Science:Math Exam Resources/Courses/MATH102/December 2011/Question 01 (b) ii

MATH102 December 2011

  •  Q1 (a) i  •  Q1 (a) ii  •  Q1 (b) i  •  Q1 (b) ii  •  Q1 (b) iii  •  Q1 (c)  •  Q2 (a) i  •  Q2 (a) ii  •  Q2 (a) iii  •  Q2 (b) i  •  Q2 (b) ii  •  Q2 (b) iii  •  Q2 (c)  •  Q3  •  Q4 (i)  •  Q4 (ii)  •  Q4 (iii)  •  Q4 (iv)  •  Q4 (v)  •  Q4 (vi)  •  Q5  •  Q6  •  Q7 (i)  •  Q7 (ii)  •  Q7 (iii)  •  Q8 (i)  •  Q8 (ii)  •  Q8 (iii)  •  Q9  •  

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Question 01 (b) ii
Short-Answer Questions. A correct answer in the box gives full marks. For partial marks work needs to be shown. Find the equation of the tangent line $y=mx+b\,$ to the graph of $f(x)=\cos(\pi x)\,$ at $x=1/2\,$.

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
You are given a point. To find the equation of a line you need either another point or the slope of the line. We can find the slope of the tangent line by using derivatives.

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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. The tangent line must have the same slope as the function $f(x)$ at $x=1/2$. Solving for the slope $m$, we get $m=f'(1/2)=-\pi \sin \left({\frac {\pi }{2}}\right)=-\pi .$ The tangent line must also touch the function at $x=1/2$. We use this to solve for the y-intercept: ${\begin{aligned}y&=mx+b\\\cos \left({\frac {\pi }{2}}\right)&=-\pi \left({\frac {1}{2}}\right)+b\\0&=-{\frac {\pi }{2}}+b\\b&={\frac {\pi }{2}}\end{aligned}}$ Therefore, the tangent line to $f(x)=\cos(\pi x)$ at $x=1/2$ is: ${\color {blue}y=-\pi x+{\frac {\pi }{2}}}$

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