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Vaselesa [24]
3 years ago
13

Help pls!

Mathematics
1 answer:
marin [14]3 years ago
3 0

Answer:

b) one distinct solution

Step-by-step explanation:

the discriminant (b² - 4ac) equals zero which means two real <em>identical </em> roots or one distinct real root

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(1 point) Find the length traced out along the parametric curve x=cos(cos(4t))x=cos⁡(cos⁡(4t)), y=sin(cos(4t))y=sin⁡(cos⁡(4t)) a
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The length of a curve C given parametrically by (x(t),y(t)) over some domain t\in[a,b] is

\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

In this case,

x(t)=\cos(\cos4t)\implies\dfrac{\mathrm dx}{\mathrm dt}=-\sin(\cos4t)(-\sin4t)(4)=4\sin4t\sin(\cos4t)

y(t)=\sin(\cos4t)\implies\dfrac{\mathrm dy}{\mathrm dt}=\cos(\cos4t)(-\sin4t)(4)=-4\sin4t\cos(\cos4t)

So we have

\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

and the arc length is

\displaystyle\int_0^1\sqrt{16\sin^24t}\,\mathrm dt=4\int_0^1|\sin4t|\,\mathrm dt

We have

\sin(4t)=0\implies4t=n\pi\implies t=\dfrac{n\pi}4

where n is any integer; this tells us \sin(4t)\ge0 on the interval \left[0,\frac\pi4\right] and \sin(4t) on \left[\frac\pi4,1\right]. So the arc length is

=\displaystyle4\left(\int_0^{\pi/4}\sin4t\,\mathrm dt-\int_{\pi/4}^1\sin4t\,\mathrm dt\right)

=-\cos(4t)\bigg_0^{\pi/4}-\left(-\cos(4t)\bigg_{\pi/4}^1\right)

=(\cos0-\cos\pi)+(\cos4-\cos\pi)=\boxed{3+\cos4}

7 0
3 years ago
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Answer:

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