find the extreme values and saddle points of f: f(x,y)=http://www.imagebam.com/image/268ae8405421087
\[f(x,y) = e^x \sin(y)\] 1) calculate the partials ...
fx=e^x * sin(y) fy=e^x*cos(y) fxx=e^x * sin(y) fyy=e^x * -sin(y) fxy=e^x
then what?
try finding stationary points, \[f_x = f_y = 0\]
if you can find any, they are the (local) extrema and/or saddle points you seek...
show me how.. thanks
there should not be any. \[f_x=e^x \sin(y)\] this is just constant* e^x as y is constant in the x direction. think about the geometry, too if you walk in the positive x direction starting at any y value, you are always walking up a hill that follows the e^x shape. if you walk in the positive y direction starting at any x value, you are always walking up and down a sinusoid but you always have a down slope on yr left and an up slope on yr right.
did this make sense?
actually i can't get it till now
in 2-D, x-y space, dy/dx = 0 is all you need to find a stationary point. in 3-D , x-y-z, where z = f(x,y), you need to find ∂z/∂x = 0 AND ∂z/∂y = 0 in this example: z = f(x,y)=e^x sin(y) ∂z/∂x is never zero, it is always increasing as you move along the x axis..... you will get peaks and troughs as you move along the y axis as ∂z/∂y=e^x*cos(y) and that will be zero when y = pi/2 + n pi. but the overall surface is always increasing in the x direction. at no point will you find ∂z/∂x = 0 AND ∂z/∂y = 0
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