Notice that any polar curver=f(θ) can be
regarded as a parametric curve with parameter θ and
coordinate functions:
x(θ)=rcosθ=f(θ)cosθy(θ)=rsinθ=f(θ)sinθ
Normally, we remove the reference to f, writing r=r(θ) and:
x(θ)=r(θ)cosθy(θ)=r(θ)sinθ
Then the position vector of the curve is
r(θ)=(x(θ),y(θ))=(r(θ)cosθ,r(θ)sinθ)
r(θ) is the position vector (Notice the arrow over the
r.) telling us the position in the plane for each value of the
parameter θ, whereas r(θ) is simply the function
defining the polar curve. Don't get confused.
We will use this formulation in two of our applications of polar curves:
slope and arclength.
c. Slope of a Polar Graph
Recall that the slope of a parametric curve
r(θ)=(x(θ),y(θ)) is:
dxdy=dθdxdθdy
In fact, applying the product rule to a polar curve gives:
dθdxdθdy=dθdrcosθ−r(θ)sinθ=dθdrsinθ+r(θ)cosθ
Consequently:
Slope of a Polar Curve
The slope of a polar curver=r(θ) is
dxdy=dθdrcosθ−r(θ)sinθdθdrsinθ+r(θ)cosθ
provided the denominator is non-zero.
Caution: Remember, dθdy is
in the numerator and dθdx is in the denominator.
It is easy to get these backwards.
Find the slope of the limacon r=1+2cosθ at:
θ=3π
θ=32π
Notice that θ=32π is a point where the curve
passes through the origin since cos32π=2−1
and r=1+2(2−1)=0.
Since dθdr=−2sinθ, the slope is
dxdy=−2sinθcosθ−(1+2cosθ)sinθ−2sinθsinθ+(1+2cosθ)cosθ
At θ=3π=60∘, sinθ=23
and cosθ=21. So:
dxdy∣∣∣∣θ=π/3=−2(23)(21)−(1+2(21))(23)−2(23)(23)+(1+2(21))(21)=−23−3−23+1=−233−21=331
At θ=32π=120∘, sinθ=23
and cosθ=2−1. So:
dxdy∣∣∣∣θ=2π/3=−2(23)(−21)−(1+2(−21))(23)−2(23)(23)+(1+2(−21))(−21)=23−0−23+0=3−3=−3
In this example, notice that tan32π=−3 which is the
slope at θ=32π. This is no coincidence. Look at the
figure. Notice that the curve comes into the origin tangent to the line
θ=32π and the slope of that line is tan32π.
If θ=θo is a point where the polar curve r=r(θ)
passes through the origin, i.e. r(θo)=0, then the slope at
θ=θo is:
dxdy∣∣∣∣θ=θo=tanθo
Proof
At θ=θo, r(θo)=0. So
dxdy∣∣∣∣θ=θo=dθdr(θo)cosθo−r(θo)sinθodθdr(θo)sinθo+r(θo)cosθo=dθdr(θo)cosθodθdr(θo)sinθo=tanθo
Since dθdr=3cos3θ, the slope is:
dxdy=3cos3θcosθ−sin3θsinθ3cos3θsinθ+sin3θcosθ
At θ=6π=30∘:
sinθ=21,cosθ=23,sin3θ=1,cos3θ=0
So:
dxdy∣∣∣∣θ=π/6=−sinθcosθ=2−123=−3
Placeholder text:
Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum
Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum
Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum
Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum
Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum Lorem ipsum