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

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!

• What value do you get and what value does your book get? Sometimes the book answer is incorrect. Oct 5 '17 at 4:50
• If I am right the arc length should be $$\int_{0}^{3}\sqrt{109+36x}\,dx = \frac{217\sqrt{217}-109\sqrt{109}}{54}.$$ Oct 5 '17 at 4:53
• @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. Oct 5 '17 at 4:59

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: $$\frac{dx}{dt} = 2t \qquad \qquad \frac{dy}{dt} = 12t^2$$
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.