Math Problem Statement

Use 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). Give your answer accurate to 4 decimal places.

Solution

We are tasked with using a tangent plane to approximate the value of the function f(x,y)=1433x2y2f(x, y) = \sqrt{143 - 3x^2 - y^2} at the point (5.1,2.1)(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)+fx(a,b)(xa)+fy(a,b)(yb)L(x, y) = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b)

where:

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

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

The function f(x,y)=1433x2y2f(x, y) = \sqrt{143 - 3x^2 - y^2}, so its partial derivatives are:

  1. Partial derivative with respect to xx: fx(x,y)=x(1433x2y2)f_x(x, y) = \frac{\partial}{\partial x} \left( \sqrt{143 - 3x^2 - y^2} \right) Applying the chain rule: fx(x,y)=121433x2y2(6x)=6x21433x2y2f_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 yy: fy(x,y)=y(1433x2y2)f_y(x, y) = \frac{\partial}{\partial y} \left( \sqrt{143 - 3x^2 - y^2} \right) Applying the chain rule: fy(x,y)=121433x2y2(2y)=2y21433x2y2f_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 fx(a,b)f_x(a, b) and fy(a,b)f_y(a, b)

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

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

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

Step 3: Apply the linearization formula

The equation of the tangent plane (linearization) is:

L(x,y)=f(5,2)+fx(5,2)(x5)+fy(5,2)(y+2)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)(x5)+0.25(y+2)L(x, y) = 8.0 + (-1.875)(x - 5) + 0.25(y + 2)

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

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

L(5.1,2.1)=8.0+(1.875)(5.15)+0.25(2.1+2)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)=1433x2y2f(x, y) = \sqrt{143 - 3x^2 - y^2} at the point (5.1,2.1)(5.1, -2.1) is approximately 7.78757.7875, 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)(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)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.

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Math Problem Analysis

Mathematical Concepts

Multivariable Calculus
Partial Derivatives
Linear Approximation

Formulas

Tangent plane equation: L(x, y) = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b)
Partial derivatives: f_x(x, y) and f_y(x, y)

Theorems

Chain Rule
Linearization in Multivariable Calculus

Suitable Grade Level

Undergraduate Calculus