Visualizing Elementary Calculus: Introduction

Recently I’ve been trying to be more geometrical when discussing elementary calculus with high school students. I don’t want to write an entire introduction to calculus, but the next few posts will outline some ways I think the geometric view can be helpful.

This series
I – Introduction
II – Trigonometry


You know about \Delta, which means “the change in”. For example, if w represents my weight, then -\Delta w represents the weight of the poop I just took.

Let’s say h is your height above sea level. \Delta h is the change in that height, but what change? The change when you climb the stairs? When you jump out of a plane? When you step on a banana peel?

When we think about change, we usually think about two things changing together. You get higher when you climb another stair on the staircase. h is changing, and so is s, the number of stairs climbed.

These two changes are related to each other. Say the stairs are 10 cm high. Then you gain 10 cm of height for each stair. We can write that as \Delta h = 10 {\rm cm} \hspace{.5em} \Delta s. We can also write it \Delta h / \Delta s = 10 \hspace{.5em}{\rm cm}. This says, “the height per stair is ten centimeters.”

This is the goal of calculus – to study the relationships between changing quantities. Let’s do a real example.

The Area of a Square

Let’s say we have a square whose sides lengths are x. Its area is x^2. What is the relationship between changes in its area and changes in the length of a side? Draw the square, then expand the sides some. The amount the sides have expanded is \Delta x. The new area that’s been added is \Delta (x^2).

We begin with the red square on the left, whose area is x^2. We add an extra amount Delta(x) to the sides, creating all the new green area.

From the picture we see

\Delta(x^2) = 2x\Delta x + (\Delta x)^2

This formula relates \Delta (x^2), the change in the area, to \Delta x, the change in the length of a side.

The Derivative of x^2

In the picture of the square, there is a little piece in the upper-right corner whose area is (\Delta x)^2. It is the smallest bit of area in the whole picture.

Look what happens when we make \Delta x even smaller.

We shrink Delta(x) and observe what happens to the different areas being added on.

In the first picture, \Delta x (no longer marked) is a quarter of x. (\Delta x)^2 is the dark green area, and it is one quarter as large as x \Delta x, the light green area. We see this because the dark patch fits inside the light one four times.

In the second picture, we shrink \Delta x to one eighth of x. All the green areas shrink, but the dark patch shrinks on two sides while the light patches shrink on only one. As a result, the dark (\Delta x)^2 is now only one eighth the size of the light x \Delta x.

If we continued to shrink \Delta x, this ratio would continue to decrease. Eventually we could tile the dark patch a million times into the light one. So, as long as \Delta x is very small, we can get a good estimate of the entire green area by ignoring the dark part (\Delta x)^2. Thus

\Delta(x^2) \approx 2x\Delta x

This approximation becomes better and better as \Delta x shrinks, becoming perfect as \Delta x becomes infinitesimally small.

When we want to indicate these infinitely small changes, we trade in the \Delta for a {\rm d} and write

\textrm{d}(x^2) = 2x \textrm{d}x

The terms \textrm{d}(x^2) and \textrm{d}x are called “differentials”. The equation expresses the relationship between two infinitely-small changes, one in x and one in x^2.

Frequently, we divide by \textrm{d}x on both sides to get

\frac{\textrm{d}(x^2)}{\textrm{d}x} = 2x

This is called “the derivative of x^2 with respect to x“.

Example 1: Estimating Squares

20^2 = 400. What is 21^2?

Here x = 20, and we’re looking at x^2. When x goes from 20 to 21, it changes by 1, so \textrm{d}x = 1. Our formula tells us

\textrm{d}(x^2) = 2x \textrm{d}x = 2*20*(1) = 40

Hence, x^2 increases by about 40, from 400 to 440.

The real value is 441. We got the change in x^2 wrong by about 2%. That’s because \textrm{d}x wasn’t infinitely small.

Let’s try again, this time estimating the square of 20.00458. Now \textrm{d}x = .00458, so

\textrm{d}(x^2) = 2 x \textrm{d}x = 2*20*.00458 = .1832

The estimate is 400.1832. The real value is 400.183221. We did much better, under-estimating the change by only 0.01% this time. Also, it was not much harder to do this problem than the last, but squaring out 20.00458 by hand would be a pain. We saved some work.

Example 2: How Far Is the Horizon?

The beach is a good place to think about calculus. If you look out at the ocean, the horizon appears perfectly flat. Nonetheless, we know the Earth is really curved. In fact, we can deduce the curvature of the Earth by standing on the beach and enlisting the help of a friend in a boat.

It works like this: You stand on the beach with your head two meters above the water. Your friend sails away until the boat begins to disappear from sight. The reason the bottom of the boat is disappearing is that it is hidden behind the curvature of Earth.

When the bottom of the boat disappears, measure the distance to some part of the boat you can still see. What’s the relationship between your height, the distance to the boat, and the radius of Earth?

A picture will help. We’ll call your height h and the distance to the horizon z.

You are the vertical stick on top, height h. The boat is the brown circle. It's at the horizon, a distance z away. The dotted line shows your line of sight. When the bottom of the boat begins disappearing, a right triangle forms.

Your height, the radius of Earth, and the distance to the horizon are related by the Pythagorean theorem to give

R^2 + z^2 = (R+h)^2

this is equivalent to

z^2 = 2Rh + h^2

As we have seen, if your height h is small compared to the size of the Earth (and it is), the term h^2 drops away and the distance to the horizon is

z = \sqrt{2Rh}

You can see about 5 {\rm km} at the beach, making the radius of Earth about 6,000 {\rm km}. (It’s actually 6378.1 {\rm km}).

Next we want to know how much further you can see if you stand on your tiptoes. That would be a small change \textrm{d}h to your height. It would let you see a small amount \textrm{d}z further. How is \textrm{d}h related to \textrm{d}z?

We already know

\textrm{d}(x^2) = 2x\textrm{d}x

So let x^2 = h, or x = \sqrt{h}, and we have

\textrm{d}h = 2\sqrt{h}\hspace{.3em}\textrm{d}(\sqrt{h})

But we also know

\sqrt{h} = \frac{z}{\sqrt{2R}}

so we can substitute that in to \textrm{d}(\sqrt{h}) and get

\textrm{d}h = 2\sqrt{h}\hspace{.3em}\textrm{d}\left(\frac{z}{\sqrt{2R}}\right)

or

\frac{\textrm{d}z}{\textrm{d}h} = \sqrt{\frac{R}{2h}}

This tells us how much further you can see if you get a little higher up. The interesting thing is it depends on h. The higher you go, the smaller \textrm{d}z. When you’re only two meters up, you get to see almost ten meters further out for every centimeter higher you go. However, if you’re 100m up on top a carousel, you get only 1 meter for each centimeter you rise.

It makes sense that the extra distance you see gets smaller and smaller the higher you go, and eventually shrinks down to zero. No matter how high you go, you can never see more than a quarter way around the globe.

(In reality, light bends due to refraction in the atmosphere, so you can sometimes see a bit further.)

Circles

Suppose we have a circle with radius r. It has a certain area (you undoubtedly know the formula already, but play along). Suppose we increase r by a small amount \textrm{d}r. What is the change \textrm{d}A in the area?

The original circle is dark blue with area A and radius R. The radius increases an amount dR, increasing the area by the light blue ring with area dA.

\textrm{d}A is the thin, light-blue ring. Imagine taking that ring and peeling it off the edge of the circle and laying it flat. We’d have a rectangle with width \textrm{d}R. Its length comes from the outside edge of the entire circle – the circumference. The circumference is 2 \pi R, so

\textrm{d}A = 2\pi R \textrm{d}R

We saw earlier that \textrm{d}(x^2) = 2x\textrm{d}x, so let x = R and we have

\textrm{d}A = \pi \textrm{d}(R^2)

Thus the quantities A and \pi R^2 change in exactly the same way. Since they also start out the same (both zero when R is zero), we have

A = \pi R^2

Next Post

We’ll look at trigonometry. Geometric arguments about the derivatives of trig functions are very simple ways of visualizing what’s going one, and are usually not introduced in a basic calculus course.

Exercises

  • Draw a cube with sides x and show that \textrm{d}(x^3) = 3x^2\textrm{d}x. Thus the derivative of x^3 with respect to x is 3x^2.
  • Draw a line with length x and show that \textrm{d}(x) = \textrm{d}x, which is of course algebraically obvious. Thus the derivative of x with respect to itself is 1.
  • Draw a rectangle with width w and length c*w and show that \textrm{d}(c*w^2) = 2cw\textrm{d}w = c\textrm{d}(w^2). Thus, whenever you have the differential of a variable multiplied by a constant, the constant can pop outside. Where was this property used implicitly in this post?
  • Now that you know \textrm{d}(x^3) = 3x^2\textrm{d}x, let x^3 = u and find the derivative of u^{1/3} with respect to u. (Answer: \frac{1}{3} u^{-2/3})
  • What is \textrm{d}(x^3)/\textrm{d}(x^2)? Let u = x^2 and find the derivative of u^{3/2} with respect to u. (Answer: \frac{3}{2}u^{1/2}).
  • Examine \textrm{d}(x^4) by letting u = x^2, so we’re looking at \textrm{d}(u^2). Find the derivative of x^4 with respect to x. (Answer: 4x^3)
  • Draw an equilateral triangle with sides of length s. Increase the sides a small amount \textrm{d}s and relate this to the change in area \textrm{d}A. Does this agree with our previous findings?
  • Draw an ellipse with a fixed with semi-major axis a and semi-minor axis b. Starting with a unit circle, argue by thinking about stretching that the area of the ellipse is \pi ab. Increase a by a small amount \textrm{d}a and increase b proportionately. This adds a small area \textrm{d}A to the ellipse. Show that this area is \pi(a^2+b^2)/b\hspace{.3em}\textrm{d}a. Does this let us find the circumference of the ellipse by the same thought process as we used for the circle? (Answer: no). Why not?
  • Draw a sphere with radius R. Use the relationship between \textrm{d}R and \textrm{d}A to find the volume of a sphere, given its surface area is 4\pi R^2. Check your answer against this post.

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2 Responses to “Visualizing Elementary Calculus: Introduction”

  1. Visualizing Elementary Calculus: Trigonometry « Arcsecond Says:

    […] Arcsecond « Visualizing Elementary Calculus: Introduction […]

  2. Visualizing Elementary Calculus: Differentiation Rules 1 « Arcsecond Says:

    […] Series I – Introduction II – Trigonometry III – Differentiation […]

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