What is the slope of r=tantheta-thetar=tanθ−θ at theta=pi/8θ=π8?

1 Answer
AltairSafir
Mar 19, 2018

Derivative with Polar Coordinates is (\partial x) / (\partial y) = ((\partial r) / (\partial theta) sin(theta)+ rcos(theta))/((\partial r) / (\partial theta) cos(theta)- rsin(theta)∂x∂y=∂r∂θsin(θ)+rcos(θ)∂r∂θcos(θ)−rsin(θ)

Explanation:

(\partial r) / (\partial theta) = 1/cos^2(theta)+1∂r∂θ=1cos2(θ)+1
r(theta = pi/8)=0.0 22r(θ=π8)=0.022
(\partial x) / (\partial y) = (tan(theta)/cos(theta)+sin(theta)+rcos(theta))/(1/cos(theta)+cos(theta)-rsin(theta))∂x∂y=tan(θ)cos(θ)+sin(θ)+rcos(θ)1cos(θ)+cos(θ)−rsin(θ)
(\partial x) / (\partial y)(theta=pi/8) =1.57 ∂x∂y(θ=π8)=1.57

Related questions
See all questions in Determining the Slope and Tangent Lines for a Polar Curve
Impact of this question
3820 views around the world
You can reuse this answer
Creative Commons License