Let f (x, y) = 1 − x cos(πy). Using differentials approximate f (1.01, 0.98). Box your final answer

To approximate \( f(1.01, 0.98) \) using differentials, we follow these steps:

### Step 1: Compute the partial derivatives of \( f(x, y) = 1 - x \cos(\pi y) \).

- The partial derivative with respect to \( x \):
  \[
  f_x(x, y) = \frac{\partial}{\partial x} \left(1 - x \cos(\pi y)\right) = -\cos(\pi y)
  \]

- The partial derivative with respect to \( y \):
  \[
  f_y(x, y) = \frac{\partial}{\partial y} \left(1 - x \cos(\pi y)\right) = x \pi \sin(\pi y)
  \]

### Step 2: Use the differentials to approximate the change in \( f \).

We want to approximate \( f(1.01, 0.98) \) near the point \( (1, 1) \), so we set \( x_0 = 1 \), \( y_0 = 1 \), \( dx = 1.01 - 1 = 0.01 \), and \( dy = 0.98 - 1 = -0.02 \).

The differential \( df \) is given by:
\[
df = f_x(x_0, y_0) dx + f_y(x_0, y_0) dy
\]

### Step 3: Evaluate the partial derivatives at \( (x_0, y_0) = (1, 1) \).

- \( f_x(1, 1) = -\cos(\pi \cdot 1) = -\cos(\pi) = 1 \)
- \( f_y(1, 1) = 1 \cdot \pi \sin(\pi \cdot 1) = \pi \cdot 0 = 0 \)

Thus, \( df \) becomes:
\[
df = (1)(0.01) + (0)(-0.02) = 0.01
\]

### Step 4: Approximate \( f(1.01, 0.98) \).

Using the differential, we approximate \( f(1.01, 0.98) \) as:
\[
f(1.01, 0.98) \approx f(1, 1) + df
\]

We know:
- \( f(1, 1) = 1 - 1 \cdot \cos(\pi \cdot 1) = 1 - 1(-1) = 2 \)

Thus:
\[
f(1.01, 0.98) \approx 2 + 0.01 = 2.01
\]

### Final Answer:
\[
\boxed{2.01}
\]

Would you like more details on any of these steps? 

Here are 5 related questions:
1. How are differentials used to approximate changes in multivariable functions?
2. What is the geometric interpretation of a partial derivative?
3. How can we improve the accuracy of approximation with second-order differentials?
4. Why does \( \sin(\pi) = 0 \) in this context?
5. What is the significance of using \( \pi \) in trigonometric functions?

**Tip:** When using differentials to approximate, remember that this method is most accurate for small changes in \( x \) and \( y \).

Let f(x, y) = 1 − x cos(πy). Using differentials approximate f(1.01, 0.98). Box your final answer

Learn how to approximate the function f(x, y) = 1 - x cos(πy) at the point (1.01, 0.98) using differentials and partial derivatives. This problem covers key concepts like linear approximation, partial derivatives, and differential calculus. Suitable for students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation