Science:Math Exam Resources/Courses/MATH110/December 2010/Question 10

MATH110 December 2010

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q2 (a)  •  Q2 (b)  •  Q2 (c)  •  Q3 (a)  •  Q3 (b)  •  Q3 (c)  •  Q4 (a)  •  Q4 (b)  •  Q5 (a)  •  Q5 (b)  •  Q5 (c)  •  Q6 (a)  •  Q6 (b)  •  Q6 (c)  •  Q7  •  Q8 (a)  •  Q8 (b)  •  Q9 (a)  •  Q9 (b)  •  Q9 (c)  •  Q9 (d)  •  Q9 (e)  •  Q9 (f)  •  Q10  •  

   Other MATH110 Exams

  •  December 2011  •  December 2010  •  April 2012  •  April 2011  •  December 2012  •  April 2013  •  December 2013  •  April 2014  •  December 2014  •  December 2015  •  April 2017  •  April 2016  •  December 2016  •  December 2017  •  April 2018  •  

Question 10
(originally for bonus marks) Prove that the curve $y={\frac {\cos x-1}{x}}$ has an infinte number of horizontal tangent lines.

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
A curve has horizontal tangent lines when its derivative is equal to zero.

Checking a solution serves two purposes: helping you if, after having used the hint, 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. As was noted in the hint, the curve has horizontal tangent lines at values where its derivative is equal to zero. Thus we first take the derivative of the curve, using the quotient rule: ${\begin{aligned}{\frac {dy}{dx}}&={\frac {(\cos x-1)'\cdot x-(\cos x-1)\cdot (x)'}{x^{2}}}\\&={\frac {-\sin x\cdot x-(\cos x-1)\cdot 1}{x^{2}}}\\&={\frac {-x\sin x-\cos x+1}{x^{2}}}\end{aligned}}$ We set this equal to zero and simplify to get: ${\begin{aligned}{\frac {-x\sin x-\cos x+1}{x^{2}}}&=0\\-x\sin x-\cos x+1&=0\\-x\sin x-\cos x&=-1\\\end{aligned}}$ So we are trying to find x-values that satisfy this final equation; if such an x-value exists, $y$ has a horizontal tangent at that point. From here, there is no specific method or surefire trick to find the answer - the only thing required is an understanding of the behavior of $\sin x$ and $\cos x$, especially considering their values as points on the unit circle. The most useful observation here is that on the unit circle, when $\cos x$ is zero, $\sin x$ is $\pm 1$ and vice versa. This is one step towards a solution, because if we can find a place where $\sin x=0$ and $\cos x=1$, we will have found a combination of $x\sin x$ and $\cos x$ that will equal -1, which is what we want. In fact, when $x=0$ we know that $\sin x=0$ and $\cos x=1$. On the right hand side of the equation above this gives -1, just like we want. Thus $x=0$ is one solution to our equation; our curve will have a horizontal tangent line here. Because $\sin x$ and $\cos x$ are periodic, we know that this solution will work anytime we are at the same angle on the unit circle, i.e. the angles $2\pi ,4\pi ,6\pi ,\dots$ and so on. So there are an infinite number of solutions, $0,2\pi ,4\pi ,\dots$ and so the curve has an infinite number of horizontal tangent lines.