The Geometric Interpretation of the Derivative and the Integral

I have been trying to learn Calculus on and off for several years. In the last 3 months I re-read about Limits, Differentiation and Applications of Differentiation and finally, reached the point where I am starting to learn about Integration. It occurred to me that it might be interesting to do a blogpost showing the differences between the two from a geometric point of view.

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Derivative

Calculus has been called the mathematics of change. Without calculus, you can determine the slope of a line; that is, the difference in \(\boldsymbol{y}\) over the difference in \(\boldsymbol{x}\) between two points: \(\boldsymbol{\frac{\Delta y}{\Delta x}}\) or rise over run. In the following graph on the left, we have a line intercepting a parabola at two points: \(\boldsymbol{\left(0 , 1\right) \text{and} \left(3 , 10\right)}\). Using the formula for the slope gives us: \(\boldsymbol{m = \frac{10 – 1}{3 – 0} = \frac{9}{3} = 3}\). That’s all well and good. But what to do when we want to know the slope of a line tangent to a point on this parabola? Notice that the slope formula requires knowledge of two points. A line tangent to a curve touches the curve at only one point. Differential calculus comes to the rescue. The derivative of a function results in a new function that enables us to obtain the slope of a tangent line at any point in the original function. Not all functions are differentiable; there are certain constrains that must be met, dealing mainly with continuity.

The graph above right shows the same parabola along with several tangent lines whose slopes were obtain using differentiation. In the vast majority of cases the calculus involved is very easy. Where it gets sometimes complicated is because of the algebra required to ‘clean up’ after differentiating. In the above right example, something called the Power Rule was used: the derivative of \(\boldsymbol{x^{n} = nx^{n-1}}\). In our example the derivative of \(\boldsymbol{x^{2} + 1 = 2x + 0}\) (the derivative of a constant is zero). For instance, at \(\boldsymbol{x = 2 \text{ the slope is: } 2 \left(2\right) = 4 \text{ and at } x = -3 \text{ the slope is } 2 \left(-3\right) = -6\,}\) (refer to the tangent lines at \(\boldsymbol{x = 2 \text{ and } x = -3}\)).

Tangent lines to a curve are fine; but what else is differentiation good for? Well, remember that a graph is just a representation of a process and a tangent line shows not the average change in that process over time, but the change in that process at a specific point time, sometimes called the instantaneous change. For examples in the use of differentiation see: Finding the Minimum Distance – an Optimization Problem, and Finding the Cheapest Route, an Optimization Problem.

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Integral

The Integral, on the other hand, is often described as the area under the curve. I have recently started learning about the Integral, so I don’t have much to say about it, yet.

In the graph below left we have the linear equation: \(\boldsymbol{y = -\frac{9}{8}x + 9}\); the area bounded by the plot of the equation and the \(\boldsymbol{\text{x- and y-axes}}\) can be easily computed because it is a triangle. \(\boldsymbol{Area = \frac{1}{2}\left(\text{base}\right)\left(\text{height}\right)\text{, corresponding to: }Area = \frac{1}{2}\left(8\right)\left(9\right) = 36\, units^{2}}\). Similarly, we can easily compute the area of the trapezoid in the right graph bounded by the equation \(\boldsymbol{y = \frac{1}{2}x + 5 \text{ and the x-axis from } x = 3 \text{ to } x = 10.\; Area = \frac{1}{2}\left(6 + 10\right)\left(7\right) = 56\, units^{2}}\).

Computing the area under the curve is more of the problem in the following graph:

Ancient Greeks worked on this problem and came up with something called the method of exhaustion. Essentially they drew rectangles and then added their areas. In the left graph the upper-left corner of the rectangles touch the curve, while in the other graph it is the upper-right corner of the rectangles that touch the curve.

It is easy to see that the left graph underestimates the area under the curve while the right graph overestimates the area. What happens if we double the number of rectangles? In the following examples I will only use the upper-left corner method.

It is quite obvious that the more rectangles, the more accurate the result. But, what do rectangles have to do with the Integral? The Integral uses, well, an infinite number of rectangles, resulting not in an approximation, but the exact value.

\[Area = \int_{0}^{10} x^{2} \,dx = \frac{x^{3}}{3}\bigg|_0^{10} = \frac{10^{3}}{3}\, – \frac{0^{3}}{3} = \frac{1000}{3}\, – 0 = \frac{1000}{3}\, \left(\approx 333.33\right)\]

In answer to the possible question: if the function is \(\boldsymbol{y = x^{2}}\), how do we get from that to using \(\boldsymbol{\frac{x^{3}}{3}}\) in the above Integral? Remember that near the beginning of the post I mentioned the Power Rule; the derivative of \(\boldsymbol{x^{n} = nx^{n-1}}\). Using the same rule, the derivative of \(\boldsymbol{\frac{x^{3}}{3} \text{ is } x^{2}}\). Integration is, mostly, differentiation backwards; sometimes called anti-differentiation. In this case, the integral of \(\boldsymbol{x^{n} \text{ is } \frac{x^{n+1}}{n+1}}\). With this information in hand, I think the above integration should make sense.

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I would like to close this blogpost with a short example on the use of integration.

Average Velocity

An object moves on a straight track with a velocity in meters per second given by \(\boldsymbol{v(t) = 8\; sin\left(\pi t\right) + 2 t}\). Find its average velocity over the time interval from \(\boldsymbol{t = 0 \text{ to } t = 10 \text{ where } t \text{ is measured in seconds.}}\)

In this problem the object is moving forwards and backwards on a track. In the accompanying graph, the x-axis is the time in seconds; the y-axis is the distance in meters. According to the graph, by the end of the 10 seconds the object has advanced about 20 meters. However, that is not what the problem is about. While not advancing much, the object was moving much faster than 2 meters per second, sometimes advancing, sometimes retracing.

To determine the average velocity, we integrate the function over the interval \(\boldsymbol{ 0 \le x \le 10}\). In other words, the answer will be the area under the curve!

\begin{align*}
\text{Average velocity }\quad &= \frac{1}{10 – 0}\int_{0}^{10}\left(8\; sin\left(\pi t\right) + 2 t\right)\, dt \\
&= \frac{1}{10} \left(-\frac{8}{\pi} cos\left(\pi t\right) + t^{2}\right)\bigg|_0^{10} \\
&= \frac{1}{10}\left[\left(-\frac{8}{\pi} + 100\right) – \left(-\frac{8}{\pi}\right)\right] \\
&= \frac{100}{10}\, = 10\; m\ /\ s
\end{align*}