Polar Coordinates/equations 10.4 coordinates (r, θ polar axis is … · 2017-01-19 · calc notes derivatives January 19, 2017 Polar Coordinates/equations 10.4 coordinates (r, θ)

calc notes derivatives January 19, 2017

Polar Coordinates/equations 10.4

coordinates (r, θ)polar axis is the "x axis"the POLE is the origin.



angle is measured from positive x -axis

and r is the distance from the pole(origin)

If the r value is described as negative,

then the point is "flipped" across the pole.


the question is how can you arrive at the location described above by reversing your direction around the circle or reflecting through the origin.

4

(4, -4 /3)

(-4, 5 / 3 )

(-4, - /3)



recall that cos = x/r and sin = y/r


x=r cos y=r sin

x=5 cos /4 y = 5 sin / 4

x=

y=


r= ?

= ?(-6, 0)


(-3,3)

tan = y/x

arctan -3/3=

arctan -1 = - / 4

but the angle must lie in quadrant 2, so the angle is 3 /4

to find r, use So r = 3 2








don't be chicken




Graph: r = 4 + 3 cos θ r= 2 - 4 sin θ

r = 3 sin 4θ r=5cos 3θ

the viewing window for the angle theta is usually

0 to 2 pi with a step of 0.1





What do you suppose the graph of

r = 1 / (cos θ + sinθ) is ?

hint: eliminate the denominator and changeto rectangular form.



Tests for symmetry





Theorem: Suppose that r is a continuous function of θ and r(θ) =0 and there is some δ>0 such that r(θ0)≠0 for0< abs(θ-θ0 )<δ. Then the graph of the polar equationr=r(θ) approaches and leaves the pole at the angle of the line, θ=θ0 as θ passes through θ0. That is the curve istangent to the line θ=θ0 at the pole.

Alternate form:If f(α)=0 and f ' (α) ≠0, then θ = α is tangent at the pole to the graph of r = f( θ ).

EX Find the tangents at the pole for r = 3 cos θ

3 cosθ = 0 ⇒ θ =π / 2

AND r ' = -3 sin θ , with -3 sin ( π / 2) ≠0then θ = π / 2 is the tangent at the pole

TRY r = 3 ( 1 - cos θ ) to find tangents at the pole

answer none



Find all points on the graph that have vertical tangent lines for r = 3 - 2 cos θ


How do you find vertical or horizontal tangent lines.

when dy/dθ = 0, horizontal

when dx/dθ = 0, vertical

r = 3 - 2cosθ

since y = r sinθ ⇒y=(3-2cosθ)sinθdy/dθ=3cosθ -2(-sinθ sinθ) + -2cosθ(cosθ)now set = 0 and solve

0=4cos2 θ-3cosθ-2 ⇒cosθ= (3±√41)/8

cos θ=-.4253 ⇒ θ=2.01

SO there is a horizontal tangent at θ=2.01 and because of symmetry, there is also a horiz. tangent at θ=-2.01





The arc length formula for a polar equation can be derived by using the arc length formula for a function in x,y or a parametric equation.

∫ √ (dx/dθ)2 + ((dy/dθ)2 dθ

or ∫ r2 + (dr / dθ )2 dθ

example: Find the arc length of the cardiod r = 1 + sin θ. (HINT GRAPH)

∫ r2 + (dr/dθ)2 dθ

∫ (1 + sin θ)2 + cos 2 θ dθ


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Polar Coordinates/equations 10.4 coordinates (r, θ polar axis is … · 2017-01-19 · calc notes derivatives January 19, 2017 Polar Coordinates/equations 10.4 coordinates (r, θ)