Final answer:
The equation of the tangent plane at the point (1,1) to the function f(x,y)=√(x³+y³) is z = (3/2)x + (3/2)y - (√2 - 3).
Step-by-step explanation:
To find the equation of the tangent plane at the point (1,1) to the function f(x,y)=√(x³+y³), we first need to compute the partial derivatives of f with respect to x and y.
The partial derivative of f with respect to x, denoted as fx, is given by:
fx(x,y) = (1/2)(x³+y³)-1/2 · 3x²
Similarly, the partial derivative of f with respect to y, denoted as fy, is:
fy(x,y) = (1/2)(x³+y³)-1/2 · 3y²
We then evaluate these derivatives at the given point (1,1):
fx(1,1) = (3/2)
fy(1,1) = (3/2)
Now we have the gradient vector of f at the point (1,1), which is ⟨3/2, 3/2⟩. The tangent plane to f at (1,1) can be expressed as:
z - f(1,1) = fx(1,1)(x - 1) + fy(1,1)(y - 1)
Since f(1,1) = √(1³+1³) = √2, the equation becomes:
z - √2 = (3/2)(x - 1) + (3/2)(y - 1)
Rearranging the terms, we obtain the equation of the tangent plane:
z = (3/2)x + (3/2)y - (√2 - 3)