Problem 27 - Calculus Early Transcendentals

Question

a. Find the linear approximation for the following functions at the given point. b. Use part (a) to estimate the given function value. $$f(x, y)=-x^{2}+2 y^{2} ;(3,-1) ; \text { estimate } f(3.1,-1.04)$$

Step-by-Step Solution

Answer

The estimated value of the function at the point $$(3.1, -1.04)$$ is approximately $$-7.44$$.

Step 1: Compute the partial derivatives of the function with respect to x and y.

To compute the partial derivatives, we will use the power rule for differentiation: $$ \frac{\partial f}{\partial x} = -2x, \quad \frac{\partial f}{\partial y} = 4y $$

Step 2: Evaluate the partial derivatives at the given point (3, -1).

Now plug in the given point into the derivatives we derived in Step 1: $$ \frac{\partial f}{\partial x}(3, -1) = -2(3) = -6, \quad \frac{\partial f}{\partial y}(3, -1) = 4(-1) = -4 $$

Step 3: Compute the tangent plane at the given point (3, -1).

Now we have the partial derivatives at the given point, we can find the equation of the tangent plane as follows: $$ L(x, y) = f(3, -1) + \frac{\partial f}{\partial x}(3, -1)(x - 3) + \frac{\partial f}{\partial y}(3, -1)(y + 1) $$ First, find the value of the function at the given point: $$ f(3, -1) = -3^2 + 2(-1)^2 = -9 + 2 = -7 $$ Now plug the values of the partial derivatives and the function at the given point into the formula for the tangent plane: $$ L(x, y) = -7 - 6(x - 3) - 4(y + 1) $$

Step 4: Estimate the value of the function at the given point (3.1, -1.04).

Use the tangent plane equation found in Step 3 to estimate the value of the given function at the point $$(3.1, -1.04)$$: $$ f(3.1, -1.04) \approx L(3.1, -1.04) = -7 - 6(3.1 - 3) - 4(-1.04 + 1) $$ Calculate the values: $$ L(3.1, -1.04) = -7 - 6(0.1) - 4(-0.04) = -7 - 0.6 + 0.16 $$ And we get the final approximation: $$ f(3.1, -1.04) \approx L(3.1, -1.04) = -7.44 $$ So, our estimate for the value of the function $$f(x, y)=-x^{2}+2 y^{2}$$ at the point $$\left(3.1, -1.04\right)$$ is approximately $$-7.44$$.

Key Concepts

Partial Derivatives

When dealing with multivariable functions, partial derivatives are a crucial concept. They help us understand how a function changes as we vary one of its variables, while fixing others.
For a function like \( f(x, y) \), which depends on variables \(x\) and \(y\), partial derivatives are written as \( \frac{\partial f}{\partial x} \) for the change along \(x\) and \( \frac{\partial f}{\partial y} \) for the change along \(y\).
To find them, imagine taking the derivative of \(f(x, y)\) with respect to one variable while treating the other as constant.

• The partial derivative \(\frac{\partial f}{\partial x}\) of our function \(-x^2 + 2y^2\) is \(-2x\). It shows how \(f\) changes as \(x\) changes.
• Similarly, \(\frac{\partial f}{\partial y}\) is \(4y\), which describes \(f\)'s rate of change with respect to \(y\).

Evaluating these partial derivatives at a specific point provides the slope of the function—for each direction—at that point, critical for creating the tangent plane.

Tangent Plane

The tangent plane is a linear surface that touches a multivariable function at a specific point, behaving much like a tangent line does for single-variable functions. It provides a great way to approximate the behavior of the function near that point.
To derive the equation of a tangent plane at a specific point, we use the function's value and its partial derivatives at that point:

• For our function, the tangent plane at point \((3, -1)\) is calculated using \( L(x, y) = f(3, -1) + \frac{\partial f}{\partial x}(3, -1)(x - 3) + \frac{\partial f}{\partial y}(3, -1)(y + 1) \).
• First, find \(f(3, -1) = -7\).
• Then, use the computed partial derivatives, \(-6\) and \(-4\), at point \((3, -1)\).
• Finally, plug these values into the tangent plane formula to get \(L(x, y) = -7 - 6(x - 3) - 4(y + 1)\).

This equation represents the plane that best approximates the original function at the vicinity of the given point.

Function Estimation

Function estimation, particularly for multivariable functions, involves using a simpler function to approximate the value of a more complex one at a specific point.
Linear approximation through the tangent plane is one common method. At a glance:

• The tangent plane at a particular point provides a straight-line approximation of the function's surface.
• By estimating \(f(3.1, -1.04)\) using the tangent plane \(L(x, y)\), we get an approximation of \(-7.44\).

This approximation works well for points in the geographic "neighborhood" of \((3, -1)\) and is particularly useful when calculating the exact value is difficult or unnecessary.

Multivariable Calculus

Multivariable calculus expands upon single-variable calculus by considering functions with more than one variable.
This field is crucial for understanding complex systems in physics, engineering, and economics.

• Core concepts include differentiating multivariable functions using partial derivatives to analyze how functions change in different dimensions.
• Tangent planes are an extension of tangent lines and provide insights into the local linear behavior of multivariable functions.
• Function estimation with linear approximations is a powerful tool to simplify complex analyses.

By grasping these concepts, students gain the ability to tackle intricate, real-world problems involving several inputs and outputs.