Respuesta :
The length of a curve C parameterized by a vector function r(t) = x(t) i + y(t) j over an interval a ≤ t ≤ b is
[tex]\displaystyle\int_C\mathrm ds = \int_a^b \sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2} \,\mathrm dt[/tex]
In this case, we have
x(t) = exp(t ) + exp(-t ) ==> dx/dt = exp(t ) - exp(-t )
y(t) = 5 - 2t ==> dy/dt = -2
and [a, b] = [0, 2]. The length of the curve is then
[tex]\displaystyle\int_0^2 \sqrt{\left(e^t-e^{-t}\right)^2+(-2)^2} \,\mathrm dt = \int_0^2 \sqrt{e^{2t}-2+e^{-2t}+4}\,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^2 \sqrt{e^{2t}+2+e^{-2t}} \,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^2\sqrt{\left(e^t+e^{-t}\right)^2} \,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^2\left(e^t+e^{-t}\right)\,\mathrm dt[/tex]
[tex]=\left(e^t-e^{-t}\right)\bigg|_0^2 = \left(e^2-e^{-2}\right)-\left(e^0-e^{-0}\right) = \boxed{e^2-\frac1{e^2}}[/tex]
The exact length of the curve when the parametric equations are x = f(t) and y = g(t) is given below.
[tex]e^2 -\dfrac{1}{e^2 }[/tex]
What is integration?
It is the reverse of differentiation.
The parametric equations are given below.
[tex]\rm x=e^t+e^{-t}, \ \ 0\leq t\leq 2\\\\y=5-2t, \ \ \ \ \ 0\leq t\leq 2[/tex]
Then the arc length of the curve will be given as
[tex]\int _0^2 \sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dx})^2}[/tex]
Then we have
[tex]\rm \dfrac{dx}{dt} = e^t-e^{-t}\\\\ \dfrac{dy}{dt} = -2[/tex]
Then
[tex]\rightarrow \int _0^2 \sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dx})^2}\ \ dt\\\\\rightarrow \int _0^2 \sqrt{(e^t-e^{-t})^2 + (-2)^2} \ dt\\\\\rightarrow \int _0^2 \sqrt{(e^t+e^{-t})^2} \ dt\\\\\rightarrow \int _0^2 (e^t+e^{-t}) \ dt\\\\\rightarrow (e^2-e^{-2}) \\\\\rightarrow e^2 - \dfrac{1}{e^2}[/tex]
More about the integration link is given below.
https://brainly.com/question/18651211
