Answer :
Length of the curve is e² - 1 .
What is function?
In arithmetic, a function from a group X to a group Y assigns to every part of X precisely one part of Y. The set X is termed the domain of the function and therefore the set Y is termed the codomain of the function
Main body:
A parametric curve is a function expressed in components form, such that
x=f(t) , y=g(t). The length of a parametric curve on the interval a ≤ t ≤ b is given by the definite integral :
L = [tex]\int\limits^a_b {\sqrt{(\frac{dx}{dt} )^2 + (\frac{dy}{dt} )^2} } \, dt[/tex]
Let's begin by getting the first derivative of the components of the function with respect to the variable t.
x = e^t - t
dx / dt = e^t−1
y = 4 e^t/2
dy /dt = 2 e^t/2
Substitute the derivatives into the following definite integral which computes the length of the parametric curve on the interval
[a,b]=[0,2].
L =[tex]\int\limits^a_b {\sqrt{(\frac{dx}{dt} )^2 + (\frac{dy}{dt} )^2} } \, dt[/tex]
L= [tex]\int\limits^2_0 {\sqrt{(e^{t}-1 )^2 + (2e^{t/2} } )^2} } \, dt[/tex]
L =[tex]\int\limits^2_0 {e^{t} -1} \, dt[/tex]
Evaluate the solution at the limits of integration to get the length.
L = (e² - 1 )- (e⁰ - 1)
L = e² - 1 - 1 + 1
L= e² - 1
Hence , length of the curve is e² - 1 .
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