Problem 42 - Calculus for Biology and Medicine

Question

Find a linear approximation to each func\mathrm{tion } \(f(x, y)\) at the indicated point. \(\mathbf{f}(x, y)=\left[\begin{array}{c}(x+y)^{2} \\ x y\end{array}\right]\) at \((-1,1)\)

Step-by-Step Solution

Answer

The linear approximation is \( \left[ \begin{array}{c} 0 \\ x - y + 1 \end{array} \right] \).

Step 1: Identify functions in vector form

We have two functions derived from the vector-valued function: \ \(f_1(x, y) = (x + y)^2\) and \(f_2(x, y) = xy\). We need approximations for each at the point \((-1, 1)\).

Step 2: Find partial derivatives for f1

The partial derivative of \(f_1\) with respect to \(x\) is \(\frac{\partial}{\partial x}(x+y)^2 = 2(x+y)\). \ The partial derivative of \(f_1\) with respect to \(y\) is \(\frac{\partial}{\partial y}(x+y)^2 = 2(x+y)\).

Step 3: Evaluate partial derivatives of f1 at the point (-1, 1)

Substitute \((-1, 1)\) into the partials: \ \(\frac{\partial f_1}{\partial x}\bigg|_{(-1, 1)} = 2(-1+1) = 0\) \ \(\frac{\partial f_1}{\partial y}\bigg|_{(-1, 1)} = 2(-1+1) = 0\).

Step 4: Linear equation for f1

The linear approximation for \(f_1\) is given by: \ \(L_1(x, y) = f_1(-1, 1) + \frac{\partial f_1}{\partial x}(-1, 1)(x + 1) + \frac{\partial f_1}{\partial y}(-1, 1)(y - 1)\). \ Since the derivative values are zero, \(L_1(x, y) = f_1(-1, 1) = (-1+1)^2 = 0\).

Step 5: Find partial derivatives for f2

The partial derivative of \(f_2\) with respect to \(x\) is simply \(\frac{\partial}{\partial x}(xy) = y\). \ The partial derivative of \(f_2\) with respect to \(y\) is \(\frac{\partial}{\partial y}(xy) = x\).

Step 6: Evaluate partial derivatives of f2 at the point (-1, 1)

Substitute \((-1, 1)\) into the partials: \ \(\frac{\partial f_2}{\partial x}\bigg|_{(-1, 1)} = 1\) \ \(\frac{\partial f_2}{\partial y}\bigg|_{(-1, 1)} = -1\).

Step 7: Linear equation for f2

The linear approximation for \(f_2\) is given by: \ \(L_2(x, y) = f_2(-1, 1) + \frac{\partial f_2}{\partial x}(-1, 1)(x + 1) + \frac{\partial f_2}{\partial y}(-1, 1)(y - 1)\). \ Calculate \(f_2(-1, 1) = (-1)(1) = -1\), so \ \(L_2(x, y) = -1 + 1(x + 1) - 1(y - 1)\).

Step 8: Simplify the linear equation for f2

Simplify \(L_2\): \ \(L_2(x, y) = -1 + x + 1 - y + 1 = x - y + 1\).

Step 9: Combine approximations

Combine the approximations for \(f_1\) and \(f_2\) into a vector: \ \(L(x, y) = \left[ \begin{array}{c} 0 \ x - y + 1 \end{array} \right]\).

Key Concepts

Vector-Valued Function

Vector-valued functions are mathematical expressions that produce vectors as output, instead of just a single scalar value. This means that when you input certain values, the output is a vector. Understanding vector-valued functions is crucial for grasping complex systems as they often describe phenomena in physics and engineering.

In our example, the vector-valued function is given by:

• \[ \mathbf{f}(x, y) = \left[ (x + y)^2, \ xy \right] \]

This function consists of two separate functions: - \(f_1(x, y) = (x + y)^2\)- \(f_2(x, y) = xy\)
These components represent how inputs \(x\) and \(y\) are transformed into outputs. We need to approximate each component at the point \((-1, 1)\).

The linear approximation becomes very useful in simplifying complex vector outputs near a particular point, making the solutions more manageable and interpretable.

Partial Derivatives

Partial derivatives are fundamental in calculus and are essential when dealing with functions of multiple variables. They show how a function changes as one of the variables change, keeping the other variables constant.

Let's break down the task of finding partial derivatives for our functions:

• For \(f_1(x, y) = (x + y)^2\), taking the partial derivative with respect to \(x\) and \(y\) yields: - \(\frac{\partial f_1}{\partial x} = 2(x+y)\) - \(\frac{\partial f_1}{\partial y} = 2(x+y)\)


• For \(f_2(x, y) = xy\), the partial derivatives are: - \(\frac{\partial f_2}{\partial x} = y\) - \(\frac{\partial f_2}{\partial y} = x\)

These derivatives provide information on how the functions change at any point based purely on individual variable adjustments. Mastery of partial derivatives enhances understanding of function behavior in multidimensional spaces.

Evaluation at a Point

Evaluating a function or its derivatives at a particular point is an essential skill in calculus, helping to calculate specific values that can aid in real-life applications. For linear approximation, this evaluation allows you to produce a local linear model of the function.

In our task, we evaluated the partial derivatives at point \((-1, 1)\):

• - For \(f_1\), both derivatives, \(\frac{\partial f_1}{\partial x}\) and \(\frac{\partial f_1}{\partial y}\), contribute as zero at \((-1, 1)\), showing that near this point, the slope is flat.
• - For \(f_2\), evaluating the derivatives: - \(\frac{\partial f_2}{\partial x}\) becomes \(1\) - \(\frac{\partial f_2}{\partial y}\) turns into \(-1\)

Finally, we combined the linear approximations of \(f_1\) and \(f_2\) at \((-1, 1)\) yielding \(L(x, y) = [0, x - y + 1]\). This linear model is a simple but powerful tool to predict behavior closely around the point of interest, making complex functions easier to handle.