Science:Math Exam Resources/Courses/MATH101/April 2005/Question 06 (b)

MATH101 April 2005

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Question 06 (b)

A tree trunk is 60 feet long and has circular cross sections. The diameters in feet measured at 10-foot intervals starting at the bottom of the tree are 10, 8, 7, 6, 5, 4, and 3. The volume of the tree, in cubic feet, is computed using the formula $V=\int _{0}^{60}A(y)\,dy,$ where $A(y)$ is the cross-sectional area of the trunk, in square feet, $y$ feet above the bottom of the tree. (b) It is known that the fourth derivative of $A(y)$ has values that lie between 3 and 30,000. How many measurements of the tree's diameter (equally spaced along the tree trunk) would be needed in order to estimate its volume using Simpson's Rule to an accuracy of 1 cubic foot? You may use the fact that the error made in approximating a definite integral $\int _{a}^{b}f(x)\,dx$ by the corresponding Simpson's Rule approximation $S_{n}$ is at most $(M(b-a)^{5})/(180n^{4})$, where $\left|f^{(4)}(x)\right|\leq M$ on the interval $[a,b]$. You must give your answer as explicit integer for full marks.

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
Justify to yourself that it suffices to find an n so that ${\frac {M(b-a)^{5}}{180n^{4}}}={\frac {30000(60-0)^{5}}{180n^{4}}}\leq 1$ Don't forget that n has to be even! (By the constraints of Simpson's rule)

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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. If the error associated with Simpson's rule is bounded by 1, then we would get that the approximation is precise to 1 cubic foot. Hence we compute when ${\frac {M(b-a)^{5}}{180n^{4}}}={\frac {30000(60-0)^{5}}{180n^{4}}}\leq 1$ Solving and using [prime] factorization techniques, we have ${\begin{aligned}{\frac {30000(60-0)^{5}}{180n^{4}}}&\leq 1\\{\frac {3\cdot 10^{4}\cdot 3^{5}\cdot 2^{5}\cdot 10^{5}}{2\cdot 3^{2}\cdot 10}}&\leq n^{4}\\{\frac {3^{6}\cdot 2^{5}\cdot 10^{9}}{2\cdot 3^{2}\cdot 10}}&\leq n^{4}\\3^{4}\cdot 2^{4}\cdot 10^{8}&\leq n^{4}\\3\cdot 2\cdot 10^{2}&\leq n\\600&\leq n\\\end{aligned}}$ Thus n should be at least 600.

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