Math Insight

Vector-valued functions and the Jacobian matrix

Math 2241, Spring 2022
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Due date: March 16, 2022, 11:59 p.m.
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Total points: 1
1. A vector-valued function is a function that outputs a vector, i.e., it outputs more than one number. We'll focus on functions that have just two outputs.
1. The function $\vc{f}(x,y) = (x+y,x-y)$ has two outputs, $x+y$ and $x-y$ in response to two inputs $x$ and $y$. For example $\vc{f}(3,1) = (3+1,3-1) = (4,2)$. Calculate the following.

$\vc{f}(0,1) = ($
$+$
,
$-$
$)$
$= ($
,
$)$

$\vc{f}(5,2) =$

$\vc{f}(a,b) =$

2. A vector-valued function can be thought of a combination of multiple functions. We could view the above function $\vc{f}=(x+y,x-y)$ as two functions, one for each of its outputs: $f_1(x,y)=x+y$ and $f_2(x,y)=x-y$. Since each function is a component of the output vector, we call $f_1$ and $f_2$ the component functions of $\vc{f}$.

What are the component functions of $\vc{g}(x,y) = (x \ln{\left (x + y \right )},x^{2} - y)$?

$g_1(x,y) =$

$g_2(x,y) =$

3. What are the component functions of $\displaystyle \vc{h}(x,y) =\left ( 2 x^{2} e^{x y} - x, \quad \frac{x^{2} - y}{y - 3}\right )$?
$h_1=$

$h_2=$

What is $\vc{h}(1,0)$?

2. Let's explore how one can take derivatives of a vector-valued function.
1. Your typical function $f(x)=x^2$ has one derivative; it is $\diff{f}{x}=2x$. How many partial derivatives does the multivariable function $f(x,y)=x^{2} - y^{2}$ have?
What are they?

$\displaystyle\pdiff{f}{x}=$

$\displaystyle\pdiff{f}{y}=$

2. A vector-valued function of the form $\vc{f}(x,y) = \left ( x^{2} y, \quad x y^{2}\right )$ has how many component functions?
What are they?
$f_1(x,y)=$

$f_2(x,y)=$
3. Each of the two component functions of $\vc{f}(x,y) =\left ( x^{2} y, \quad x y^{2}\right )$ has how many partial derivatives?
This means that the vector-valued function $\vc{f}$ has
$\times$
=
partial derivatives.
4. What are the partial derivatives of the function $\vc{f}(x,y) = \left ( x^{2} y, \quad x y^{2}\right )$?

$\displaystyle \pdiff{f_1}{x} =$
, $\displaystyle \pdiff{f_1}{y} =$

$\displaystyle \pdiff{f_2}{x} =$
, $\displaystyle \pdiff{f_2}{y} =$

5. What are the partial derivatives of $\vc{g}(x,y) =\left ( x^{2} + y^{2}, \quad x - y\right )$?

$\displaystyle \pdiff{g_1}{x} =$
, $\displaystyle \pdiff{g_1}{y} =$

$\displaystyle \pdiff{g_2}{x} =$
, $\displaystyle \pdiff{g_2}{y} =$

3. A vector-valued function $\vc{f}(x,y)$ with two inputs and two outputs has four partial derivatives. We will organize these partial derivatives into a matrix.

A matrix is array of numbers or expressions. A $2 \times 2$ matrix has two rows and two columns. We could write a $2 \times 2$ matrix as \begin{gather*} \begin{pmatrix} 1 & 2\\ 3 & 4 \end{pmatrix} \quad \text{or} \quad \begin{bmatrix} 1 & 2\\ 3 & 4 \end{bmatrix}. \end{gather*}

We can write the four partial derivatives of $\vc{f}(x,y)$ into a $2 \times 2$ matrix. We call this matrix either the Jacobian matrix or the matrix of partial derivatives (two names for the same matrix).

1. Let $\vc{f}(x,y) = (x^{2} + y,3 y^{2})$. What are the four partial derivatives of $\vc{f}$?

$\displaystyle \pdiff{f_1}{x} =$
, $\displaystyle \pdiff{f_1}{y} =$

$\displaystyle \pdiff{f_2}{x} =$
, $\displaystyle \pdiff{f_2}{y} =$

2. When we form the matrix of partial derivatives of $\vc{f}$ (or Jacobian matrix), we write the derivatives in the same order as above. The matrix is $$D\vc{f}(x,y) = \begin{bmatrix}\textstyle\pdiff{f_1}{x} & \textstyle\pdiff{f_1}{y}\\ \textstyle\pdiff{f_2}{x} & \textstyle\pdiff{f_2}{y}\end{bmatrix}.$$ The first row contains the derivatives of the first component $f_1$; the second row contains the derivatives of $f_2$. The columns correspond to the variables, $x$ or $y$, that we differentiate with respect to.

Write the matrix of partial derivatives of $\vc{f}(x,y) = (x^2+y, 3y^2)$.

$D\vc{f}(x,y)=$
3. Calculate the Jacobian matrix of $\vc{g}(x,y) =(x^{2} y^{3},5 x - 3 y^{2} - 1)$.

$D\vc{g}(x,y)=$

4. Calculate the Jacobian matrix of the following functions. In some cases, you may need to use the product rule or chain rule to calculate the partial derivatives.
1. $\vc{f}(x,y) = \left ( e^{x} + e^{y}, \quad e^{x + y}\right )$

$D\vc{f}(x,y)=$
2. $\vc{g}(x,y) = \left ( x \ln{\left (y \right )}, \quad y \ln{\left (y \right )}\right )$

$D\vc{g}(x,y)=$
3. $\vc{h}(x,y) = \left ( \ln{\left (x y \right )}, \quad x e^{x y}\right )$

$D\vc{h}(x,y)=$