anonymous
  • anonymous
Find the local maximum and minimum values and saddle point(s) of the function f(x,y) = e^x cos(y)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@amistre64
anonymous
  • anonymous
I worked out fx and fy so\[f_x=e^x\cos(y)\text{ }f_y=-e^x\sin(y)\] So now?
amistre64
  • amistre64
fxx fyy - (fxy)^2 rings a bell

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anonymous
  • anonymous
Yes, I have to determine that as well. But first I need to find the critical points, right? I'm a little lost... hah
amistre64
  • amistre64
critical points are equal to 0 or undefined
amistre64
  • amistre64
fx is just ignoring the y direction and focusing on how it curves along x
anonymous
  • anonymous
Does it work like that every single time? I.e. when you look at fx=0, you ignore all the y's if there are any in the fx? Same for the fy?
amistre64
  • amistre64
f = e^x cos(y) fx = e^x cos(y) fy = -e^x sin(y) fx = 0 when e^x = -cos(y) by ignoring the ys, i mean that y is simply some constant during a derivative
amistre64
  • amistre64
fxx tells us curvative
amistre64
  • amistre64
fxx = e^x cos(y) fyy = -e^x cos(y) fxy = -e^x sin(y)
anonymous
  • anonymous
Oh haha! so how do I get the values when I say fx and fy = 0? Im struggling with them in problems like this one
amistre64
  • amistre64
i think we should get it to this point, then compare the rules: D = fxx fyy - (fxy)^2 if D>0 and fxx > 0, then min at a,b if D>0 and fxx < 0, then max at a,b if D<0, then saddle point if D=0, then its still undetermined
anonymous
  • anonymous
Okay
amistre64
  • amistre64
also, there is something about a system of equations for fx fy fx = 0 fy = 0 solve the system of equations
amistre64
  • amistre64
e^x cos(y) = -e^x sin(y) cos(y) = -sin(y) ; this is a 45 degree angle in Q2 and Q4 if my minds eye is not blind
anonymous
  • anonymous
Okay so e^x cos(y) = 0 -e^(x) sin(y) = 0 How to I get the values of x and y? In fx we know that either/both of e^x or cos y makes the answer be 0. It cannot be e^x, right? So we are left with cos y = 0, right?
anonymous
  • anonymous
Oh I see! I can say 0 = -sin(y)/cos(y) = -tan(y) and solve for y plug that in to any eq to get x?
amistre64
  • amistre64
sounds like a winner to me
anonymous
  • anonymous
Was my first attempt wrong?
amistre64
  • amistre64
i would say -1 = tan(y) is a solution, not 0=tan(y)
amistre64
  • amistre64
y = 3pi/4 or -pi/4
anonymous
  • anonymous
I didn't get that with -tan(y) = 0. I got y = 0
anonymous
  • anonymous
Ah I see, I made a stupid mistake haha
amistre64
  • amistre64
thats because its not 0 :) e^x cos(y) = -e^x sin(y) -cos(y) = sin(y) -1 = tan(y) sin(-pi/4) = -sqrt(2)/2 cos(-pi/4) = sqrt(2)/2 sin(3pi/4) = sqrt(2)/2 cos(3pi/4) = -sqrt(2)/2
anonymous
  • anonymous
Thanks @amistre64 ! I got the correct answer now!
amistre64
  • amistre64
yay!! :)

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