Question

16. Find the slope of the tangent line to the given polar curve at the point specified by the value of θ.

Answer 1

Given:

`\begin{gathered} r=\sin\theta+5\cos\theta \\ \theta=(\pi)/(2) \end{gathered}`

To find: The slope of the tangent line

Step-by-step explanation:

Let us take,

`\begin{gathered} x=r\cos\theta \\ y=r\sin\theta \end{gathered}`

Substituting the r-value in the above equations, we get,

`x=(\sin\theta+5\cos\theta)\cos\theta`

Differentiating using the product rule,

`\begin{gathered} (dx)/(d\theta)=(\sin\theta+5\cos\theta)(-\sin\theta)+\cos\theta(\cos\theta-5\sin\theta) \\ =-\sin^2\theta-5\sin\theta\cos\theta+\cos^2\theta-5\sin\theta\cos\theta \\ =\cos^2\theta-\sin^2\theta..........(1) \end{gathered}`

And we have,

`y=(\sin\theta+5\cos\theta)\sin\theta`

Differentiating using the product rule,

`\begin{gathered} (dy)/(d\theta)=(\sin\theta+5\cos\theta)(cos\theta)+\sin\theta(\cos\theta-5\sin\theta) \\ =\sin\theta\cos\theta+5\cos^2\theta+\sin\theta\cos\theta-5\sin^2\theta \\ =2\sin\theta\cos\theta+5(\cos^2\theta-\sin^2\theta)..........(2) \end{gathered}`

Dividing equation (2) by (1), we get

`\begin{gathered} (dy)/(dx)=((dy)/(d\theta))/((dx)/(d\theta)) \\ =(2\sin\theta\cos\theta+5(\cos^2\theta-\sin^2\theta))/(\cos^2\theta-sin^2\theta) \end{gathered}`

The slope of the tangent line at the given angle is,

`\begin{gathered} [(dy)/(dx)]_{(\pi)/(2)}=(2(1)(0)+5(0-1))/(0-1) \\ =-(5)/(-1) \\ =5 \end{gathered}`

Final answer:

The slope of the tangent line to the given polar curve at the point specified by the value of θ is 5.