Welcome to Physics Forums.day1ok said:Find the tanget plane using partial derivatives.
z=ycos(x-y), at point (2,2,2)attempted: fx(x,y)=ysin(x-y)*1
fx(2,2)=0
fy(x,y)=1*sin(x-y)*1
fx(2,2)=0
z-2=0(x-2)+0(x-2)
z=2 Incorrect
The tangent plane of a function at a given point is a flat surface that touches the function at that point and has the same slope as the function at that point. In this case, the tangent plane of z=ycos(x-y) at point (2,2,2) would be a flat surface that touches the function z=ycos(x-y) at the point (2,2,2) and has the same slope as the function at that point.
To find the equation of the tangent plane at a given point, you need to first find the partial derivatives of the function with respect to each variable (x, y, z). Then, you can use the point-slope form of a line to write the equation of the tangent plane, with the partial derivatives as the slope and the given point as a point on the plane.
The normal vector of a plane is a vector that is perpendicular to the plane. In this case, the normal vector of the tangent plane at point (2,2,2) will be perpendicular to the slope of the function at that point, which is determined by the partial derivatives.
As the point moves along the function, the tangent plane will also change. This is because the slope of the function at each point is different, which means the slope of the tangent plane will also be different. Additionally, the normal vector of the tangent plane will also change as the point moves along the function.
No, the tangent plane can never be parallel to the function. This is because the tangent plane must have the same slope as the function at the point of tangency, and if it were parallel, the slopes would not be the same.