Need help, posted this a few times. Please don\'t copy and paste the solution so

Question

Need help, posted this a few times. Please don't copy and paste the solution someone posted on the internet (don't think its correct)! I have part A figured out and if you need it for any reason I could post it.  Need help with B C and D.

Any help would be appreciated.

Thanks!

Here is the image link:

i.imgur.com/BB6pTvN.jpg?1

Also, these notes I took will help hopefully.

i.imgur.com/GvVTT6k.jpg

Need help, posted this a few times. Please don't copy and paste the solution someone posted on the internet (don't think its correct)! I have part A figured out and if you need it for any reason I could post it. Need help with B C and D.

Explanation / Answer

Here is B and C

first need to find the derivative

dH/dx = -1/(1 + x^2 + 2y^2)^2 * 2x

dH/dy = -1/(1 + x^2 + 2y^2)^2 * 4y

so del H = dH/dx i + dH/dy j

at (1,0)
dH/dx = -1/(1 + 1)^2*2 = -1/2
dH/dy = 0

so direction is (-1,0)

mag = del f * direction = 1/2
so angle = arcos(1/2) = 60 degrees

side direction is (0,1)
del H* (0,1) = 0 so sidehill slope is 0

at (0,1)
dH/dx = 0
dH/dy = -1/(1+2)^2*4 = -4/9

so direction of max slope is (0,-1)

mag = delf*direction = 4/9
angle = arcos(4/9)=63.61 degrees

sidehill direction is (1,0)
del*side direction = 0


at (sqrt(2)/2, sqrt(2)/2)

dHdx/ = -1/( 1 + 1/2 + 1)^2*2*sqrt(2)/2 = - 4 sqrt(2)/25

dH/dy = -1/( 1 + 1/2 + 1)^2*4*sqrt(2)/2 = - 8 sqrt(2)/25

so direction is (-1/sqrt(5), -2/sqrt(5) )

mag = 4 sqrt(2)/(25*sqrt(5)) + 16 sqrt(2)/(25*sqrt(5))

angle = arcos(4 sqrt(2)/(25*sqrt(5)) + 16 sqrt(2)/(25*sqrt(5))) = 59.6 degrees
side direction is (2/sqrt(5), -1/sqrt(5))
then delH* side direction = 0

C)
since uphill direction points in the direction of del H
and side direction is perp to taht direction
so del H* side = mag delH*1*cos(angle)
so angle = 90 degrees so perp del H* side = 0
so angle of inclination is zero

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