Finding the exact length of the curve: $y^2=16(x+3)^3,\;\;0\le x\le3,\;\;y\gt0$

Question

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!

Answer 1

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*}

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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*}

Answer 2

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.