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 $y$.