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

Question

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

Evelyn Agard · Accepted Answer

#5\ ext{units# Given that #x=3t+1\implies dx/dt=3# #y=2-4t\implies dy/dt=-4# # herefore \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\ ext{units#

Adam Adkins · Answer

undefined To find the length of the curve defined by $ x = 3t + 1 $ and $ y = 2 - 4t $ for $ 0 \leq t \leq 1 $, you can use the formula for the arc length of a parametric curve:

$$ s = \int_{a}^{b} \sqrt{\left(\frac{{dx}}{{dt}}\right)^2 + \left(\frac{{dy}}{{dt}}\right)^2} \, dt $$

where $ a $ and $ b $ are the limits of the parameter $ t $. In this case, $ a = 0 $ and $ b = 1 $.

Plugging in the given parametric equations:

$$ \frac{{dx}}{{dt}} = 3 $$
$$ \frac{{dy}}{{dt}} = -4 $$

Substitute these derivatives into the formula and integrate from 0 to 1:

$$ s = \int_{0}^{1} \sqrt{(3)^2 + (-4)^2} \, dt $$

$$ s = \int_{0}^{1} \sqrt{9 + 16} \, dt $$

$$ s = \int_{0}^{1} \sqrt{25} \, dt $$

$$ s = \int_{0}^{1} 5 \, dt $$

$$ s = 5t \Bigg|_{0}^{1} $$

$$ s = 5(1) - 5(0) $$

$$ s = 5 $$

Therefore, the length of the curve is $ 5 $ units.

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

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