Answer
$$L = \frac 1 2 ({\sqrt 2} + \ln ({\sqrt 2 + 1})) $$
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 $\frac{dy}{dt}$
$\frac{dx}{dt} = \frac{d(tsin(t))}{dt} = sin(t) + tcos(t)$
$\frac{dy}{dt} = \frac{d(tcos(t))}{dt} = cos(t) - tsin(t)$
3. Substitute the values into the definite integral:
$L = \int_{\alpha}^{\beta} \sqrt{(sin(t) + tcos(t))^2 + (cos(t) - tsin(t))^2} \space dt$
$L = \int_{\alpha}^{\beta} \sqrt{(sin^2(t) +2tsin(t)cos(t) + t^2cos^2(t)) + (cos^2(t) -2tsin(t)cos(t) + t^2sin^2(t))} \space dt$
$L = \int_{\alpha}^{\beta} \sqrt{1 + t^2} \space dt$
Since $t$ goes from 0 to 1:
$L = \int_{0}^{1} \sqrt{1 + t^2} \space dt$
$L = (\frac 1 2 t \sqrt {1 + t^2} + \frac 1 2\ln ({\sqrt {1 + t^2} + t}))]^1_0$
$L = (\frac 1 2 (1) \sqrt {1 + 1^2} + \frac 1 2\ln ({\sqrt {1 + 1^2} + 1})) - (\frac 1 2 (0) \sqrt {1 + 0^2} + \frac 1 2\ln ({\sqrt {1 + 0^2} + 0}))$
$L = \frac {\sqrt 2} 2 + \frac 1 2 \ln ({\sqrt 2 + 1}) + \ln (1)$
$L = \frac {\sqrt 2} 2 + \frac 1 2 \ln ({\sqrt 2 + 1}) $
$L = \frac 1 2 ({\sqrt 2} + \ln ({\sqrt 2 + 1})) $
