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

MATH110 December 2010

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

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

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[show] 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.