9. Find the linear approximation of the function
f(x,y) = √1+ x^2+ y^2 at the point (x,y,z) = (2, -2, 3).
assuming you mean
z = f(x,y) = √(1+ x^2+ y^2) then we have the equation of the tangent plane as
z-f(2,-2,3) = ∂f/∂x (x-2) + ∂f/∂y (y+2)
z-3 = 2/√5 (x-2) - 2/√5 (y+2)
To find the linear approximation of a function f(x, y) at a point (x, y, z), we can use the first-order partial derivatives.
First, we need to calculate the first-order partial derivatives of f(x, y) with respect to x and y. Let's denote these derivatives as f_x and f_y, respectively.
f_x = d(f(x, y))/dx
= d(√(1 + x^2 + y^2))/dx
= (1/2)(1 + x^2 + y^2)^(-1/2)(d(1 + x^2 + y^2))/dx
= (1/2)(1 + x^2 + y^2)^(-1/2)(2x)
= x/(√(1 + x^2 + y^2))
Similarly, we can find the partial derivative with respect to y:
f_y = d(f(x, y))/dy
= d(√(1 + x^2 + y^2))/dy
= (1/2)(1 + x^2 + y^2)^(-1/2)(d(1 + x^2 + y^2))/dy
= (1/2)(1 + x^2 + y^2)^(-1/2)(2y)
= y/(√(1 + x^2 + y^2))
Next, we evaluate the derivatives at the given point (2, -2, 3):
f_x(2, -2) = 2/(√(1 + 2^2 + (-2)^2))
= 2/(√(1 + 4 + 4))
= 2/(√9)
= 2/3
f_y(2, -2) = -2/(√(1 + 2^2 + (-2)^2))
= -2/(√(1 + 4 + 4))
= -2/(√9)
= -2/3
Now, using the first-order partial derivatives, we can write the linear approximation of f(x, y) as:
L(x, y) = f(2, -2) + f_x(2, -2)(x - 2) + f_y(2, -2)(y + 2)
Let's substitute the values from above:
L(x, y) = √(1 + (2)^2 + (-2)^2) + (2/3)(x - 2) + (-2/3)(y + 2)
= √(1 + 4 + 4) + (2/3)(x - 2) + (-2/3)(y + 2)
= √9 + (2/3)(x - 2) - (2/3)(y + 2)
= 3 + (2/3)(x - 2) - (2/3)(y + 2)
Thus, the linear approximation of the function f(x, y) = √(1 + x^2 + y^2) at the point (2, -2, 3) is L(x, y) = 3 + (2/3)(x - 2) - (2/3)(y + 2).