Final answer:
The unit normal vector for the function f(x, y) = x³ at point P(-3,-2,-27) is found by evaluating the partial derivatives to obtain the gradient vector grad f(P) = <27, 0, 0>, and then normalizing it to get the unit normal vector n = <1, 0, 0>.
Step-by-step explanation:
To find a unit normal vector for the function f(x, y) = x³ at the point P(-3,-2,-27), we first need to find the gradient of the function. The gradient vector gives the direction of the steepest ascent at a point on the surface of the function, which is orthogonal to the level curve at that point.
The gradient of the function f(x, y) is calculated by taking the partial derivatives with respect to x and y. Since f(x, y) does not explicitly depend on y, its partial derivative with respect to y is zero. Therefore, we have:
- Partial derivative with respect to x: fx(x, y) = 3x²
- Partial derivative with respect to y: fy(x, y) = 0
At the point P(-3,-2,-27), we evaluate the partial derivative with respect to x to get:
Thus, the gradient vector at P is:
grad f(P) = <27, 0, 0>
To find the unit normal vector, we divide the gradient vector by its magnitude:
Magnitude of grad f(P) = √(27² + 0² + 0²) = √(729) = 27
Hence, the unit normal vector n is:
n = <1, 0, 0>
Please note that the normal vector here is simply a unit vector along the x-axis, as the function is a perfect cube with no variation in the y-dimension.