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We are tasked with using a tangent plane to approximate the value of the function \( f(x, y) = \sqrt{143 - 3x^2 - y^2} \) at the point \( (5.1, -2.1) \). To do this, we need to use the linearization formula, which is the equation of the tangent plane:

\[
L(x, y) = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b)
\]

where:
- \( (a, b) \) is the point of tangency, here \( (5, -2) \),
- \( f_x(a, b) \) and \( f_y(a, b) \) are the partial derivatives of \( f(x, y) \) evaluated at \( (a, b) \),
- \( L(x, y) \) will give us the linear approximation of \( f(x, y) \) near \( (a, b) \).

### Step 1: Compute \( f_x(x, y) \) and \( f_y(x, y) \)
The function \( f(x, y) = \sqrt{143 - 3x^2 - y^2} \), so its partial derivatives are:

1. **Partial derivative with respect to \( x \):**
   \[
   f_x(x, y) = \frac{\partial}{\partial x} \left( \sqrt{143 - 3x^2 - y^2} \right)
   \]
   Applying the chain rule:
   \[
   f_x(x, y) = \frac{1}{2\sqrt{143 - 3x^2 - y^2}} \cdot (-6x) = \frac{-6x}{2\sqrt{143 - 3x^2 - y^2}}
   \]

2. **Partial derivative with respect to \( y \):**
   \[
   f_y(x, y) = \frac{\partial}{\partial y} \left( \sqrt{143 - 3x^2 - y^2} \right)
   \]
   Applying the chain rule:
   \[
   f_y(x, y) = \frac{1}{2\sqrt{143 - 3x^2 - y^2}} \cdot (-2y) = \frac{-2y}{2\sqrt{143 - 3x^2 - y^2}}
   \]

### Step 2: Evaluate \( f_x(a, b) \) and \( f_y(a, b) \)
Now we evaluate the partial derivatives at the point \( (a, b) = (5, -2) \).

Let me calculate these values.We have evaluated the following at the point \( (5, -2) \):

- \( f_x(5, -2) = -1.875 \)
- \( f_y(5, -2) = 0.25 \)
- \( f(5, -2) = 8.0 \)

### Step 3: Apply the linearization formula
The equation of the tangent plane (linearization) is:

\[
L(x, y) = f(5, -2) + f_x(5, -2)(x - 5) + f_y(5, -2)(y + 2)
\]

Substituting the values we computed:

\[
L(x, y) = 8.0 + (-1.875)(x - 5) + 0.25(y + 2)
\]

### Step 4: Approximate \( f(5.1, -2.1) \)
We now approximate the function value at \( (5.1, -2.1) \) using the linearization:

\[
L(5.1, -2.1) = 8.0 + (-1.875)(5.1 - 5) + 0.25(-2.1 + 2)
\]

Let me calculate this next.The linear approximation of the function \( f(x, y) = \sqrt{143 - 3x^2 - y^2} \) at the point \( (5.1, -2.1) \) is approximately \( 7.7875 \), accurate to four decimal places.

Would you like more details or have any questions?

Here are some related questions to explore:
1. How do we interpret the meaning of a tangent plane in this context?
2. How would the approximation change if we used a different nearby point, such as \( (4.9, -1.9) \)?
3. What are other ways to approximate the value of a function near a point besides using a tangent plane?
4. How would the result differ if we considered a function with higher degrees of nonlinearity?
5. Can we find the exact value of \( f(5.1, -2.1) \) and compare it with this approximation?

**Tip:** The tangent plane is a powerful tool for approximating the behavior of a function near a given point, especially when dealing with smooth surfaces.

Use a tangent plane to approximate the value of the function f(x, y) = sqrt(143 - 3x^2 - y^2) at the point (5.1, -2.1). Give your answer accurate to 4 decimal places.

This problem demonstrates how to use the tangent plane to approximate the value of the multivariable function f(x, y) = sqrt(143 - 3x^2 - y^2) at the point (5.1, -2.1). The solution involves applying partial derivatives and the linearization formula. A detailed step-by-step explanation is provided, including the tangent plane equation.

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