Partial Derivatives

[miss_b]

I have two questions that I am having trouble with.

The first is the easy one:

A surface on a 3D cartesian grid is described by the equation \(\displaystyle z = sin(6.56x - 5.23y)\)

Calculate the value of z (I know that one), the slope in the x-direction and the slope in the y-direction at the point where (x,y) are (-1.3, -1.5) **I'm guessing that you apply the chain rule, but I can't get it to work out**

Answer (two decimal places):
f[sub:1w8zqiz6]x[/sub:1w8zqiz6]= 5.09
f[sub:1w8zqiz6]y[/sub:1w8zqiz6] = -4.06


The second question is horrible, mainly because I cannot visualize what it is supposed to look like, if I could get some help starting, that would be brilliant

A flat plane on a 3D cartesian grid is described by the equation \(\displaystyle z = -8.11x - 1.82y - 5.67\)

** If you were standing at (x,y) = (4.7, -2) in which horizontal direction is the steepest slope downwards?
** In that direction, what is the absolute angle between the surface and the horizontal plane? ~~I don't even understand this bit! Something about the slope when it hits 'ground zero'?? I have no idea!

Answer (two decimal places):
Direction of steepest descent (degrees): 12.65
Absolute angle of steepest descent (degrees): 83.14

Ugh!!

Thank you in advance
Beckie

 

[galactus]

The direction of steepest descent for problem #2:

\(\displaystyle \sqrt{(-8.11)^{2}+(-1.82)^{2}}=8.3117\)

\(\displaystyle cos^{-1}(\frac{8.11}{8.3117})\approx 12.648\)

 

[BigGlenntheHeavy]

!st one.

\(\displaystyle z \ = \ f(x,y) \ = \ sin(6.56x-5.23y)\)

\(\displaystyle f_x(x,y) \ = \ 6.56cos(6.56x-5.23y)\)

\(\displaystyle f_x(-1.3,-1.5) \ = \ 6.56cos[(6.56)(-1.3)+(-5.23)(-1.5)] \ = \ 5.09\)

\(\displaystyle f_y(x,y) \ = \ -5.23cos(6.56x-5.23y)\)

\(\displaystyle f_y(-1.3,-1.5) \ = \ -5.23cos[(6.56)(-1.3)+(-5.23)(-1.5)] \ = \ -4.06\)

 

[miss_b]

Thank you to both of you for your quick replies. I appologise for being rude and not responding, I haven't had internet for a week