The simplest path from (0, 0, 0) to (1, 1, 1) is a straight line, denoted
, which we can parameterize by the vector-valued function,
![\mathbf r(t)=(1-t)(\mathbf i+\mathbf j+\mathbf k)](https://tex.z-dn.net/?f=%5Cmathbf%20r%28t%29%3D%281-t%29%28%5Cmathbf%20i%2B%5Cmathbf%20j%2B%5Cmathbf%20k%29)
for
, which has differential
![\mathrm d\mathbf r=-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt](https://tex.z-dn.net/?f=%5Cmathrm%20d%5Cmathbf%20r%3D-%28%5Cmathbf%20i%2B%5Cmathbf%20j%2B%5Cmathbf%20k%29%5C%2C%5Cmathrm%20dt)
Then with
, we have
![\displaystyle\int_{\mathcal C}\nabla f(x,y,z)\cdot\mathrm d\mathbf r=\int_{t=0}^{t=1}\nabla f(x(t),y(t),z(t))\cdot\mathrm d\mathbf r](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_%7B%5Cmathcal%20C%7D%5Cnabla%20f%28x%2Cy%2Cz%29%5Ccdot%5Cmathrm%20d%5Cmathbf%20r%3D%5Cint_%7Bt%3D0%7D%5E%7Bt%3D1%7D%5Cnabla%20f%28x%28t%29%2Cy%28t%29%2Cz%28t%29%29%5Ccdot%5Cmathrm%20d%5Cmathbf%20r)
![=\displaystyle\int_{t=0}^{t=1}\left(2(1-t)^3e^{(1-t)^2}\,\mathbf i+(1-t)e^{(1-t)^2}\,\mathbf j+(1-t)e^{(1-t)^2}\,\mathbf k\right)\cdot-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint_%7Bt%3D0%7D%5E%7Bt%3D1%7D%5Cleft%282%281-t%29%5E3e%5E%7B%281-t%29%5E2%7D%5C%2C%5Cmathbf%20i%2B%281-t%29e%5E%7B%281-t%29%5E2%7D%5C%2C%5Cmathbf%20j%2B%281-t%29e%5E%7B%281-t%29%5E2%7D%5C%2C%5Cmathbf%20k%5Cright%29%5Ccdot-%28%5Cmathbf%20i%2B%5Cmathbf%20j%2B%5Cmathbf%20k%29%5C%2C%5Cmathrm%20dt)
![\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)(t^2-2t+2)\,\mathrm dt](https://tex.z-dn.net/?f=%5Cdisplaystyle%3D-2%5Cint_%7Bt%3D0%7D%5E%7Bt%3D1%7De%5E%7B%281-t%29%5E2%7D%281-t%29%28t%5E2-2t%2B2%29%5C%2C%5Cmathrm%20dt)
Complete the square in the quadratic term of the integrand:
, then in the integral we substitute
:
![\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)((1-t)^2+1)\,\mathrm dt](https://tex.z-dn.net/?f=%5Cdisplaystyle%3D-2%5Cint_%7Bt%3D0%7D%5E%7Bt%3D1%7De%5E%7B%281-t%29%5E2%7D%281-t%29%28%281-t%29%5E2%2B1%29%5C%2C%5Cmathrm%20dt)
![\displaystyle=-2\int_{u=0}^{u=1}e^{u^2}u(u^2+1)\,\mathrm du](https://tex.z-dn.net/?f=%5Cdisplaystyle%3D-2%5Cint_%7Bu%3D0%7D%5E%7Bu%3D1%7De%5E%7Bu%5E2%7Du%28u%5E2%2B1%29%5C%2C%5Cmathrm%20du)
Make another substitution of
:
![\displaystyle=-\int_{v=0}^{v=1}e^v(v+1)\,\mathrm dv](https://tex.z-dn.net/?f=%5Cdisplaystyle%3D-%5Cint_%7Bv%3D0%7D%5E%7Bv%3D1%7De%5Ev%28v%2B1%29%5C%2C%5Cmathrm%20dv)
Integrate by parts, taking
![r=v+1\implies\mathrm dr=\mathrm dv](https://tex.z-dn.net/?f=r%3Dv%2B1%5Cimplies%5Cmathrm%20dr%3D%5Cmathrm%20dv)
![\mathrm ds=e^v\,\mathrm dv\implies s=e^v](https://tex.z-dn.net/?f=%5Cmathrm%20ds%3De%5Ev%5C%2C%5Cmathrm%20dv%5Cimplies%20s%3De%5Ev)
![\displaystyle=-e^v(v+1)\bigg|_{v=0}^{v=1}+\int_{v=0}^{v=1}e^v\,\mathrm dv](https://tex.z-dn.net/?f=%5Cdisplaystyle%3D-e%5Ev%28v%2B1%29%5Cbigg%7C_%7Bv%3D0%7D%5E%7Bv%3D1%7D%2B%5Cint_%7Bv%3D0%7D%5E%7Bv%3D1%7De%5Ev%5C%2C%5Cmathrm%20dv)
![\displaystyle=-(2e-1)+(e-1)=-e](https://tex.z-dn.net/?f=%5Cdisplaystyle%3D-%282e-1%29%2B%28e-1%29%3D-e)
So, we have by the fundamental theorem of calculus that
![\displaystyle\int_C\nabla f(x,y,z)\cdot\mathrm d\mathbf r=f(1,1,1)-f(0,0,0)](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_C%5Cnabla%20f%28x%2Cy%2Cz%29%5Ccdot%5Cmathrm%20d%5Cmathbf%20r%3Df%281%2C1%2C1%29-f%280%2C0%2C0%29)
![\implies-e=f(1,1,1)-2](https://tex.z-dn.net/?f=%5Cimplies-e%3Df%281%2C1%2C1%29-2)
![\implies f(1,1,1)=2-e](https://tex.z-dn.net/?f=%5Cimplies%20f%281%2C1%2C1%29%3D2-e)
In the problem -5+x=0, x=5
Answer: C. 45 degrees and 135 degrees
Answer:option C is the correct answer.
Step-by-step explanation:
When the ladder leans against the wall, it forms a right angle triangle with the wall. The length of the ladder becomes the hypotenuse of the right angle triangle.
Since the length of the ladder is 12 foot, then
Hypotenuse = 12 foot
The angle formed by the ladder with the ground is 75.5 degrees. Therefore, the height, y which is the distance from the point where the ladder touches the wall to the foot of the wall becomes the opposite side. It would be determined by applying trigonometric ratio
Sin θ = opposite side/hypotenuse
Sin 75.5 = y/12
Answer:
sorry i cant I really need the points tho