Home > Calculus > Determining the Length of a Parametric Curve (Parametric Form)

What is the arc length of #f(t)=(3te^t,t-e^t) # over #t in [2,4]#?

Answer 1

William Abdalla

# 612.530 # (3dp)

We have:

# f(t) = (3te^t, t-e^t ) # where #t in [2,4]#

The parametric arc-length is given by:

# L = int_(alpha)^(beta) \ sqrt((dx/dt)^2 + (dy/dt)^2 ) \ dt #

We can differentiate the parameters:

# x(t) = 3te^t => dx/dt = 3te^t + 3e^t #

# y(t) = t-e^t => dy/dt = 1-e^t #

Then the arc-length is given by:

# L = int_2^4 \ sqrt( (3te^t + 3e^t)^2 + (1-e^t)^2 ) \ dt #

This integral dos not have a trivial anti-derivative, and so is evacuated using numerical methods to give:

# L = 612.530 # (3dp)

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Answer 2

Emma Aldridge

The arc length of the curve defined by ( f(t) = (3te^t, t - e^t) ) over the interval ( t ) in ([2, 4]) is given by the formula:

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

Substituting the given function ( f(t) = (3te^t, t - e^t) ), we find ( \frac{dx}{dt} = 3e^t + 3te^t ) and ( \frac{dy}{dt} = 1 - e^t ).

Plugging these into the arc length formula and integrating over the interval ([2, 4]) will give us the arc length.

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