Problem 26 - Calculus for Biology and Medicine

Question

Find the linear approximation of $$f(x, y)=\sin (x+2 y)$$ at \((0,0)\), and use it to approximate \(f(-0.1,0.2)\). Using a calculator, compare the approximation with the exact value of \(f(-0.1,0.2)\).

Step-by-Step Solution

Answer

The linear approximation at \((-0.1,0.2)\) is 0.3, which is close to the exact value \(\approx 0.2955\).

Step 1: Identify the function and point of approximation

We have the function \(f(x, y) = \sin(x + 2y)\) and we wish to find the linear approximation at the point \((0,0)\). This is known as the tangent plane at \((0,0)\).

Step 2: Compute the partial derivatives

Find the partial derivatives of \(f(x, y)\) with respect to \(x\) and \(y\). - \(f_x(x, y) = \frac{\partial}{\partial x} \sin(x + 2y) = \cos(x + 2y)\)- \(f_y(x, y) = \frac{\partial}{\partial y} \sin(x + 2y) = 2\cos(x + 2y)\).

Step 3: Evaluate the partial derivatives at the given point

Evaluate the partial derivatives at the point \((0,0)\): - \(f_x(0,0) = \cos(0 + 2 \cdot 0) = \cos(0) = 1\)- \(f_y(0,0) = 2 \cos(0 + 2 \cdot 0) = 2 \cdot 1 = 2\).

Step 4: Determine the linearization formula

The linear approximation (or linearization) of \(f\) at \((0,0)\) is given by:\[L(x, y) = f(0,0) + f_x(0,0)(x - 0) + f_y(0,0)(y - 0)\]Since \(f(0,0) = \sin(0) = 0\), we have:\[L(x, y) = 1x + 2y\].

Step 5: Use the linear approximation to estimate the function value

Substitute \(x = -0.1\) and \(y = 0.2\) into the linear approximation:\[L(-0.1, 0.2) = 1 \times (-0.1) + 2 \times 0.2\]Calculate:\[-0.1 + 0.4 = 0.3\].

Step 6: Calculate the exact value using the original function

Compute \(f(-0.1, 0.2)\) using the original function:\[f(-0.1, 0.2) = \sin(-0.1 + 2 \times 0.2) = \sin(0.3)\approx 0.2955\].

Step 7: Compare the approximation with the exact value

The linear approximation \(L(-0.1, 0.2) = 0.3\) is quite close to the exact value \(f(-0.1, 0.2) \approx 0.2955\). This shows that the linear approximation is a good estimate at points near \((0,0)\).

Key Concepts

Tangent Plane

A tangent plane is a flat surface that just "touches" the curve of a function at a specific point. It's like placing a sheet of paper on the curve so it only touches at that point with the same slope as the curve there.
Imagine a mountain slope: the tangent plane would look like a super tiny flat piece of board just resting exactly on the slope at a particular spot.
This is mostly useful for understanding the behavior of functions close to a particular point. In our example, we are interested in the tangent plane of the function \( f(x, y) = \sin(x + 2y) \) at the point \((0,0)\). The equation of the tangent plane here follows the formula for the linear approximation, which matches the concept of linearization. You're basically making a simpler, linear version of the function.
By calculating the partial derivatives, we ensure our tangent matches the curve exactly at the point of interest.

Partial Derivatives

Partial derivatives are like regular derivatives, but for multivariable functions. Just as derivatives tell us how a function changes as we change one variable, partial derivatives focus on how the function changes with each individual variable, while holding the others constant.
In the linear approximation example given, our function \(f(x, y) = \sin(x + 2y)\) depends on both \(x\) and \(y\).
For each variable, we compute a partial derivative:

• With respect to \(x\), noted as \(f_x(x, y)\), it shows how \(f\) changes as \(x\) changes, given by \(\cos(x + 2y)\).
• With respect to \(y\), noted as \(f_y(x, y)\), it represents how the function changes as \(y\) changes, given by \(2\cos(x + 2y)\). The factor of 2 arises because of the \(2y\) in the original sine function.

Evaluating these at a specific point, like \((0,0)\), gives the slope of our tangent plane, which is crucial for forming a linear approximation.

Function Evaluation

Function evaluation involves calculating the exact function value at a given point. It’s the process of substituting numbers into the function's variables to find a result.
For the function \(f(x, y) = \sin(x + 2y)\), we evaluate it at specific points to make accurate comparisons between the function's behavior and our linear approximation.
In our example:

• We first use the linear approximation by plugging in \(x = -0.1\) and \(y = 0.2\) into \(L(x, y) = 1x + 2y\), yielding \(0.3\).
• Then, we calculate the exact value \(f(-0.1, 0.2)\). By substituting into the original sine function, we get \(\sin(0.3) \approx 0.2955\).

This step not only allows us to see the accuracy of our approximation, but also the usefulness of linear approximations. Near the point \((0,0)\), our approximation is quite close to the actual value of the function. Understanding this process helps establish when such approximations can be trusted.