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To find the normal vector to the tangent plane of the surface \( f(x, y) = x^2 - y^2 \) at the point \( (2, 1, 3) \), we need to compute the gradient of \( f \), which gives the normal vector.

The gradient of \( f \), denoted as \( \nabla f \), is given by:
\[
\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right).
\]

Let's compute these partial derivatives:
1. \( \frac{\partial f}{\partial x} = 2x \).
2. \( \frac{\partial f}{\partial y} = -2y \).

Now, we evaluate these derivatives at the given point \( (x, y) = (2, 1) \):
\[
\frac{\partial f}{\partial x} \bigg|_{(2,1)} = 2(2) = 4,
\]
\[
\frac{\partial f}{\partial y} \bigg|_{(2,1)} = -2(1) = -2.
\]

Thus, the gradient vector (and therefore the normal vector to the tangent plane) at the point \( (2, 1) \) is:
\[
\nabla f = \langle 4, -2 \rangle.
\]

Since the normal vector to the tangent plane should also include the z-component, which corresponds to the partial derivative with respect to \( z \), and given that \( z \) is not explicitly in the equation (making it effectively zero), the full normal vector is:
\[
\langle 4, -2, -1 \rangle.
]

The solution yields \( \text{Normal Vector to the Tangent plane}?

Which of the following vectors is a normal vector to the tangent plane of f(x, y) = x^2 - y^2 at the point (2, 1, 3)?

Find the normal vector to the tangent plane of the surface f(x, y) = x² - y² at the point (2, 1, 3). This problem involves computing the gradient and using it to determine the normal vector. Suitable for college-level calculus students.

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