Problem 25 - Calculus for Biology and Medicine

Question

Find the linear approximation of $$f(x, y)=e^{x+y}$$ at \((0,0)\), and use it to approximate \(f(0.1,0.05)\). Using a calculator, compare the approximation with the exact value of \(f(0.1,0.05)\).

Step-by-Step Solution

Answer

The approximation using linearization is 1.15, while the exact value is approximately 1.1618.

Step 1: Determine the linear approximation formula

To find the linear approximation of a function, we use the formula: \[ L(x, y) = f(a, b) + f_x(a, b)(x-a) + f_y(a, b)(y-b) \] where \((a, b)\) is the point at which we are approximating the function, and \(f_x\) and \(f_y\) are the partial derivatives of \(f\) with respect to \(x\) and \(y\), respectively.

Step 2: Calculate Partial Derivatives

Compute the partial derivatives of \(f(x, y) = e^{x+y}\). The partial derivative with respect to \(x\) is \(f_x(x, y) = \frac{\partial}{\partial x} e^{x+y} = e^{x+y}\). Similarly, the partial derivative with respect to \(y\) is \(f_y(x, y) = \frac{\partial}{\partial y} e^{x+y} = e^{x+y}\).

Step 3: Evaluate Function and Derivatives at the Point

Evaluate \(f(x, y), f_x(x, y),\) and \(f_y(x, y)\) at the point \((0, 0)\). We have \(f(0, 0) = e^{0+0} = 1\), \(f_x(0, 0) = e^{0+0} = 1\), and \(f_y(0, 0) = e^{0+0} = 1\).

Step 4: Write the Linear Approximation

Insert the evaluated values into the linear approximation formula to get: \[ L(x, y) = 1 + 1(x-0) + 1(y-0) = 1 + x + y \].

Step 5: Approximate the Function at (0.1, 0.05)

Use the linear approximation \(L(x, y) = 1 + x + y\) to approximate \(f(0.1, 0.05)\): \[ L(0.1, 0.05) = 1 + 0.1 + 0.05 = 1.15 \].

Step 6: Calculate the Exact Value

Find the exact value of \(f(0.1, 0.05)\) by calculating \(e^{0.1 + 0.05} = e^{0.15}\), which approximately equals 1.1618. Use a calculator for this step.

Step 7: Compare the Values

Compare the approximated value of 1.15 with the exact value of 1.1618. The linear approximation is close to the exact value, indicating that the approximation is reasonable for small changes near the point \((0, 0)\).

Key Concepts

Linear approximation

Linear approximation is a technique used to estimate the value of a function near a certain point using the tangent plane at that point. Think of it as simplifying a curve into a straight line that closely hugs the curve right around a chosen spot.

The formula used for the linear approximation of a function of two variables, say \(f(x, y)\), at a point \((a, b)\) is given by:

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

This equation forms a plane that touches the surface defined by the function at point \((a, b)\). The terms \(f_x\) and \(f_y\) represent the slopes of the function at that point concerning the respective axes.

In simpler terms, you're estimating what the function, which might be quite complicated, is doing near a point using straight lines to approximate those changes. It's much like saying, "Close to you, this smooth curve feels almost like a straight line." This method is mainly used when you're interested in small variations, making complex functions easier to handle practically. By using the linear approximation, we can understand how a small change in \(x\) and \(y\) will affect the function's value.

Partial derivatives

Partial derivatives are a cornerstone in calculus when dealing with functions of several variables, such as \(f(x, y)\). They measure how the function changes as one chosen variable is tweaked, while all other variables are kept constant. This makes them extremely useful for understanding multi-variable contexts.

For the function \(f(x, y) = e^{x+y}\), the partial derivative with respect to \(x\), denoted \(f_x(x, y)\), reveals how \(f\) changes as \(x\) changes while \(y\) stays fixed. Solving this gives you \(f_x(x, y) = e^{x+y}\), similarly to the function itself.

Similarly, the partial derivative concerning \(y\), which is \(f_y(x, y)\), does the same for variations in \(y\) alone, resulting in \(f_y(x, y) = e^{x+y}\) as well.

These derivatives not only tell us about the function's 'steepness' or slope in different directions but are also crucial for the linear approximation formula. The values of these derivatives at a specific point, like \((0, 0)\), indicate how rapidly the function is increasing or decreasing in response to changes in the \(x\) or \(y\). That's why they're so important: they unlock the mystery of how multifaceted variables work together in calculus.

Function evaluation

Function evaluation involves determining the output of a function for specific input values. This is a simple yet crucial skill in applying calculus concepts.

For instance, consider the function \(f(x, y) = e^{x+y}\). To evaluate this function at a point \((0, 0)\), you substitute the variables with these values:

• The solution is \(f(0, 0) = e^{0+0} = 1\).

This means that when both \(x\) and \(y\) are zero, the value of the function is 1. Once the function and derivatives are evaluated at a certain point, these values become components of the linear approximation.

To approximate \(f(0.1, 0.05)\), you use the linear formula:

• \(L(0.1, 0.05) = 1 + 0.1 + 0.05 = 1.15\).

The exact calculation of the function at these new inputs, \(f(0.1, 0.05) = e^{0.15}\), yields approximately 1.1618. You can often use a calculator here to find this exact value by directly computing the exponential function. By comparing our calculated approximation and the exact answer, you can see how close the approximation was, giving a sense of the accuracy of linear approximation in real-life computations.