The temperature at any point in the plane is given by function f(x,y) = 100/(x^(2)+y^(2)+1). Find a direction at the point (3,2) in which the temperature does not increase or decrease.
Help, please? Thank you.
To find a direction at a given point where the temperature does not increase or decrease, we need to find the gradient of the temperature function and then determine the direction of the gradient vector at that point.
The gradient of a function f(x, y) is a vector that points in the direction of the steepest increase of the function and its magnitude gives the rate of increase. In our case, the temperature function is f(x, y) = 100/(x^2 + y^2 + 1).
To find the gradient vector ∇f(x, y), we need to compute the partial derivatives of f with respect to x and y, respectively.
∂f/∂x = -200x/(x^2 + y^2 + 1)^2
∂f/∂y = -200y/(x^2 + y^2 + 1)^2
Now, let's evaluate the gradient at the point (3, 2):
∂f/∂x = -200(3)/((3^2) + (2^2) + 1)^2 = -600/14400 = -1/24
∂f/∂y = -200(2)/((3^2) + (2^2) + 1)^2 = -400/14400 = -1/36
So, the gradient at the point (3, 2) is ∇f(3, 2) = (-1/24, -1/36).
To find a direction where the temperature does not increase or decrease, we need to find a vector perpendicular to the gradient vector (∇f). We can achieve this by swapping the components and changing the sign of one component.
Therefore, a direction at the point (3, 2) in which the temperature does not increase or decrease is given by the vector (1/36, -1/24).
Please note that this vector represents a magnitude and direction, it is not a unique solution, and any scalar multiple of this vector would also satisfy the condition.