Science:Math Exam Resources/Courses/MATH110/December 2010/Question 08 (b)

MATH110 December 2010

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Question 08 (b)
Find the equation of the line tangent to the curve $\displaystyle y=4\sin x\cos x$ at $x={\frac {\pi }{3}}$.

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 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
To find the equation of the line you'll need a slope and a point on the line. The slope of a tangent line is its derivative at a point; one of the points on the tangent line is also a point on the graph of the original function.

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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 a tangent line we need both the slope and a point on the line. To find the slope, we simply take the derivative of the function using the product rule and plug in the value $x={\frac {\pi }{3}}$ ${\frac {dy}{dx}}=4((\sin x)'\cos x+\sin x(\cos x)')=4(\cos ^{2}x-\sin ^{2}x)$ We know from part (a) that $\cos \left({\frac {\pi }{3}}\right)={\frac {1}{2}}$ and $\sin \left({\frac {\pi }{3}}\right)={\frac {\sqrt {3}}{2}}$. So the value of the derivative at $x={\frac {\pi }{3}}$ is: $\displaystyle 4\left(\left({\frac {1}{2}}\right)^{2}-\left({\frac {\sqrt {3}}{2}}\right)^{2}\right)=4\left({\frac {1}{4}}-{\frac {3}{4}}\right)=-{\frac {4}{2}}=-2$. So the slope of the tangent line at $x={\frac {\pi }{3}}$ is $-2$. In order to find a point on the line, we will use the point on the graph of the original function at $x={\frac {\pi }{3}}$. We already have the x-coordinate, so we only need to find the y-coordinate by plugging the x-value into the original function. This gives: $y=4\sin \left({\frac {\pi }{3}}\right)\cos \left({\frac {\pi }{3}}\right)=4\cdot {\frac {\sqrt {3}}{2}}\cdot {\frac {1}{2}}={\sqrt {3}}$ Now that we have the slope $-2$ and the point $\left({\frac {\pi }{3}},{\sqrt {3}}\right)$, we use the point-slope formula to get the equation of the tangent line. $y-{\sqrt {3}}=-2\left(x-{\frac {\pi }{3}}\right)$ Note that this would be a sufficient answer, however, if simplified to $y=mx+b$ form, the equation would be: $y=-2x+{\frac {2\pi }{3}}+{\sqrt {3}}$

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