How do you find the length of the curve y^2 = 16(x+1)^3 where x is between [0,3] and y>0?

Feb 24, 2015

The answer is: $L = \frac{1}{54} \left(\sqrt{145} - 37 \sqrt{37}\right)$.

The curve can be written:

$y = \pm 4 \sqrt{{\left(x + 1\right)}^{3}}$, but we need only the branch with $y > 0$, so:

$y = 4 \sqrt{{\left(x + 1\right)}^{3}}$.

To find the lenght of a curve written as a function and in cartesian coordinates we have to remember this formula:

$L = {\int}_{a}^{b} \sqrt{1 + {\left[f ' \left(x\right)\right]}^{2}} \mathrm{dx}$.

So, we have first of all calculate the derivative of:

$y = 4 {\left(x + 1\right)}^{\frac{3}{2}} \Rightarrow y ' = 4 \cdot \frac{3}{2} {\left(x + 1\right)}^{\frac{1}{2}} = 6 \sqrt{x + 1}$

$L = {\int}_{0}^{3} \sqrt{1 + 36 \left(x + 1\right)} \mathrm{dx} = {\int}_{0}^{3} \sqrt{36 x + 37} \mathrm{dx} =$

$= \frac{1}{36} {\int}_{0}^{3} 36 \sqrt{36 x + 37} \mathrm{dx} = \frac{1}{36} {\left[{\left(36 x + 37\right)}^{\frac{1}{2} + 1} / \left(\frac{1}{2} + 1\right)\right]}_{0}^{3} =$

$= \frac{1}{36} \cdot \frac{2}{3} {\left[\sqrt{{\left(36 x + 37\right)}^{3}}\right]}_{0}^{3} = \frac{1}{54} \left(\sqrt{145} - \sqrt{{37}^{3}}\right) = \frac{1}{54} \left(\sqrt{145} - 37 \sqrt{37}\right)$.