Science:Math Exam Resources/Courses/MATH110/April 2013/Question 06

MATH110 April 2013

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[hide] Question 06
At what point on the interval $(0,{\frac {\pi }{2}})$ does the tangent line to the curve $y=3x+{\sqrt {3}}\cos(x)-\sin(x)$ have the smallest slope?

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.

[show] Hint 1
See the word "smallest"? -- this is an optimization question. But read carefully; what is being minimized here?

[show] Hint 2
This question is asking you to find a minimum slope of a tangent line. This means you first need to find a formula for the slope of the tangent line (the derivative); then find the minimum value of this expression.

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[show]

Solution

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First we find an formula for the quantity we're minimizing: the slope of the tangent line. This formula is the derivative of the function:

\displaystyle f'(x)=3-{\sqrt {3}}\sin x-\cos x

This is the formula we are minimizing, which we will now call $\displaystyle g(x)=3-{\sqrt {3}}\sin x-\cos x$ . Following our usual procedure for finding maxima/minima, we take the derivative of g(x), set it equal to zero, and solve for the critical points.

g'(x)=-{\sqrt {3}}\cos x+\sin x

Setting equal to zero and solving gives:

{\sqrt {3}}\cos x=\sin x

So we are looking for an angle where $\sin(x)$ is equal to ${\sqrt {3}}\cos(x)$ . There is no "formula" to solve this equation; instead you have to use trig ratios and special triangles, or consult the unit circle. On a 30-60-90 degree triangle, or at the angles 30 (in radians: x = π/6) and 60 (in radians: x = π/3) on the unit circle, $\sin(x)$ and $\cos(x)$ are related in a ratio of $1:{\sqrt {3}}$ so these angles are our candidates. At these values we have

\cos \left({\frac {\pi }{6}}\right)={\frac {\sqrt {3}}{2}},\quad \sin \left({\frac {\pi }{6}}\right)={\frac {1}{2}}

\cos \left({\frac {\pi }{3}}\right)={\frac {1}{2}},\quad \sin \left({\frac {\pi }{3}}\right)={\frac {\sqrt {3}}{2}}

Hence x = π/3 is the solution of equation and hence our critical point. We check if this point is indeed a minimum by performing the first derivative test: