Science:Math Exam Resources/Courses/MATH103/April 2012/Question 02 (b)

MATH103 April 2012

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Question 02 (b)

Short answer question. Show in detail how you arrive at your answer (work will be considered for this problem). Use the formula arc length to find an integral of the form $\int _{\alpha }^{\beta }h(x)\,dx,$ with suitable $\alpha$, $\beta$, and h(x), that represents the circumference of the ellipse ${\frac {y^{2}}{a^{2}}}+{\frac {x^{2}}{b^{2}}}=1,$ where a and b are positive constants. Do not evaluate this integral!

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
The arc length formula states that if y(x) is a curve, then its length between x = a and x = b is given by the integral $\int _{a}^{b}{\sqrt {1+(y'(x))^{2}}}\,dx$

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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. We solve for y to obtain $y=\pm a{\sqrt {1-{\frac {x^{2}}{b^{2}}}}}.$ It suffices to take the values y ≥ 0 and then multiply the resulting arc-length by 2. Then the intersection points with y = 0 are x = -b and x = b. Taking the derivative of $y=a{\sqrt {1-{\frac {x^{2}}{b^{2}}}}}$ yields $y'(x)={\frac {1}{2}}a\left(1-{\frac {x^{2}}{b^{2}}}\right)^{-1/2}\left(-{\frac {2x}{b^{2}}}\right)=-{\frac {ax}{b^{2}{\sqrt {1-{\frac {x^{2}}{b^{2}}}}}}}$ Plugging into the formula for the arc length we obtain our final answer (don't forget the 2) $L=2\int _{-b}^{b}{\sqrt {1+(y'(x))^{2}}}\,dx=\int _{-b}^{b}2{\sqrt {1+{\frac {a^{2}x^{2}}{b^{4}(1-{\frac {x^{2}}{b^{2}}})}}}}\,dx.$

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