Answer
$L = 4 \sqrt 2 - 2 $
Work Step by Step
1. Write the formula for the length "$L$"of a curve:
$$L = \int_{\alpha}^{\beta} \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \space dt$$
2. Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$
$\frac{dx}{dt} = \frac{d(1 + 3t^2)}{dt} = 6t$
$\frac{dy}{dt} = \frac{d(4 + 2t^3)}{dt} = 6t^2$
3. Substitute the values into the definite integral:
$L = \int_{\alpha}^{\beta} \sqrt{(6t)^2 + (6t^2)^2} \space dt$
$L = \int_{\alpha}^{\beta} \sqrt{36t^2 + 36t^4} \space dt$
$L = \int_{\alpha}^{\beta} \sqrt{36t^2( 1 + t^2)} \space dt$
$L = \int_{\alpha}^{\beta} 6t \sqrt{( 1 + t^2)} \space dt$
$u = 1 + t^2$
$\frac{du}{dt} = 2t \longrightarrow \frac {du}{2 \space dt} =t$
$L = \int_{\alpha}^{\beta} 6(\frac{du}{2 \space dt}) \sqrt u \space dt$
$L = \int_{\alpha}^{\beta} 3 \sqrt u \space du$
$L = 3 \int_{\alpha}^{\beta} \sqrt u \space du$
- Since $t$ goes from 0 to 1:
$u_1 = 1 + 0^2 = 1$
$u_2 = 1 + 1^2 = 2$
$u$ goes from 1 to 2
$L = 3 \int_{1}^{2} \sqrt u \space du$
$L = 3 (\frac 2 3 u^{3/2})]^2_1$
$L = (2 u^{3/2})]^2_1$
$L = ( (2 \times 2^{3/2}) - (2 \times 1^{3/2})) $
$L = 4 \sqrt 2 - 2 $
