# C.4 Multivariable Calculus

C.4 Multivariable Calculus

In this review, we present a couple of the more important Multivariable Calculus methods commonly used in STAT 414, mainly for Exam 4 and the Final Exam. While this is not a complete review, you should use this to refresh your memory and guide you to where you need to spend time reviewing. As always, practice is key!

First, multivariable calculus involves functions of several variables. For simplicity, we focus on functions of two variables. You can find information on the web or in other text to review in more detail, if you need.

#### Partial Derivatives

Let's begin with Partial Derivatives. Suppose we have the function $f(x,y)$. The partial derivative with respect to x would be

$f_x(x,y)=\lim_{h\rightarrow 0} \frac{f(x-h, y)}{x-h}$

Similarly, the partial derivative of $f(x,y)$ with respect to $y$ would be

$f_y(x,y)=\lim_{h\rightarrow 0} \frac{f(x, y-h)}{y-h}$

The notation for partial derivatives is not the same for all texts. You should be able to recognize the different forms. The notation, for example, for the partial derivative of $f(x,y)$, with respect to $x$, could be denoted as:

$f_x(x,y)=\frac{\partial}{\partial x}f(x, y)=\frac{\partial f}{\partial x}$

Derivatives of Multivariable Functions Video, (Khan Academy)

#### Double Integrals

Integrating over regions will be important in STAT 414. Suppose we have the function $f(x,y)$, the over the region R, would be:

$\int \int_R f(x,y)\; dx dy$

Consider the rectangular region defined by $a\le x\le b$ and $c\le y\le d$, or $R=[a,b]\times[c, d]$. Then the iterated integral would be:

$\int_c^d \left[\int_a^b f(x,y) dx\right] dy=\int_a^b \left[\int_c^d f(x,y) dy\right] dx$

When the region is not rectangular, things can get complicated. It is important to draw out the support space and consider the region when building these double integrals.

Double Integrals Video, (Khan Academy)

## Example C.4.1

Consider the function $f(x,y)=\dfrac{1}{x^2}+6\sqrt{y}-xy$. Find both partial derivatives.

First, let's find the partial of x. To do this, we consider y as a constant.

$\dfrac{\partial f}{\partial x}=-\dfrac{2}{x^3}-y$

Now, let's find $\dfrac{\partial f}{\partial y}$.

$\dfrac{\partial f}{\partial y}=\dfrac{3}{\sqrt{y}}-x$

## Example C.4.2

Integrate $f(x,y) =24xy$. For $0< x < 1, 0 <y<1$ and $x+y < 1$ over the space where $x+y<\dfrac{1}{2}$.

\begin{align*}
\int_0^{1/2}\int_{0}^{1/2-y} 24xy\; dx \; dy &=\int_0^{1/2} 12y^3-12y^2+3y\; dy\\
& = 3y^4-4y^3+\dfrac{3}{2}y^2|_0^{1/2}\\&=3\left(\dfrac{1}{2}\right)^4-4\left(\dfrac{1}{2}\right)^3+\dfrac{3}{2}\left(\dfrac{1}{2}\right)^2\\
& = \dfrac{1}{16}
\end{align*}

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