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I'm trying to solve the arc length for the following. $c(t) = (2, 6t^2, 4t^3)$ from $0\le t\le 1.$

I've checked on WolframAlpha and I get the answer $8\sqrt2 - 4$ but when I work it out I get $16\sqrt2$ where am I going wrong?

\begin{align*} L&= \int_0^1 \sqrt{(12t)^2 + (12t^2)^2}\, dt\\ &= \int_0^1 \sqrt{144t^2 + 144t^4}\, dt\\ &= \int_0^1 12t\sqrt{1 + t^2}\,dt \end{align*}

Then I do $$ 8t\sqrt{(1+t^2)^3} = 16\sqrt2.$$

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  • $\begingroup$ How did you get to $16\sqrt{2}$? $\endgroup$
    – user12345
    May 10, 2017 at 0:22

3 Answers 3

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From $$\int_{0}^{1}12t\sqrt{1+t^2}dt$$ Use $u=1+t^2\implies \frac{1}{2}du=tdt$ to get $$\int_{0}^{1}12t\sqrt{1+t^2}dt=\int12\sqrt{u}\cdot\frac{1}{2}du=6u^{3/2}\cdot\frac{2}{3}=4(1+t^2)^{3/2}\bigg\vert_{0}^{1}=4\cdot2^{3/2}-4\cdot1$$ $$=4\cdot2\cdot2^{1/2}-4\cdot1=8\sqrt{2}-4$$

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From $$\int_0^1 12t\sqrt{1 + t^2}\,dt,$$ let \begin{align*} u &= 1+t^{2}\\ du &= 2t\,dt. \end{align*}

This yields $$\int_{1}^{2}6 u^{1/2}\,du =4u^{3/2}\bigg|_{1}^{2}=4\sqrt{8}-4 = 8\sqrt{2}-4.$$

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I'll just add a solution in the complex plane for contrast. Let

$$z=6t^2+i4t^3\\ \dot z=12t+i12t^2\\ |\dot z|=12t\sqrt{1+t^2}\\ s=\int_0^1|\dot z|dt=6\int_0^12t\sqrt{1+t^2}dt=6\int_0^1\sqrt{1+u}\ du,\quad u=t^2, du=2tdt\\ s=4(1+u)^{3/2}\big|_0^1=8\sqrt{2}-4 $$

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