If g(x, y) = x2 + y2 − 6x, find the gradient vector ∇g(2, 6) and use it to find the tangent line to the level curve g(x, y) = 28 at the point (2, 6). gradient vector

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.

If g(x, y) = x^2 + y^2 − 6x, find the gradient vector ∇g(2, 6) and use it to find the tangent line to the level curve g(x, y) = 28 at the point (2, 6).

Explore how to find the gradient vector ∇g(2, 6) for g(x, y) = x^2 + y^2 - 6x and use it to determine the tangent line to the level curve g(x, y) = 28 at the point (2, 6). This solution demonstrates key concepts in multivariable calculus, including gradient vectors and level curve analysis, suitable for students in Grades 11-12.

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