The length of a curve [tex]C[/tex] given parametrically by [tex](x(t),y(t))[/tex] over some domain [tex]t\in[a,b][/tex] 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,
[tex]x(t)=\cos(\cos4t)\implies\dfrac{\mathrm dx}{\mathrm dt}=-\sin(\cos4t)(-\sin4t)(4)=4\sin4t\sin(\cos4t)[/tex]
[tex]y(t)=\sin(\cos4t)\implies\dfrac{\mathrm dy}{\mathrm dt}=\cos(\cos4t)(-\sin4t)(4)=-4\sin4t\cos(\cos4t)[/tex]
So we have
[tex]\displaystyle\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2=16\sin^24t\sin^2(\cos4t)+16\sin^24t\cos^2(\cos4t)=16\sin^24t[/tex]
and the arc length is
[tex]\displaystyle\int_0^1\sqrt{16\sin^24t}\,\mathrm dt=4\int_0^1|\sin4t|\,\mathrm dt[/tex]
We have
[tex]\sin(4t)=0\implies4t=n\pi\implies t=\dfrac{n\pi}4[/tex]
where [tex]n[/tex] is any integer; this tells us [tex]\sin(4t)\ge0[/tex] on the interval [tex]\left[0,\frac\pi4\right][/tex] and [tex]\sin(4t)<0[/tex] on [tex]\left[\frac\pi4,1\right][/tex]. So the arc length is
[tex]=\displaystyle4\left(\int_0^{\pi/4}\sin4t\,\mathrm dt-\int_{\pi/4}^1\sin4t\,\mathrm dt\right)[/tex]
[tex]=-\cos(4t)\bigg_0^{\pi/4}-\left(-\cos(4t)\bigg_{\pi/4}^1\right)[/tex]
[tex]=(\cos0-\cos\pi)+(\cos4-\cos\pi)=\boxed{3+\cos4}[/tex]
In this exercise we have to use the knowledge of the integral to calculate the length of the arc:
The length of a curve is given by [tex]3+ cos( 4)[/tex]
The length of a curve given parametrically by [tex](x(t), y(t))[/tex] over some domain is [tex]t \in [a, b][/tex]
[tex]\int\limits_C \, ds = \int\limits^a_b {\sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 } } \, dt[/tex]
Now performing the derivatives with respect to X and Y, we find that:
[tex]x(t) = cos(cos4t) \rightarrow \frac{dx}{dt} = -sin(cos4t)(-sin4t)(4)= 4sin4tsin(cos4t)\\y(t) = sin(cos4t) \rightarrow \frac{dy}{dt} = cos(cos4t)(-sin4t)(4)=-4sin4tcos(cos4t)[/tex]
So we have:
[tex](\frac{dx}{dt} )^2+(\frac{dy}{dt} )^2= 16sin^24tsin^2(cos4t)+16sin^24tcos^2(cos4t)= 16sin^24t[/tex]
And the arc length is:
[tex]\int\limits^1_0 {\sqrt{16sin^24t} } \, dt= 4\int\limits^1_0 {sin4t} \, dt\\sin(4t)=0 \rightarrow 4t=n\pi \rightarrow t= \frac{n\pi }{4}[/tex]
where n is any integer; this tells us [tex]sin(4t)\geq 0[/tex] on the interval [tex][0,\pi/4][/tex] and [tex]sin(4t)<0[/tex] on [tex][\pi/4, 1][/tex]. So the arc length is:
[tex]=4(\int\limits^{\pi/4}_0 {sin(4t)} \, dt - \int\limits^1_{\pi/4} {sin(4t)} \, dt)\\=(cos(0)-cos(\pi))+(cos(4)-cos(\pi))= 3+cos(4)[/tex]
See more about integral at brainly.com/question/18651211
