Given:
[tex]\begin{gathered} r=\sin\theta+5\cos\theta \\ \theta=\frac{\pi}{2} \end{gathered}[/tex]To find: The slope of the tangent line
Explanation:
Let us take,
[tex]\begin{gathered} x=r\cos\theta \\ y=r\sin\theta \end{gathered}[/tex]Substituting the r-value in the above equations, we get,
[tex]x=(\sin\theta+5\cos\theta)\cos\theta[/tex]Differentiating using the product rule,
[tex]\begin{gathered} \frac{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}[/tex]And we have,
[tex]y=(\sin\theta+5\cos\theta)\sin\theta[/tex]Differentiating using the product rule,
[tex]\begin{gathered} \frac{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}[/tex]Dividing equation (2) by (1), we get
[tex]\begin{gathered} \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \\ =\frac{2\sin\theta\cos\theta+5(\cos^2\theta-\sin^2\theta)}{\cos^2\theta-sin^2\theta} \end{gathered}[/tex]The slope of the tangent line at the given angle is,
[tex]\begin{gathered} [\frac{dy}{dx}]_{\frac{\pi}{2}}=\frac{2(1)(0)+5(0-1)}{0-1} \\ =-\frac{5}{-1} \\ =5 \end{gathered}[/tex]Final answer:
The slope of the tangent line to the given polar curve at the point specified by the value of θ is 5.