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To solve the problem, we need to approximate the value of the function \( f(x, y) = \sqrt{67 - 2x^2 - y^2} \) using a tangent plane at the point \( (-0.9, 0.9) \).

The equation of the tangent plane at a point \( (x_0, y_0) \) for a function \( f(x, y) \) is given by:

\[
L(x, y) = f(x_0, y_0) + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)
\]

Where:
- \( f_x \) is the partial derivative of \( f \) with respect to \( x \).
- \( f_y \) is the partial derivative of \( f \) with respect to \( y \).

### Step 1: Find the partial derivatives of \( f(x, y) \).

1. **Partial derivative with respect to \( x \) ( \( f_x(x, y) \)):**
   
   \[
   f(x, y) = \sqrt{67 - 2x^2 - y^2}
   \]
   Using the chain rule, we get:

   \[
   f_x(x, y) = \frac{d}{dx}\left( \sqrt{67 - 2x^2 - y^2} \right) = \frac{-4x}{2\sqrt{67 - 2x^2 - y^2}} = \frac{-2x}{\sqrt{67 - 2x^2 - y^2}}
   \]

2. **Partial derivative with respect to \( y \) ( \( f_y(x, y) \)):**

   Similarly, we compute \( f_y(x, y) \) using the chain rule:

   \[
   f_y(x, y) = \frac{d}{dy}\left( \sqrt{67 - 2x^2 - y^2} \right) = \frac{-2y}{2\sqrt{67 - 2x^2 - y^2}} = \frac{-y}{\sqrt{67 - 2x^2 - y^2}}
   \]

### Step 2: Evaluate the function and partial derivatives at the point \( (-0.9, 0.9) \).

We will now calculate \( f(-0.9, 0.9) \), \( f_x(-0.9, 0.9) \), and \( f_y(-0.9, 0.9) \).

1. **\( f(-0.9, 0.9) \):**

   \[
   f(-0.9, 0.9) = \sqrt{67 - 2(-0.9)^2 - (0.9)^2}
   \]

2. **\( f_x(-0.9, 0.9) \):**

   \[
   f_x(-0.9, 0.9) = \frac{-2(-0.9)}{\sqrt{67 - 2(-0.9)^2 - (0.9)^2}}
   \]

3. **\( f_y(-0.9, 0.9) \):**

   \[
   f_y(-0.9, 0.9) = \frac{-0.9}{\sqrt{67 - 2(-0.9)^2 - (0.9)^2}}
   \]

Let's compute these values.We have the following values at the point \( (-0.9, 0.9) \):

1. \( f(-0.9, 0.9) = 8.0355 \)
2. \( f_x(-0.9, 0.9) = 0.2240 \)
3. \( f_y(-0.9, 0.9) = -0.1120 \)

### Step 3: Equation of the tangent plane

The equation of the tangent plane is:

\[
L(x, y) = f(-0.9, 0.9) + f_x(-0.9, 0.9)(x + 0.9) + f_y(-0.9, 0.9)(y - 0.9)
\]

Substituting the values:

\[
L(x, y) = 8.0355 + 0.2240(x + 0.9) - 0.1120(y - 0.9)
\]

This equation can now be used to approximate the value of \( f(x, y) \) near the point \( (-0.9, 0.9) \).

Would you like more details on how to use this tangent plane for approximation or clarification on any step?

Here are 5 related questions for further practice:
1. How do you compute the tangent plane for a function at a different point?
2. What is the geometric interpretation of the tangent plane in 3D space?
3. How do partial derivatives represent rates of change in multivariable calculus?
4. How would the result change if we used a second-degree Taylor expansion instead?
5. How do we apply the tangent plane method in real-life problems like linear approximations?

**Tip:** Understanding partial derivatives in multivariable functions is crucial for analyzing how each variable affects the function independently.

Use a tangent plane to approximate the value of the following function at the point (-0.9, 0.9). Give your answer accurate to 4 decimal places. f(x, y) = sqrt(67 - 2x^2 - y^2)

Learn how to use the tangent plane method to approximate the value of the function f(x, y) = sqrt(67 - 2x^2 - y^2) at the point (-0.9, 0.9). This problem covers multivariable calculus concepts like partial derivatives and tangent plane approximation. Suitable for undergraduate calculus students.

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