let f(x,y)=sqrt (1-x^2-y^2) where x^2+y^2 <1. (positive sqrt) a.) what is the graph of f in geometric terms? b.) find the equation of the tangent plane to the graph (surface) at a point (x,y) with x^2 +y^2 <1
a) To visualize the graph of the function f(x, y) = sqrt(1 - x^2 - y^2) where x^2 + y^2 < 1, we need to consider the unit disk in the xy-plane since the condition x^2 + y^2 < 1 restricts the domain of the function within this region.
The graph of f(x, y) will be the upper half of the unit sphere centered at (0, 0, 1), which is located above the xy-plane. As for the lower half of the sphere, it is not included in the graph since we consider the positive square root in the function.
b) To find the equation of the tangent plane to the graph (surface) at a point (x, y) with x^2 + y^2 < 1, we need to compute the partial derivatives of f(x, y) with respect to x and y.
The partial derivative of f(x, y) with respect to x can be obtained by treating y as a constant and differentiating with respect to x. Similarly, the partial derivative of f(x, y) with respect to y can be obtained by treating x as a constant and differentiating with respect to y.
Let's calculate the partial derivatives:
Partial derivative of f(x, y) with respect to x:
∂f/∂x = (∂/∂x) sqrt(1 - x^2 - y^2)
= (-2x) / (2 * sqrt(1 - x^2 - y^2))
= -x / sqrt(1 - x^2 - y^2)
Partial derivative of f(x, y) with respect to y:
∂f/∂y = (∂/∂y) sqrt(1 - x^2 - y^2)
= (-2y) / (2 * sqrt(1 - x^2 - y^2))
= -y / sqrt(1 - x^2 - y^2)
Next, we evaluate the partial derivatives at the given point (x, y). With this information, we can construct the equation of the tangent plane.
Let's denote the point (x, y) as (a, b) for simplicity. So, we have:
Partial derivative of f(x, y) with respect to x at (a, b):
∂f/∂x = -a / sqrt(1 - a^2 - b^2)
Partial derivative of f(x, y) with respect to y at (a, b):
∂f/∂y = -b / sqrt(1 - a^2 - b^2)
Using the equation of a plane:
z - f(a, b) = (∂f/∂x)(x - a) + (∂f/∂y)(y - b),
we substitute the known values f(a, b) = sqrt(1 - a^2 - b^2), (∂f/∂x) = -a/sqrt(1 - a^2 - b^2), (∂f/∂y) = -b/sqrt(1 - a^2 - b^2), and (x, y) = (a, b).
Finally, we simplify the equation, and the equation of the tangent plane becomes:
z = a^2 + b^2 + sqrt(1 - a^2 - b^2) - 1.