It depends, not exactly, because the techtonic plates aren't connected.
![\vec F(x,y,z)=(x^2+y)\,\vec\imath+(y^2+x)\,\vec\jmath+ze^z\,\vec k](https://tex.z-dn.net/?f=%5Cvec%20F%28x%2Cy%2Cz%29%3D%28x%5E2%2By%29%5C%2C%5Cvec%5Cimath%2B%28y%5E2%2Bx%29%5C%2C%5Cvec%5Cjmath%2Bze%5Ez%5C%2C%5Cvec%20k)
We want to find
such that
. This means
![\dfrac{\partial f}{\partial x}=x^2+y](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20x%7D%3Dx%5E2%2By)
![\dfrac{\partial f}{\partial y}=y^2+x](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20y%7D%3Dy%5E2%2Bx)
![\dfrac{\partial f}{\partial z}=ze^z](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20z%7D%3Dze%5Ez)
Integrating both sides of the latter equation with respect to
tells us
![f(x,y,z)=e^z(z-1)+g(x,y)](https://tex.z-dn.net/?f=f%28x%2Cy%2Cz%29%3De%5Ez%28z-1%29%2Bg%28x%2Cy%29)
and differentiating with respect to
gives
![x^2+y=\dfrac{\partial g}{\partial x}](https://tex.z-dn.net/?f=x%5E2%2By%3D%5Cdfrac%7B%5Cpartial%20g%7D%7B%5Cpartial%20x%7D)
Integrating both sides with respect to
gives
![g(x,y)=\dfrac{x^3}3+xy+h(y)](https://tex.z-dn.net/?f=g%28x%2Cy%29%3D%5Cdfrac%7Bx%5E3%7D3%2Bxy%2Bh%28y%29)
Then
![f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+h(y)](https://tex.z-dn.net/?f=f%28x%2Cy%2Cz%29%3De%5Ez%28z-1%29%2B%5Cdfrac%7Bx%5E3%7D3%2Bxy%2Bh%28y%29)
and differentiating both sides with respect to
gives
![y^2+x=x+\dfrac{\mathrm dh}{\mathrm dy}\implies\dfrac{\mathrm dh}{\mathrm dy}=y^2\implies h(y)=\dfrac{y^3}3+C](https://tex.z-dn.net/?f=y%5E2%2Bx%3Dx%2B%5Cdfrac%7B%5Cmathrm%20dh%7D%7B%5Cmathrm%20dy%7D%5Cimplies%5Cdfrac%7B%5Cmathrm%20dh%7D%7B%5Cmathrm%20dy%7D%3Dy%5E2%5Cimplies%20h%28y%29%3D%5Cdfrac%7By%5E3%7D3%2BC)
So the scalar potential function is
![\boxed{f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+\dfrac{y^3}3+C}](https://tex.z-dn.net/?f=%5Cboxed%7Bf%28x%2Cy%2Cz%29%3De%5Ez%28z-1%29%2B%5Cdfrac%7Bx%5E3%7D3%2Bxy%2B%5Cdfrac%7By%5E3%7D3%2BC%7D)
By the fundamental theorem of calculus, the work done by
along any path depends only on the endpoints of that path. In particular, the work done over the line segment (call it
) in part (a) is
![\displaystyle\int_L\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(4,0,0)=\boxed{1+3e^4}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_L%5Cvec%20F%5Ccdot%5Cmathrm%20d%5Cvec%20r%3Df%284%2C0%2C4%29-f%284%2C0%2C0%29%3D%5Cboxed%7B1%2B3e%5E4%7D)
and
does the same amount of work over both of the other paths.
In part (b), I don't know what is meant by "df/dt for F"...
In part (c), you're asked to find the work over the 2 parts (call them
and
) of the given path. Using the fundamental theorem makes this trivial:
![\displaystyle\int_{L_1}\vec F\cdot\mathrm d\vec r=f(0,0,0)-f(4,0,0)=-\frac{64}3](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_%7BL_1%7D%5Cvec%20F%5Ccdot%5Cmathrm%20d%5Cvec%20r%3Df%280%2C0%2C0%29-f%284%2C0%2C0%29%3D-%5Cfrac%7B64%7D3)
![\displaystyle\int_{L_2}\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(0,0,0)=\frac{67}3+3e^4](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_%7BL_2%7D%5Cvec%20F%5Ccdot%5Cmathrm%20d%5Cvec%20r%3Df%284%2C0%2C4%29-f%280%2C0%2C0%29%3D%5Cfrac%7B67%7D3%2B3e%5E4)
For this case, we have the following expression:
![-3x + x + 5](https://tex.z-dn.net/?f=-3x%20%2B%20x%20%2B%205)
We simplify the expression:
If we add similar terms, taking into account that different signs are subtracted and the sign of the greater one is placed, we have that ![-3x + x = -2x](https://tex.z-dn.net/?f=-3x%20%2B%20x%20%3D%20-2x)
So, we have to:
![-3x + x + 5 = -2x + 5](https://tex.z-dn.net/?f=-3x%20%2B%20x%20%2B%205%20%3D%20-2x%20%2B%205)
Answer:
-3x + x + 5 = -2x + 5
Answer:
im guessing?
Step-by-step explanation:
each=4921
Answer:
x^2 - 2x + 6
Explanation:
(x^2 + 1) - (2x - 5)
x^2 + 1 - 2x + 5
x^2 - 2x + 6