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:

$\begin{array}{rlr}u(x)& =\cos (x)& u^{\prime }(x)=-\sin (x)\\ v(x)& =\sin (x)& v^{\prime }(x)=\cos (x)\end{array}$

This gives:

$\begin{array}{rl}\frac{\mathrm{d}y}{\mathrm{d}x}& =\frac{v{u}^{\prime }-u{v}^{\prime }}{v^2}\\ & =\frac{\sin (x)\cdot (-\sin (x)-\cos (x)\cdot \cos (x)}{[\sin (x){]}^2}\\ & =\frac{-{\sin }^2(x)-{\cos }^2(x)}{{\sin }^2(x)}\\ & =\frac{-({\sin }^2(x)+{\cos }^2(x))}{{\sin }^2(x)}\\ & =\frac{-1}{{\sin }^2(x)}\\ \end{array}$

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

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

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

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

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

This means our tangent line is:

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