Science:Math Exam Resources/Courses/MATH110/April 2012/Question 02 (a)

MATH110 April 2012

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Question 02 (a)
Use a suitable linear approximation to estimate $\ln(1.1)$.

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 linear approximation can also be thought of as the equation of a tangent line. For which x-value should you find your linear approximation/tangent line?

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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. In order to estimate $\ln(1.1)$ we need to find a linear approximation of $\ln(x)$ at some point a. Because I know that $\ln(1)=0$ and 1 is close to 1.1, x = 1 would be a good choice for a. The linear approximation of $f(x)$ at a is the same as finding the equation of the tangent line to $f(x)$ at a. Our function is $f(x)=\ln x$ and we chose a = 1. We find the slope of the tangent line/linear approximation by taking the derivative $f'(x)=1/x$ and then calculating $f'(1)=1$. The tangent line/linear approximation runs though the point $(1,f(1))$ or (1,0). Using our slope of 1 and the point (1,0), we find the equation of the tangent line/linear approximation to be L(x) = x - 1. To complete the approximation of $\ln(1.1)$, we plug 1.1 into L(x) to get: $L(1.1)=1.1-1=0.1$ So $\ln(1.1)\approx 0.1$.

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