and vertical tangents
Horizontal and Vertical tangents

A horizontal tangent occurs when dy/dt = 0 but dx/dt0.

A vertical tangent occurs when dx/dt = 0 but dy/dt0.

Vertical tangents

Horizontal tangent

polar coordinates
Polar Coordinates

Pole

Polar axis

Figure 9.37.

polar rectangular equivalences
Polar/Rectangular Equivalences

θ)

x2 + y2 = r2

tan θ = y/x

x = r cos θ

y = r sin θ

symmetries
Symmetries

Figure 9.40(a-c).

slide12

Figure 9.42(a-b).

Graph r2 = 4 cos θ

finding points of intersection
Finding points of intersection

Figure 9.45.

Third point does not show up.

On r = 1, point is (1, π)

On r = 1-2 cos θ, point is (-1, 0)

slope of a polar curve
Where x = r cos θ = f(θ) cos θ

And y = r sin θ = f(θ) sin θ

Slope of a polar curve

Horizontal tangent where dy/dθ = 0 and dx/dθ≠0

Vertical tangent where dx/dθ = 0 and dy/dθ≠0

finding slopes and horizontal and vertical tangent lines
For r = 1 – cosθ

(a) Find the slope at θ = π/6

(b) Find horizontal tangents

(c) Find vertical tangents

Finding slopes and horizontal and vertical tangent lines
find vertical tangents
Find Vertical Tangents

Horizontal tangents at:

Vertical tangents at:

finding tangent lines at the pole
Finding Tangent Lines at the pole

Figure 9.47.

r = 2 sin 3θ

r = 2 sin 3θ = 0

3θ = 0, π, 2 π, 3 π

θ = 0, π/3, 2 π/3, π

area in the plane
Area in the Plane

Figure 9.48.

area of region
Area of region

Figure 9.49.

length of a curve in polar coordinates
Length of a Curve in Polar Coordinates

Find the length of the arc for r = 2 – 2cosθ

sin2A =(1-cos2A)/2

2 sin2A =1-cos2A

2 sin2 (1/2θ) =1-cosθ

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