Science:Math Exam Resources/Courses/MATH101/April 2006/Question 02 (c)

MATH101 April 2006

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Question 02 (c)
Full-Solution Problems. Justify your answers and show all your work. Determine the length of the curve $y=(2/3)x^{3/2}$ on $0\leq x\leq 3$. Evaluate and completely simplify your answer.

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 formula for the arc length of this curve on the interval from $0\leq x\leq 3$ is $\int _{0}^{3}{\sqrt {1+\left({\frac {dy}{dx}}\right)^{2}}}\,dx.$

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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 first compute the derivative of $y=(2/3)x^{3/2}$ and see that $\displaystyle {\frac {dy}{dx}}={\sqrt {x}}$ Plugging this into the arc length formula gives $\int _{0}^{3}{\sqrt {1+\left({\frac {dy}{dx}}\right)^{2}}}\,dx=\int _{0}^{3}{\sqrt {1+{\sqrt {x}}^{2}}}\,dx=\int _{0}^{3}{\sqrt {1+x}}\,dx$ Now, let $u=1+x$ so that $du=dx$ and $u(0)=1$ and $u(3)=4$. Then ${\begin{aligned}\int _{0}^{3}{\sqrt {1+x}}\,dx&=\int _{1}^{4}{\sqrt {u}}\,du\\&=\left.{\frac {2u^{3/2}}{3}}\right|_{1}^{4}\\&={\frac {2}{3}}\left(4^{3/2}-1\right)\\&={\frac {2}{3}}\left(8-1\right)\\&={\frac {14}{3}}\end{aligned}}$ completing the question.

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