Slope of Tangent line to Polar curve

[JProgrammer]

I am trying to find the slope of the tangent line of this polar equation:

r = 4 + sin theta, (4,0)

I put the equation into wolfram alpha and it gives me a 3D plot.

If someone could help me find the slope of the tangent line, I would really appreciate it.
Thank you.

 

[johng1]

Here's the graph and a way to compute:

View attachment 5902

 

[soroban]

I am trying to find the slope of the tangent line of this polar equation:

[tex]r \:=\: 4 + \sin\theta\;\; (4,0)[/tex]

Here is the derivation of the formula you want.

[tex]\begin{array}{cccccc}
y \;=\;r\sin\theta & \Rightarrow & \dfrac{dy}{d\theta} \;=\;r\cos\theta + r'\sin\theta \\
x \;=\;r\cos\theta & \Rightarrow & \;\;\dfrac{dx}{d\theta} \;=\;-r\sin\theta + r'\cos\theta \end{array}[/tex]

[tex]\text{Therefore: }\;\frac{dy}{dx} \;=\;\dfrac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}} \;=\; \dfrac{r\cos\theta + r'\sin\theta}{-r\sin\theta + r'\cos\theta}[/tex][tex]\text{For this problem: }\;r \:=\:4 + \sin\theta,\;r' \:=\:\cos\theta[/tex]

[tex]\text{We have: }\;\dfrac{dy}{dx} \;=\;\dfrac{(4+\sin\theta)\cos\theta + \cos\theta\sin\theta}{-(4+\sin\theta)\sin\theta + \cos\theta\cos\theta}[/tex]

Now substitute [tex]\theta = 0.[/tex]