I'm pretty sure its the 3rd one
Answer:
The third line
Step-by-step explanation:
let's see all choices by solving them
1) The first one
2) The second one
3) The third one
4) The fourth one
I've attached a plot of the intersection (highlighted in red) between the parabolic cylinder (orange) and the hyperbolic paraboloid (blue).
The arc length can be computed with a line integral, but first we'll need a parameterization for
![C](https://tex.z-dn.net/?f=C)
. This is easy enough to do. First fix any one variable. For convenience, choose
![x](https://tex.z-dn.net/?f=x)
.
Now,
![x^2=2y\implies y=\dfrac{x^2}2](https://tex.z-dn.net/?f=x%5E2%3D2y%5Cimplies%20y%3D%5Cdfrac%7Bx%5E2%7D2)
, and
![3z=xy\implies z=\dfrac{x^3}6](https://tex.z-dn.net/?f=3z%3Dxy%5Cimplies%20z%3D%5Cdfrac%7Bx%5E3%7D6)
. The intersection is thus parameterized by the vector-valued function
![\mathbf r(x)=\left\langle x,\dfrac{x^2}2,\dfrac{x^3}6\right\rangle](https://tex.z-dn.net/?f=%5Cmathbf%20r%28x%29%3D%5Cleft%5Clangle%20x%2C%5Cdfrac%7Bx%5E2%7D2%2C%5Cdfrac%7Bx%5E3%7D6%5Cright%5Crangle)
where
![0\le x\le 4](https://tex.z-dn.net/?f=0%5Cle%20x%5Cle%204)
. The arc length is computed with the integral
![\displaystyle\int_C\mathrm dS=\int_0^4\|\mathbf r'(x)\|\,\mathrm dx=\int_0^4\sqrt{x^2+\dfrac{x^4}4+\dfrac{x^6}{36}}\,\mathrm dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_C%5Cmathrm%20dS%3D%5Cint_0%5E4%5C%7C%5Cmathbf%20r%27%28x%29%5C%7C%5C%2C%5Cmathrm%20dx%3D%5Cint_0%5E4%5Csqrt%7Bx%5E2%2B%5Cdfrac%7Bx%5E4%7D4%2B%5Cdfrac%7Bx%5E6%7D%7B36%7D%7D%5C%2C%5Cmathrm%20dx)
Some rewriting:
![\sqrt{x^2+\dfrac{x^4}4+\dfrac{x^6}{36}}=\sqrt{\dfrac{x^2}{36}}\sqrt{x^4+9x^2+36}=\dfrac x6\sqrt{x^4+9x^2+36}](https://tex.z-dn.net/?f=%5Csqrt%7Bx%5E2%2B%5Cdfrac%7Bx%5E4%7D4%2B%5Cdfrac%7Bx%5E6%7D%7B36%7D%7D%3D%5Csqrt%7B%5Cdfrac%7Bx%5E2%7D%7B36%7D%7D%5Csqrt%7Bx%5E4%2B9x%5E2%2B36%7D%3D%5Cdfrac%20x6%5Csqrt%7Bx%5E4%2B9x%5E2%2B36%7D)
Complete the square to get
![x^4+9x^2+36=\left(x^2+\dfrac92\right)^2+\dfrac{63}4](https://tex.z-dn.net/?f=x%5E4%2B9x%5E2%2B36%3D%5Cleft%28x%5E2%2B%5Cdfrac92%5Cright%29%5E2%2B%5Cdfrac%7B63%7D4)
So in the integral, you can substitute
![y=x^2+\dfrac92](https://tex.z-dn.net/?f=y%3Dx%5E2%2B%5Cdfrac92)
to get
![\displaystyle\frac16\int_0^4x\sqrt{\left(x^2+\frac92\right)^2+\frac{63}4}\,\mathrm dx=\frac1{12}\int_{9/2}^{41/2}\sqrt{y^2+\frac{63}4}\,\mathrm dy](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cfrac16%5Cint_0%5E4x%5Csqrt%7B%5Cleft%28x%5E2%2B%5Cfrac92%5Cright%29%5E2%2B%5Cfrac%7B63%7D4%7D%5C%2C%5Cmathrm%20dx%3D%5Cfrac1%7B12%7D%5Cint_%7B9%2F2%7D%5E%7B41%2F2%7D%5Csqrt%7By%5E2%2B%5Cfrac%7B63%7D4%7D%5C%2C%5Cmathrm%20dy)
Next substitute
![y=\dfrac{\sqrt{63}}2\tan z](https://tex.z-dn.net/?f=y%3D%5Cdfrac%7B%5Csqrt%7B63%7D%7D2%5Ctan%20z)
, so that the integral becomes
![\displaystyle\frac1{12}\int_{9/2}^{41/2}\sqrt{y^2+\frac{63}4}\,\mathrm dy=\frac{21}{16}\int_{\arctan(3/\sqrt7)}^{\arctan(41/(3\sqrt7))}\sec^3z\,\mathrm dz](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cfrac1%7B12%7D%5Cint_%7B9%2F2%7D%5E%7B41%2F2%7D%5Csqrt%7By%5E2%2B%5Cfrac%7B63%7D4%7D%5C%2C%5Cmathrm%20dy%3D%5Cfrac%7B21%7D%7B16%7D%5Cint_%7B%5Carctan%283%2F%5Csqrt7%29%7D%5E%7B%5Carctan%2841%2F%283%5Csqrt7%29%29%7D%5Csec%5E3z%5C%2C%5Cmathrm%20dz)
This is a fairly standard integral (it even has its own Wiki page, if you're not familiar with the derivation):
![\displaystyle\int\sec^3z\,\mathrm dz=\frac12\sec z\tan z+\frac12\ln|\sec x+\tan x|+C](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint%5Csec%5E3z%5C%2C%5Cmathrm%20dz%3D%5Cfrac12%5Csec%20z%5Ctan%20z%2B%5Cfrac12%5Cln%7C%5Csec%20x%2B%5Ctan%20x%7C%2BC)
So the arc length is
Answer:
5
Step-by-step explanation:
you add up all the numbers, then divide but how many numbers there are:
all add up to 40, divide the 40 by 8 to get 5
<span>Can 49/90 be simplified?
No it cannot be simplified </span>