Science:Math Exam Resources/Courses/MATH100/December 2010/Question 01 (i)

MATH100 December 2010

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Question 01 (i)

Short-Answer Questions. Each question is worth 3 marks, but not all questions are of equal difficulty. Full marks will be given for correct answers placed in the box, but at most 1 mark will be given for incorrect answers. Unless otherwise stated, it is not necessary to simplify your answers in this question. Use a suitable linear approximation to estimate (30)1/3. Give your answer as a fraction with integer numerator and denominator.

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
The question is asking for an approximation of the cube root of 30. For what number near 30 is it easy to calculate an exact cube root?

Hint 2
The formula for a linear approximation of a function $f(x)$ at a point $x=a$ is given by: $\displaystyle L(x)=f(a)+f'(a)(x-a)$

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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 ƒ (x)=x1/3. First we need a number a  as close to 30 as possible but whose cube root we know. A good candidate would be 27, because 33=27 and so $27^{1/3}=3$. We know that the general formula for a linear approximation is given by $\displaystyle L(x)=f(a)+f'(a)(x-a)$ We can find the derivative: $\displaystyle f'(x)={\frac {x^{-2/3}}{3}}$ Knowing these values, we can plug them in to the linear approximation equation and find the approximation for x=30, using a=27: $\displaystyle {\begin{aligned}L(x)&=(27)^{1/3}+{\frac {27^{-2/3}}{3}}(x-27)\\L(30)&=(27)^{1/3}+{\frac {27^{-2/3}}{3}}(30-27)\\&=3+{\frac {1/9}{3}}(3)\\&=3+{\frac {1}{9}}\\&={\frac {28}{9}}\end{aligned}}$ So using linear approximation, $(30)^{1/3}\approx 28/9$

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