We are tasked with using a tangent plane to approximate the value of the function $f(x, y) = \sqrt{143 - 3x^2 - y^2}$ at the point $(5.1, -2.1)$. To do this, we need to use the linearization formula, which is the equation of the tangent plane:

$L(x, y) = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b)$

where:

• $(a, b)$ is the point of tangency, here $(5, -2)$,
• $f_x(a, b)$ and $f_y(a, b)$ are the partial derivatives of $f(x, y)$ evaluated at $(a, b)$,
• $L(x, y)$ will give us the linear approximation of $f(x, y)$ near $(a, b)$.

Step 1: Compute $f_x(x, y)$ and $f_y(x, y)$

The function $f(x, y) = \sqrt{143 - 3x^2 - y^2}$, so its partial derivatives are:

1. Partial derivative with respect to $x$: $f_x(x, y) = \frac{\partial}{\partial x} \left( \sqrt{143 - 3x^2 - y^2} \right)$ Applying the chain rule: $f_x(x, y) = \frac{1}{2\sqrt{143 - 3x^2 - y^2}} \cdot (-6x) = \frac{-6x}{2\sqrt{143 - 3x^2 - y^2}}$

2. Partial derivative with respect to $y$: $f_y(x, y) = \frac{\partial}{\partial y} \left( \sqrt{143 - 3x^2 - y^2} \right)$ Applying the chain rule: $f_y(x, y) = \frac{1}{2\sqrt{143 - 3x^2 - y^2}} \cdot (-2y) = \frac{-2y}{2\sqrt{143 - 3x^2 - y^2}}$

Step 2: Evaluate $f_x(a, b)$ and $f_y(a, b)$

Now we evaluate the partial derivatives at the point $(a, b) = (5, -2)$.

Let me calculate these values.We have evaluated the following at the point $(5, -2)$:

• $f_x(5, -2) = -1.875$
• $f_y(5, -2) = 0.25$
• $f(5, -2) = 8.0$

Step 3: Apply the linearization formula

The equation of the tangent plane (linearization) is:

$L(x, y) = f(5, -2) + f_x(5, -2)(x - 5) + f_y(5, -2)(y + 2)$

Substituting the values we computed:

$L(x, y) = 8.0 + (-1.875)(x - 5) + 0.25(y + 2)$

Step 4: Approximate $f(5.1, -2.1)$

We now approximate the function value at $(5.1, -2.1)$ using the linearization:

$L(5.1, -2.1) = 8.0 + (-1.875)(5.1 - 5) + 0.25(-2.1 + 2)$

Let me calculate this next.The linear approximation of the function $f(x, y) = \sqrt{143 - 3x^2 - y^2}$ at the point $(5.1, -2.1)$ is approximately $7.7875$, accurate to four decimal places.

Would you like more details or have any questions?

Here are some related questions to explore:

1. How do we interpret the meaning of a tangent plane in this context?
2. How would the approximation change if we used a different nearby point, such as $(4.9, -1.9)$?
3. What are other ways to approximate the value of a function near a point besides using a tangent plane?
4. How would the result differ if we considered a function with higher degrees of nonlinearity?
5. Can we find the exact value of $f(5.1, -2.1)$ and compare it with this approximation?

Tip: The tangent plane is a powerful tool for approximating the behavior of a function near a given point, especially when dealing with smooth surfaces.