# How do you find the length of the curve y=(2x+1)^(3/2), 0<=x<=2?

Apr 29, 2018

Length is 10.38 units. See details below

#### Explanation:

The length of a curve between a and b values for x is given by

L=int_a^bsqrt(1+y´^2)dx

y´^2=18x+9

L=int_0^2sqrt(18x+10)dx=1/27(18x+10)^(3/2)]_0^2=10.38

Apr 29, 2018

$\frac{1}{27} \left\{46 \sqrt{46} - 10 \sqrt{10}\right\} \approx 10.384 \setminus \left(3 \mathrm{dp}\right)$

#### Explanation:

The Arc Length of a curve $y = f \left(x\right)$ from $x = a$ to $x = b$ is given by:

$L = {\int}_{a}^{b} \setminus \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \setminus \mathrm{dx}$

Sop, for the given curve $y = {\left(2 x + 1\right)}^{\frac{3}{2}}$ for x in [0,2, we form the derivative using the power rule for differentiation in conjunction with the chain rule:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{3}{2}\right) {\left(2 x + 1\right)}^{\frac{3}{2} - 1} \left(2\right)$

$\setminus \setminus \setminus \setminus \setminus \setminus = 3 {\left(2 x + 1\right)}^{\frac{1}{2}}$

So then, the arc length is:

$L = {\int}_{0}^{2} \setminus \sqrt{1 + {\left(3 {\left(2 x + 1\right)}^{\frac{1}{2}}\right)}^{2}} \setminus \mathrm{dx}$

$\setminus \setminus = {\int}_{0}^{2} \setminus \sqrt{1 + 9 \left(2 x + 1\right)} \setminus \mathrm{dx}$

$\setminus \setminus = {\int}_{0}^{2} \setminus \sqrt{1 + 18 x + 9} \setminus \mathrm{dx}$

$\setminus \setminus = {\int}_{0}^{2} \setminus \sqrt{18 x + 10} \setminus \mathrm{dx}$

And using the power rule for integration, we can integrate to get:

$L = {\left[\frac{{\left(18 x + 10\right)}^{\frac{3}{2}}}{\frac{3}{2}} \cdot \frac{1}{18}\right]}_{0}^{2}$

$\setminus \setminus = {\left[\frac{1}{27} {\left(18 x + 10\right)}^{\frac{3}{2}}\right]}_{0}^{2}$

$\setminus \setminus = \frac{1}{27} \left\{{\left(36 + 10\right)}^{\frac{3}{2}} - {10}^{\frac{3}{2}}\right\}$

$\setminus \setminus = \frac{1}{27} \left\{46 \sqrt{46} - 10 \sqrt{10}\right\}$

$\setminus \setminus \approx 10.384 \setminus \left(3 \mathrm{dp}\right)$