Tangent Plane to f(x,y) = sin(x)cos(y) at (π/3,π/2)

[Feodalherren]

Homework Statement

3a) Find the equation of the tangent plane to the function f(x,y) = sin(x)cos(y) at the point (∏/3,∏/2).

The Attempt at a Solution

There is quite clearly a z in the definition. What's going on?

 

[SteamKing]

It's the way you have g defined. According to the function definition, g := z - sin (x)*cos (y), so you need 3 arguments to specify a value for g.

 

[Feodalherren]

So if I evaluate the function f(x,y) = sin(x)cos(y) at the point (∏/3,∏/2) and then use that point to find my value of Z it should work?

 

[SteamKing]

Hey, give it a shot.

 

[rezeew]

The equation of the tangent plane to the function f(x,y) = sin(x)cos(y) at the point (π/3,π/2) is given by z = f(π/3,π/2) + f_x(π/3,π/2)(x-π/3) + f_y(π/3,π/2)(y-π/2), where f_x and f_y represent the partial derivatives of f with respect to x and y, respectively. In this case, f_x(π/3,π/2) = -cos(π/3)sin(π/2) = 0 and f_y(π/3,π/2) = -sin(π/3)cos(π/2) = -1/2. Therefore, the equation of the tangent plane becomes z = 1/2 + (-1/2)(y-π/2) = 1/2 - 1/2y. This result may seem strange since there is no z in the original function, but it is important to remember that the tangent plane is a two-dimensional surface that is tangent to the graph of the function f(x,y) at the given point. It is not a graph of a new function, but rather a linear approximation of the original function at that point.