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#