To find the gradient vector of $g(x, y) = x^2 + y^2 - 6x$ at the point $(2, 6)$, and then use it to find the tangent line to the level curve $g(x, y) = 28$ at $(2, 6)$, we will proceed in steps.

Step 1: Compute the Gradient Vector $\nabla g(x, y)$

The gradient vector $\nabla g(x, y)$ is given by the partial derivatives of $g$ with respect to $x$ and $y$: $\nabla g(x, y) = \left( \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y} \right)$

1. Compute $\frac{\partial g}{\partial x}$: $\frac{\partial g}{\partial x} = 2x - 6$

2. Compute $\frac{\partial g}{\partial y}$: $\frac{\partial g}{\partial y} = 2y$

So, the gradient vector is: $\nabla g(x, y) = (2x - 6, 2y)$

Step 2: Evaluate the Gradient Vector at $(2, 6)$

Substitute $x = 2$ and $y = 6$ into the gradient vector: $\nabla g(2, 6) = (2(2) - 6, 2(6)) = (4 - 6, 12) = (-2, 12)$

Thus, the gradient vector at $(2, 6)$ is: $\nabla g(2, 6) = (-2, 12)$

Step 3: Tangent Line to the Level Curve $g(x, y) = 28$ at $(2, 6)$

The level curve $g(x, y) = 28$ represents points $(x, y)$ where the function $g(x, y)$ equals 28. Since the gradient vector $\nabla g(2, 6)$ is perpendicular to the level curve at $(2, 6)$, it gives the direction of the normal line. The direction vector of the tangent line, which is perpendicular to the normal, is therefore $(12, 2)$.

Equation of the Tangent Line

The tangent line at $(2, 6)$ can be written as: $y - 6 = m(x - 2)$ where $m$ is the slope of the tangent line. Since the tangent direction vector is $(12, 2)$, the slope $m$ is: $m = \frac{2}{12} = \frac{1}{6}$

Thus, the equation of the tangent line is: $y - 6 = \frac{1}{6}(x - 2)$ or, simplifying, $y = \frac{1}{6}x + \frac{34}{6} = \frac{1}{6}x + \frac{17}{3}$

Final Answers:

• Gradient vector at $(2, 6)$: $\nabla g(2, 6) = (-2, 12)$
• Equation of the tangent line to the level curve $g(x, y) = 28$ at $(2, 6)$: $y = \frac{1}{6}x + \frac{17}{3}$

Would you like further details or have any questions?

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1. What are the applications of the gradient vector in optimization problems?
2. How does the gradient vector help in determining the direction of the steepest ascent?
3. What is a level curve, and why is the gradient vector perpendicular to it?
4. How can the tangent line equation be used to approximate values near $(2, 6)$?
5. How would the tangent line equation change if $g(x, y)$ had different coefficients?

Tip: The gradient vector points in the direction of the greatest rate of increase of a function.