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

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Answer 1 · 2018-07-27T20:59:48

$5\ \text{units$

Given that

$x=3t+1\implies dx/dt=3$

$y=2-4t\implies dy/dt=-4$

$\therefore \frac{dy}{dx}=\frac{dy/dt}{dx/dt}$

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

$=-4/3$

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

$\int ds$

$=\int \sqrt{(dx)^2+(dy)^2}$

$=\int \sqrt{1+(dy/dx)^2}\ dx$

$=\int_0^1 \sqrt{1+(-4/3)^2}\ (3dt)$

$=\int_0^1 5/3 (3dt)$

$=5\int_0^1 dt$

$=5[t]_0^1$

$=5\ \text{units$

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