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I need to find the exact length of the curve in the title. I'm mostly confused about how to set up y. Would y equal the square root of the other side? I've tried that and my answer comes out wrong.

Here's what I'm trying: $$y=\sqrt{16(x+3)^3}$$ And then I take the derivative of that and put it into the arc length formula and solve, but I don't get the value that my book gets.

Thanks for any help!

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  • $\begingroup$ What value do you get and what value does your book get? Sometimes the book answer is incorrect. $\endgroup$ Oct 5 '17 at 4:50
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    $\begingroup$ If I am right the arc length should be $$\int_{0}^{3}\sqrt{109+36x}\,dx = \frac{217\sqrt{217}-109\sqrt{109}}{54}.$$ $\endgroup$ Oct 5 '17 at 4:53
  • $\begingroup$ @JackD'Aurizio It was the book that was wrong! Somehow, somewhere, they got 105 instead of 109... I was correct in the first place. Must have been a typo. It was driving me crazy trying to find where I was wrong. Thank you. $\endgroup$
    – JustHeavy
    Oct 5 '17 at 4:59
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Personally, I would approach the problem by parametrizing the curve as follows: $$ x = t^2 - 3 \qquad \qquad y = 4t^3 \qquad (\sqrt{3} \leq t \leq \sqrt{6})$$

In which case we see that: $$ \frac{dx}{dt} = 2t \qquad \qquad \frac{dy}{dt} = 12t^2 $$

and so the arc length is expressed as: \begin{align*} L &= \int_C \sqrt{\left(\frac{dx}{dt} \right)^2 + \left(\frac{dy}{dt} \right)^2} \, dt \\ &= \int_\sqrt{3}^\sqrt{6} \sqrt{(2t)^2 + (12t^2)^2} \, dt \\ &= \int_\sqrt{3}^\sqrt{6} \sqrt{144t^4 + 4t^2} \, dt \\ &= \int_\sqrt{3}^\sqrt{6} 2t \sqrt{36t^2 + 1} \, dt \\ &= \int_3^6 \sqrt{36u + 1} \, du \\ &= \left[ \frac{1}{54}(36u+1)^\frac{3}{2} \right]_3^6 \\ &= \frac{217\sqrt{217} - 109\sqrt{109}}{54} \approx 38.1225 \end{align*}


Though I come to see that the problem also works out nicely in Cartesian coordinates rather than parametric coordinates: \begin{align*} y &= 4(x+3)^{3/2} \\ \frac{dy}{dx} &= 6\sqrt{x+3} \end{align*} So the arc length is expressed as: \begin{align*} L &= \int_C \sqrt{1 + \left(\frac{dy}{dx} \right)^2} \, dx \\ &= \int_0^3 \sqrt{1 + 36(3+x)} \, dx \\ &= \int_0^3 \sqrt{36x + 109} \, dx \\ &= \left[\frac{1}{54}(36x+109)^\frac{3}{2} \right]_0^3 \\ &= \frac{217\sqrt{217} - 109\sqrt{109}}{54} \approx 38.1225 \end{align*}

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You want: $$ \begin{align} \int_0^3\sqrt{1+\left(y'\right)^2}\,dx \end{align}$$ And with implicit differentiation you have that: $$2yy'=48(x+3)^2$$ $$y'=\frac{24(x+3)^2}{y}$$ So you want: $$\begin{align} \int_0^3\sqrt{1+\left(\frac{24(x+3)^2}{y}\right)^2}\,dx &=\int_0^3\sqrt{1+\frac{24^2(x+3)^4}{y^2}}\,dx\\ &=\int_0^3\sqrt{1+\frac{24^2(x+3)^4}{16(x+3)^3}}\,dx\\ &=\int_0^3\sqrt{1+36(x+3)}\,dx\\ &=\int_0^3\sqrt{36x+109}\,dx\\ &=\frac{1}{36}\int_{109}^{217}\sqrt{u}\,du\\ \end{align}$$ which should be easy to finish.

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