Answer:
It can be Degree 2 or Degree 3
Explanation:
This is because of the roots of the equation, where the y values are equal to zero. So the only possible roots can be under 3 so that is why it is only Degree 2 and Degree 3.
Answer:
Step-by-step explanation:
Line 1 to Line 2: Associative Property(?)
Line 2 to Line 3: Associative Property
The numbers in the parenthesis are switched, which is associative property.
Answer:
ur down 24 pts.
Step-by-step explanation:
3×8=24
...............
What are
![p](https://tex.z-dn.net/?f=p)
and
![q](https://tex.z-dn.net/?f=q)
supposed to be? Can't answer a/b without that information.
I'll come back to part (c) in a moment. If we can show
![\mathbf f](https://tex.z-dn.net/?f=%5Cmathbf%20f)
is conservative, this part will be a breeze.
For part (d), to show whether
![\mathbf f](https://tex.z-dn.net/?f=%5Cmathbf%20f)
is conservative, we have to show that there is a scalar function
![f](https://tex.z-dn.net/?f=f)
such that
![\nabla f=\mathbf f](https://tex.z-dn.net/?f=%5Cnabla%20f%3D%5Cmathbf%20f)
. It appears that you've written
![\mathbf f=-y\,\mathbf i+x\,\mathbf j](https://tex.z-dn.net/?f=%5Cmathbf%20f%3D-y%5C%2C%5Cmathbf%20i%2Bx%5C%2C%5Cmathbf%20j)
I'm not sure what to make of the
![x^2+y^2](https://tex.z-dn.net/?f=x%5E2%2By%5E2)
that follows.
![\nabla f=\mathbf f](https://tex.z-dn.net/?f=%5Cnabla%20f%3D%5Cmathbf%20f)
means that
![\dfrac{\partial f}{\partial x}=-y](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20x%7D%3D-y)
![\dfrac{\partial f}{\partial y}=x](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20y%7D%3Dx)
Integrating the first PDE with respect to
![x](https://tex.z-dn.net/?f=x)
gives
![f(x,y)=-xy+g(y)](https://tex.z-dn.net/?f=f%28x%2Cy%29%3D-xy%2Bg%28y%29)
Differentiating with respect to
![y](https://tex.z-dn.net/?f=y)
gives
![\dfrac{\partial f}{\partial y}=-x+\dfrac{\mathrm dg}{\mathrm y}=x](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20y%7D%3D-x%2B%5Cdfrac%7B%5Cmathrm%20dg%7D%7B%5Cmathrm%20y%7D%3Dx)
![\implies\dfrac{\mathrm dg}{\mathrm dy}=2x](https://tex.z-dn.net/?f=%5Cimplies%5Cdfrac%7B%5Cmathrm%20dg%7D%7B%5Cmathrm%20dy%7D%3D2x)
But we assumed that
![g(y)](https://tex.z-dn.net/?f=g%28y%29)
is a function of
![y](https://tex.z-dn.net/?f=y)
alone, which means there is no solution for
![g](https://tex.z-dn.net/?f=g)
, and therefore no solution for
![f](https://tex.z-dn.net/?f=f)
. Hence
![f](https://tex.z-dn.net/?f=f)
is not conservative.
Back to part (c).
![\mathbf f](https://tex.z-dn.net/?f=%5Cmathbf%20f)
is not conservative, so we have to compute the line integral the "long" way. Replacing
![x=\cos t](https://tex.z-dn.net/?f=x%3D%5Ccos%20t)
and
![y=\sin t](https://tex.z-dn.net/?f=y%3D%5Csin%20t)
, we have
![\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=\int_{t=0}^{t=2\pi}(-\sin t\,\mathbf i+\cos t\,\mathbf j)\cdot(-\sin t\,\mathbf i+\cos t\,\mathbf j)\,\mathrm dt](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_%7B%5Cmathcal%20C%7D%5Cmathbf%20f%5Ccdot%5Cmathrm%20d%5Cmathbf%20r%3D%5Cint_%7Bt%3D0%7D%5E%7Bt%3D2%5Cpi%7D%28-%5Csin%20t%5C%2C%5Cmathbf%20i%2B%5Ccos%20t%5C%2C%5Cmathbf%20j%29%5Ccdot%28-%5Csin%20t%5C%2C%5Cmathbf%20i%2B%5Ccos%20t%5C%2C%5Cmathbf%20j%29%5C%2C%5Cmathrm%20dt)