Multivariable calculus and vector analysis

Introduction to partial derivatives

Ordinary derivatives in one-variable calculus

Your heating bill depends on the average temperature outside. If all other factors remain constant, then the heating bill will increase when temperatures drop. Let’s denote average temperature by $T$, and define a function $h:\mathbf{R}\to \mathbf{R}$ so that $h\left(T\right)$ gives the heating bill as a function of $T$.

We can then interpret the ordinary derivative (i.e., the derivative you learned about in first semester calculus) as indicating how much the heating bill will change as you change the temperature:

$$\begin{array}{lll}\hfill \frac{\mathrm{d}h}{\mathrm{d}T}\left(a\right)=\frac{\text{ change in }h}{\text{ change in }T}\left(\text{ at }T=a\right).& & \hfill \end{array}$$

If we graph $h$ as a function of $T$, then $\frac{\mathrm{d}h}{\mathrm{d}T}\left(a\right)$ gives the slope of the graph at the point where $T=a$. We say that $\frac{\mathrm{d}h}{\mathrm{d}T}$ is the derivative of $h$ with respect to $T$. If $T$ is given in degrees Celsius, then $\frac{\mathrm{d}h}{\mathrm{d}T}\left(a\right)$ is change in heating cost per degree Celsius of temperature increase when the temperature is $a$. Since $h$ decreases as $T$ increases, we expect $\frac{\mathrm{d}h}{\mathrm{d}T}$ to be negative. (The rate of change in heating cost per degree of Celsius of temperature decrease is positive. But this positive rate is equal to $-\frac{\mathrm{d}h}{\mathrm{d}T}$.)

Although I don’t know what $h\left(T\right)$ should really look like, pretend it looks like thick green curve graphed in the below CVT. The point $\left(a,h\left(a\right)\right)$ is shown in red, which you can change by dragging with your mouse. We can visualize the derivative by drawing a (thin blue) line tangent to the curve at the point $\left(a,h\left(a\right)\right)$. The slope of the line is equal to the slope of the graph when $T=a$; hence, the slope of the line is equal to the derivative $\frac{\mathrm{d}h}{\mathrm{d}T}\left(a\right)$.

Partial derivatives are analogous to ordinary derivatives

Clearly, writing the heating bill as a function of temperature is a gross oversimplification. The heating bill will depend on other factors, not least of which is the amount of insulation in your house, which we’ll denote by $I$. We can define a new function $h:{\mathbf{R}}^2\to \mathbf{R}$ so that $h\left(T,I\right)$ gives the heating bill as function of both temperature $T$ and insulation $I$.

Who knows, maybe this function would look something like the below graph.

Suppose you aren’t changing the amount of insulation in your house, so that we view $I$ as a fixed number. Then, if we look at how the heating bill changes as temperature changes, we’re back to our first case above. The only difference is that we now view $h$ as a function of both $T$ and $I$, and we are explicitly leaving one of the variables ($I$) constant. In this case, we call the change in $h$ the partial derivative of $h$ with respect to $T$, a term that reflects the fact some variables remain constant. We also change our notation by writing the $d$ as a $\partial$, so that

$$\begin{array}{lll}\hfill \frac{\partial h}{\partial T}\left(a,b\right)=\frac{\text{ change in }h}{\text{ change in }T}\left(\text{ at }T=a\text{ while holding }I\text{ constant at }b\right).& & \hfill \end{array}$$

If $T$ is given in degrees Celsius, then $\frac{\partial h}{\partial T}\left(a,b\right)$ is change in heating cost per degree Celsius of temperature increase when the outside temperature is $a$ and the amount of insulation is $b$.

Now, imagine you are considering the possibility of lowering your heating bill by installing additional insulation. To help you decide if it will be worth your money, you may want to know how much adding insulation will decrease the heating bill, assuming the temperature remains constant. In other words, you want to know the partial derivative of $h$ with respect to $I$:

$$\begin{array}{lll}\hfill \frac{\partial h}{\partial I}\left(a,b\right)=\frac{\text{ change in }h}{\text{ change in }I}\left(\text{ at }I=b\text{ while holding }T\text{ constant at }a\right).& & \hfill \end{array}$$

If $I$ is given in centimeters of insulation, then $\frac{\partial h}{\partial I}\left(a,b\right)$ is change in heating cost per added centimeter of insulation when the outside temperature is $a$ and the amount of insulation is $b$.

The partial derivative $\frac{\partial h}{\partial I}$ indicates how much effect additional insulation will have on the heating bill. Since additional insulation will presumably lower the heating bill, $\frac{\partial h}{\partial I}$ will be negative. If additional insulation will have a large effect, then $\frac{\partial h}{\partial I}$ will be a large, negative number. If, for your house, $\frac{\partial h}{\partial I}$ is large and negative, you may be inclined to add insulation to save money.

In the graph of $h\left(T,I\right)$, the partial derivatives can be viewed as the slopes of the graphs in the $T$ direction and in the $I$ direction. In the below CVT, the partial derivative $\frac{\partial h}{\partial T}$ corresponds to the slope of the dark blue line, and the partial derivative $\frac{\partial h}{\partial I}$ corresponds to the slope of the light green line. (The numerical values of these partial derivatives are also displayed. However, I wrote them like $dh∕dT$ because I couldn’t get a $\partial$ character to display in the graph.)

You can drag the red point around to change the values of $T$ and $I$ see how, for example, the partial derivative $\frac{\partial h}{\partial I}$ depends on both temperature and insulation. Consequently, your decision to add insulation will be affected by what temperatures you expect and how much insulation your home has already. You might expect that additional insulation will have a larger effect (i.e., $\frac{\partial h}{\partial I}$ will be larger negative number) for lower temperatures and smaller amounts of insulation. So if you live in Minnesota and have an old, poorly insulated house, it’s likely that $\frac{\partial h}{\partial I}$ will be a very large, negative number so that adding a moderate amount of insulation could dramatically decrease your heating bill.

As mentioned above, a heating bill depends on many more factors than temperature and insulation. We could define a function $h$ of many variables to give a more accurate estimate of heating costs. However, the math is the same no matter how many variables $h$ depends on, as long as it depends on two or more. So, if you want to include the effect of the size $S$ of a house, you can define $h\left(T,I,S\right)$ to be the heating bill as a function of temperature, insulation, and size. Then $\frac{\partial h}{\partial S}\left(T,I,S\right)$ would tell you how much the heating costs change as you change the size, leaving temperature and insulation constant. (Even though I cannot plot a graph of $h\left(T,I,S\right)$, I hope that the concept of the partial derivatives of $h$ still makes sense.) I imagine the value of $\frac{\partial h}{\partial S}$ would be useful only if you are planning to move since you probably don’t plan to cut a room off your house to save heating costs (though I suppose you could just not heat a room and effectively reduce the size).

I hope you stay warm this winter.

You can find examples of calculating partial derivatives by following this link.

Contents by topic

1. Geometry in higher dimensions

2. Differential Calculus

2a. The partial derivative

• Introduction to partial derivatives
• Partial derivatives by limit definition
• Partial derivative examples

2b. The derivative and differentiability

2c. The chain rule, directional derivative and gradient

2d. Applications of differential calculus

3. Integral calculus

4. Vector calculus

5. The fundamental theorems of vector calculus

6. Review material