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

MATH110 December 2013

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Question 05 (b)
Find the equation of the line tangent to the curve $y={\dfrac {\cos x}{\sin x}}$ at $x={\dfrac {5\pi }{3}}$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
To compute the tangent line at a point, what important concept in this course do we need to utilize?

Hint 2
You will need to compute the derivative (which is the slope of the tangent line to the curve at a point). This can be done using the quotient rule.

Hint 3
Don't forget to find the y value of the point, plug in the x coordinate into the original equation.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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: $y=-{\frac {\sqrt {3}}{3}}-{\frac {4}{3}}\left(x-{\frac {5\pi }{3}}\right)$

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