Science:Math Exam Resources/Courses/MATH102/December 2016/Question B 05 (b)

MATH102 December 2016

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Question B 05 (b)
Use linear approximation at a suitable close value to estimate $\arctan(0.9)$. Your answer may be left in terms of fractions.

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
Note that for any function $f(x)$, the linear approximation around $x=a$ is given by $L(x)=f(a)+f'(a)(x-a)$ , and for points close to $a,\ f(x)\approx L(x)$ . For this question, what is $a$? what is $x$?

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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. Let $f(x)=\arctan x$. Then, we'd like to find $f(0.9)=\arctan(0.9)$. To estimate $f(0.9)$, we use the linear approximation of $f$ around the point $a=1$. Here, we choose $a=1$ because it is the closest point to $0.9$ that we know its function value, $\arctan 1={\frac {\pi }{4}}$. First, we find its derivative at $a=1$. $$f(x)=\arctan x\Rightarrow f'(x)={\frac {1}{1+x^{2}}}\Rightarrow f'(a)=f'(1)={\frac {1}{1+1}}={\frac {1}{2}},$$ and then find its function value at $a=1$ $$f(a)=f(1)=\arctan 1={\frac {\pi }{4}}.$$ By the formula in the Hint, we have ${\begin{aligned}f(x)\approx L(x)=f(a)+f'(a)(x-a)&\Rightarrow f(x)\approx L(x)=f(1)+f'(1)(x-1)={\frac {\pi }{4}}+{\frac {1}{2}}(x-1)\\&\Rightarrow f(0.9)\approx L(0.9)={\frac {\pi }{4}}+{\frac {1}{2}}(0.9-1)=\color {blue}{\frac {\pi }{4}}-0.05\end{aligned}}$