Evaluate 3/2 y + -3 + 5/3 z when y = 6 and z = 3.

Mathematics

1 answer:

Ket [755]4 years ago

7 0

Answer:

Step-by-step explanation:

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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

Mazyrski [523]

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