Use the linear approximation of  
f
(
x
,
y
)
=
3
cos
(
x
)
+
2
sin
(
x
+
y
)
  at the point  
(
0
,
0
)
  to approximate the value of  
f
(
0.002
,
−
0.4
)
 . Round your answer to the nearest hundredth if necessary.

We are tasked with approximating the value of the function \( f(x, y) = 3\cos(x) + 2\sin(x + y) \) at the point \( (0.002, -0.4) \), using the linear approximation at the point \( (0, 0) \).

### Step 1: Linear Approximation Formula
The linear approximation (or tangent plane approximation) for a function \( f(x, y) \) at the point \( (a, b) \) is given by:

\[
f(x, y) \approx f(a, b) + f_x(a, b) \cdot (x - a) + f_y(a, b) \cdot (y - b)
\]

where:
- \( f(a, b) \) is the function value at the point \( (a, b) \),
- \( f_x(a, b) \) is the partial derivative of \( f \) with respect to \( x \),
- \( f_y(a, b) \) is the partial derivative of \( f \) with respect to \( y \).

### Step 2: Calculate \( f(0, 0) \)
First, evaluate \( f(x, y) \) at \( (0, 0) \):

\[
f(0, 0) = 3\cos(0) + 2\sin(0 + 0)
\]
\[
f(0, 0) = 3(1) + 2(0) = 3
\]

### Step 3: Calculate the Partial Derivatives

#### Partial Derivative with Respect to \( x \):
\[
f_x(x, y) = \frac{\partial}{\partial x} \left( 3\cos(x) + 2\sin(x + y) \right)
\]
Using the chain rule:
\[
f_x(x, y) = -3\sin(x) + 2\cos(x + y)
\]
Now, evaluate at \( (0, 0) \):
\[
f_x(0, 0) = -3\sin(0) + 2\cos(0 + 0) = -3(0) + 2(1) = 2
\]

#### Partial Derivative with Respect to \( y \):
\[
f_y(x, y) = \frac{\partial}{\partial y} \left( 3\cos(x) + 2\sin(x + y) \right)
\]
\[
f_y(x, y) = 2\cos(x + y)
\]
Now, evaluate at \( (0, 0) \):
\[
f_y(0, 0) = 2\cos(0 + 0) = 2(1) = 2
\]

### Step 4: Apply the Linear Approximation Formula
Now that we have the necessary values:
- \( f(0, 0) = 3 \),
- \( f_x(0, 0) = 2 \),
- \( f_y(0, 0) = 2 \),

we can use the linear approximation at \( (0, 0) \) to approximate \( f(0.002, -0.4) \):

\[
f(0.002, -0.4) \approx f(0, 0) + f_x(0, 0) \cdot (0.002 - 0) + f_y(0, 0) \cdot (-0.4 - 0)
\]
\[
f(0.002, -0.4) \approx 3 + 2 \cdot 0.002 + 2 \cdot (-0.4)
\]
\[
f(0.002, -0.4) \approx 3 + 0.004 - 0.8
\]
\[
f(0.002, -0.4) \approx 2.204
\]

### Final Answer:
The linear approximation of \( f(0.002, -0.4) \) is approximately \( 2.20 \) (rounded to the nearest hundredth).

---

Would you like a more detailed breakdown of any part of the process? Here are some related questions for further practice:
1. How would the linear approximation change if we used a different point, like \( (1, 1) \)?
2. What happens to the approximation as the values of \( (x, y) \) get further from \( (0, 0) \)?
3. How do we interpret the result of the linear approximation in terms of actual error?
4. Can you derive the second-order approximation for this function at the same point?
5. What would happen if we approximated \( f(x, y) \) using a different method, such as polynomial interpolation?

**Tip**: Linear approximations work best when the point of approximation is close to the target values. As you move farther from the point, the linear model may become less accurate.

Use the linear approximation of f(x, y) = 3cos(x) + 2sin(x + y) at the point (0, 0) to approximate the value of f(0.002, -0.4). Round your answer to the nearest hundredth if necessary.

Learn how to use linear approximation to estimate the value of the function f(x, y) = 3cos(x) + 2sin(x + y) at a nearby point. This solution covers partial derivatives and the linear approximation formula, providing a clear, step-by-step breakdown. Suitable for students in Grades 11-12.

Math Problem Statement

Use the linear approximation of
f ( x , y )

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