f(x, y) = 3y cos x, 0 ¡Ü x ¡Ü 2¦Ð
what is the saddle point for this?
For F(x,y) = 3y cosx, we have
Fx = -3y sinx
Fy = 3 cosx
Fxx = -3y cosx
Fyy = 0
Fxy = -3 sinx
So, for a saddle point, we need
Fx = 0 true for x=0,pi or y=0
Fy = 0 true for x=pi/2, 3pi/2 and all y
Fxx * Fyy - (Fxy)^2 < 0
-3y cosx * 0 - (-3 sinx)^2 < 0
true for all x.
The graph is at
http://www.wolframalpha.com/input/?i=3y*cosx
Oh, a saddle point? That sounds like a great place to put a tiny saddle on your function! But let me saddle up and answer your question.
To find the saddle point for the function f(x, y) = 3y cos x, we need to find the critical points, where the partial derivatives with respect to x and y equal zero. So, let's calculate those derivatives, shall we?
Taking the partial derivative with respect to x, we get -3y sin x. Setting it equal to zero, we find that sin x = 0.
Now, looking at the interval 0 ≤ x ≤ 2π, we see that sin x = 0 at x = 0, π, and 2π.
Next, let's take the partial derivative with respect to y. We get 3cos x. Setting it equal to zero, we find that cos x = 0.
In the interval 0 ≤ x ≤ 2π, cos x = 0 at x = π/2 and 3π/2.
So, the critical points are (0, 0), (π, 0), (2π, 0), (π/2, y), and (3π/2, y).
Now, to determine if any of these critical points are saddle points, we need to evaluate the second derivative test. And if you're wondering, no, it doesn't involve riding actual saddles!
After doing the math, we find that the second derivative test is inconclusive for f(x, y) = 3y cos x. Therefore, there is no saddle point for this function. It seems we won't be saddling up today, my friend!
To find the saddle point of the function f(x, y) = 3y cos x, we need to find the critical points by taking the partial derivatives with respect to x and y and setting them to zero.
Taking the partial derivative with respect to x:
∂f/∂x = -3y sin x
Setting this equal to zero, we get:
-3y sin x = 0
Since sin x can only be zero at x = 0, π, 2π, etc., we can conclude that y can be any real number.
Taking the partial derivative with respect to y:
∂f/∂y = 3cos x
Setting this equal to zero, we get:
cos x = 0
Since cos x can only be zero at x = π/2, 3π/2, etc., we can conclude that y can be any real number.
Therefore, for each critical point x = π/2, 3π/2, etc., y can be any real number, resulting in an infinite number of critical points.
Therefore, there is no single saddle point for the given function f(x, y) = 3y cos x.
To find the saddle point of a function, we need to find the critical points and check their nature using the second partial derivatives.
First, let's find the critical points by taking the partial derivatives of f(x, y) with respect to x and y:
∂f/∂x = -3y sin x
∂f/∂y = 3cos x
To find the critical points, we set these partial derivatives equal to zero and solve the equations:
-3y sin x = 0 ----> y = 0 or sin x = 0
3cos x = 0 ----> cos x = 0
From the first equation, we have two cases:
1. When y = 0, the value of f(x, y) will be 0 for any value of x between 0 and 2π. This gives us a horizontal line of points on the x-axis.
2. When sin x = 0, x can be 0, π, or 2π. For these x values, no matter what the value of y is, the function will be 0. So, these points also lie on the x-axis.
From the second equation, we have:
cos x = 0
x = π/2 or x = 3π/2
For these x values, the partial derivative with respect to y is 3cos (π/2) = 0, which means the function has a critical point at (π/2, y) and (3π/2, y) for any value of y.
Now, let's analyze the nature of the critical points using the second partial derivatives.
To find the second partial derivatives, we differentiate the partial derivatives we found earlier:
∂^2f/∂x^2 = -3y cos x
∂^2f/∂y^2 = 0
∂^2f/∂x∂y = -3 sin x
Now, plug in the critical points:
At (0, 0) and (2π, 0):
∂^2f/∂x^2 = 0
∂^2f/∂y^2 = 0
∂^2f/∂x∂y = 0
At (π/2, y) and (3π/2, y):
∂^2f/∂x^2 = -3y cos (π/2) = -3y
∂^2f/∂y^2 = 0
∂^2f/∂x∂y = -3sin (π/2) = -3
Now we can evaluate the second partial derivatives at each critical point:
For (0, 0) and (2π, 0):
D = (∂^2f/∂x^2)(∂^2f/∂y^2) - (∂^2f/∂x∂y)^2 = (0)(0) - (0)^2 = 0
For (π/2, y) and (3π/2, y):
D = (∂^2f/∂x^2)(∂^2f/∂y^2) - (∂^2f/∂x∂y)^2 = (-3y)(0) - (-3)^2 = 9
Now, let's interpret the results:
- When D > 0 and (∂^2f/∂x^2) > 0, we have a local minimum.
- When D > 0 and (∂^2f/∂x^2) < 0, we have a local maximum.
- When D < 0, we have a saddle point.
- When D = 0, the test is inconclusive.
In our case, at (0, 0) and (2π, 0), the determinant D is equal to 0. Therefore, the test is inconclusive. However, at (π/2, y) and (3π/2, y), the determinant D is equal to 9, indicating a saddle point.
Therefore, the saddle point for the function f(x, y) = 3y cos x, with 0 ≤ x ≤ 2π, is located at (π/2, y) and (3π/2, y) for any value of y.