Question

What is the slope of the tangent line of `r=2theta^2-3thetacos(2theta-(pi)/3)` at `theta=(-5pi)/3`?

Answer 1

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)`

Explanation:

`(\partial r) / (\partial theta) = 4 theta-3*cos(2 theta-(pi)/(3))-3 theta*(-sin(2 theta - pi/3))*2=`
`=4 theta-3*cos(2 theta-(pi)/(3))+6 theta*(sin(2 theta - pi/3))`
The numerator:
`(\partial r) / (\partial theta)*sin(theta) =`
`=((4 theta-3*cos(2 theta-(pi)/(3))+6 theta*(sin(2 theta - pi/3)))*sin(theta)`
`r*cos(theta)=(2 theta^2-3 theta cos(2 theta-(pi)/3))*cos(theta)`

The denominator:
`(\partial r) / (\partial theta)*cos(theta) =`
`4 theta-3*cos(2 theta-(pi)/(3))+6 theta*(sin(2 theta - pi/3))*cos(theta)`
`r*sin(theta)=(2 theta^2-3 theta cos(2 theta-(pi)/3))*sin(theta)`
You are calculating the slope at `theta = -(5 pi)/3`, insert `theta = -(5 pi)/3` in the formulae above:

The numerator:
`(\partial r) / (\partial theta)*sin(theta) = -42.999`
`r*cos(theta)=31.343`

The denominator:
`(\partial r) / (\partial theta)*cos(theta) = -24.825`
`r*sin(theta)=54.287`

`(\partial x) / (\partial y)(theta=-(5 pi)/3)=`
`=(-42.999+31.343)/(-24.825-54.287)=0.15`

For more click here.