4.7: Derivatives in Space

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So far when we have thought about applying derivatives, we’ve always thought in terms of time. So we’ve focused on rates like meters per second, degrees Celsius per second, liters per minute, and so on, everything per unit time. However, we don’t always get derivatives with respect to time.

For example, consider this box:

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And let’s consider the temperature of the box at every point. We can show this using red to mean hot and blue to mean cold:

Note that the temperature changes, but not in time. Instead, the temperature changes as you move down the box — that is, the temperature changes in space. We can show the same thing in the graph:

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And just as before, the slope of this graph at any point is the derivative. What are the units? Well, they would be $\frac{\text{C$^\circ$}}{\text{cm}}$.

Let’s continue with this example.

Temp in a box

Suppose the equation $T(x) = -\frac{1}{16} x^2 + 100$ described the temperature, in Celsius, of the box from the previous diagrams at a distance of $x$ centimeters from the back.

What is the derivative of the temperature halfway down the box?
What is the derivative of the temperature three-quarters of the way down the box?
Let’s go back to thinking about time. According to the heat equation, a point in space will tend to get hotter if the second derivative in space is positive, and colder if the second derivative in space is negative. Based on this statement, is this box going to get hotter or colder? What does that mean regarding its derivative in time?

We can see from the previous graph that the box is $40$ centimeters long, so half way would be $x = 20$ cm. We know $T'(x) = -\frac{1}{16} (2x) = -\frac{1}{8}x$. So if we just plug in $x = 20$, we get
$T'(20) = -\frac{1}{8}(20) = -2.5$. Again, the units are $\frac{\text{C$^\circ$}}{\text{cm}}$. So that means for every cm you travel, the box gets $2.5^\circ$ C colder.
Now we just plug in $x = 30$ into the derivative we already found $T'(x) = -\frac{1}{8}x$, and we see $T'(30) = -\frac{1}{8}(30) = -3.75$. This means, at this point in the box, the box is getting colder faster. You can see this in the graph as well.
To answer this question, we need the second derivative $T''(x)$. This is just the derivative of $T'(x) = -\frac{1}{8}x$, and so we just drop the $x$ and get $T''(x) = -\frac{1}{8}$. Since the second derivative with respect to space is negative, the heat equation says the first derivative with respect to time is also negative. So if we just leave this box alone, it would tend to get colder. That is, the $T'(t)$ is negative as well.