Science:Math Exam Resources/Courses/MATH110/December 2013/Question 05 (b)/Solution 1

To find the equation of the tangent line, we need to find the slope via the derivative. We can either apply the quotient rule or the power rule (by recognising $y=\tan(x)^{-1}$. Using the quotient rule:

${\begin{aligned}u(x)&=\cos(x)&u'(x)=-\sin(x)\\v(x)&=\sin(x)&v'(x)=\cos(x)\end{aligned}}$

This gives:

${\begin{aligned}{\frac {{\rm {d}}{y}}{{\rm {d}}x}}&={\frac {vu'-uv'}{v^{2}}}\\&={\frac {\sin(x)\cdot (-\sin(x)-\cos(x)\cdot \cos(x)}{[\sin(x)]^{2}}}\\&={\frac {-\sin ^{2}(x)-\cos ^{2}(x)}{\sin ^{2}(x)}}\\&={\frac {-\left(\sin ^{2}(x)+\cos ^{2}(x)\right)}{\sin ^{2}(x)}}\\&={\frac {-1}{\sin ^{2}(x)}}\\\end{aligned}}$

At $x={\frac {5\pi }{3}}$, we have $\sin \left({\frac {5\pi }{3}}\right)=-{\frac {\sqrt {3}}{2}}$ (from part (a)), and so that means that the slope of the curve at $x={\frac {5\pi }{3}}$ is

${\begin{aligned}y'&={\frac {-1}{\sin ^{2}\left(x\right)}}\\&={\frac {-1}{\sin ^{2}\left({\frac {5\pi }{3}}\right)}}\\&={\frac {-1}{\left[-{\frac {\sqrt {3}}{2}}\right]^{2}}}\\&={\frac {-1}{\frac {3}{4}}}\\&={\frac {-4}{3}}\\\end{aligned}}$

We will also need the $y$-value at the point

${\begin{aligned}y&={\frac {\cos(x)}{\sin(x)}}\\&={\frac {1}{\tan \left({\frac {5\pi }{3}}\right)}}\\&={\frac {1}{\tan \left({\frac {2\pi }{3}}\right)}}\\&={\frac {1}{-{\sqrt {3}}}}\\&=-{\frac {\sqrt {3}}{3}}\end{aligned}}$

(here we used the fact that the function $\displaystyle \tan(x)$ is $\pi$ periodic and then applied the computation from part (a)).

This means our tangent line is:

$y=-{\frac {\sqrt {3}}{3}}-{\frac {4}{3}}\left(x-{\frac {5\pi }{3}}\right)$