Science:Math Exam Resources/Courses/MATH104/December 2016/Question 04 (a)

MATH104 December 2016

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Question 04 (a)
Use linear approximation to estimate ${\sqrt {69}}$ .

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
A simple way of approximating ${\sqrt {69}}$ is to construct the linear approximation of (i.e., tangent line to) $f(x)={\sqrt {x}}$ at some point $(a,f(a))$ for which $f(a)$ is known exactly. Therefore, the point needs to satisfy two things: have a pretty straightforward square root and also close to ${\sqrt {69}}$ . Thus we choose $a=64.$

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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. Note that The linear approximation of $f(x)$ at a point $x=a$ is given by $T(x)=f(a)+f'(a)\cdot (x-a)$ To estimate ${\sqrt {69}}$ , we first find the linear approximation of $f(x)={\sqrt {x}}$ at $a=64$ . (The reason we choose $64$ is explained in the Hint) For this purpose, we need $f'(x)={\frac {1}{2{\sqrt {x}}}}$ , thus $f'(64)={\frac {1}{2\cdot 8}}={\frac {1}{16}}$ . Therefore the linear approximation line is: ${\begin{aligned}T(x)=&f(a)+f'(a)\cdot (x-a)\\=&f(64)+f'(64)\cdot (x-64)\\=&8+{\frac {1}{16}}\cdot (x-64)\end{aligned}}$ Since $f(x)\sim T(x)$ around $x$ near $64$ , we have ${\sqrt {69}}=f(69)\sim T(69)=8+{\frac {1}{16}}\cdot (69-64)=8+{\frac {5}{16}}={\frac {133}{16}}$ . Answer: $\color {blue}{\frac {133}{16}}$