Use the linear approximation of
f ( x , y )

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).

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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.