# How do you find the length of the curve x=3t+1, y=2-4t, 0<=t<=1?

5\ \text{units

#### Explanation:

Given that

$x = 3 t + 1 \setminus \implies \frac{\mathrm{dx}}{\mathrm{dt}} = 3$

$y = 2 - 4 t \setminus \implies \frac{\mathrm{dy}}{\mathrm{dt}} = - 4$

$\setminus \therefore \setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \setminus \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

$= \setminus \frac{- 4}{3}$

$= - \frac{4}{3}$

hence the length of curve $f \left(x\right)$ is given as

$\setminus \int \mathrm{ds}$

$= \setminus \int \setminus \sqrt{{\left(\mathrm{dx}\right)}^{2} + {\left(\mathrm{dy}\right)}^{2}}$

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

$= \setminus {\int}_{0}^{1} \setminus \sqrt{1 + {\left(- \frac{4}{3}\right)}^{2}} \setminus \left(3 \mathrm{dt}\right)$

$= \setminus {\int}_{0}^{1} \frac{5}{3} \left(3 \mathrm{dt}\right)$

$= 5 \setminus {\int}_{0}^{1} \mathrm{dt}$

$= 5 {\left[t\right]}_{0}^{1}$

=5\ \text{units