What do you get when you equate the first partials to zero?
I need to calculate the stationary points and determine their nature of f(x,y) = x + ysin(x).
I'm a little confused by setting δf/δx and δf/δy equal to zero and by that finding the stationary points...
Can someone solve this for me? Thanks!
f(x,y) = x + ysin(x)
∂/∂x = 1 + ycos(x)
∂/∂y = sin(x)
1 + ycos(x) = 0 -> y = 1 when even and -1 when odd
sin(x) = 0 -> x = kπ
-> Critical points (kπ,1) and (kπ, -1)
Second derivative test for P(kπ,1) :
∂²/∂x² = ysin(x) -> in P: -1sin(π) = 0
∂²/∂y² = 0 -> in P: 0
∂²/∂x∂y = cos(x) -> in P: cos(kπ) = -1 or 1
So Δ = (∂²/∂x²)(∂²/∂y²) - (∂²/∂x∂y)² = (0)(0) - (±1)² = 0 - 1 = -1
Thus Δ < 0 -> P(kπ,1) is a saddle point.
I agree with your first partials:
However, from these I get the critical points:
The sign of the y-coordinate is reversed from what you stated, but this doesn't actually affect the conclusion drawn from the second partials test.
(note we need the negative sign as .
Hence, none of the critical points are extrema.