Word Problems – 3

This one is a little different. Recently, I have had a few students struggling with literal equations. These are equations with many letters or symbols. For science and engineering wannabes, you need to develop the skill to work with these. Below is an example from orbital dynamics, but first some background.

Two Body Problem

Accurately determining orbits in the real world, requires computers. However, a good approximation that makes orbital calculations possible by hand (OK, using calculators), is to assume that the only two bodies that exist in the universe are the bodies orbiting each other. The shape of an orbit of a body in orbit around another in this universe can have 1 of 4 shapes: circular, elliptical, parabolic, or hyperbolic. I won’t talk about the last three, but let’s consider the circular orbit.

Here is a picture of an circular orbit:

In this kind of orbit, the earth is at the centre and the satellite follows an circular path around the earth.

In an orbit, such as a satellite orbiting the earth, we want to know a lot of things about the orbit, but two primary things are the satellite’s distance from the earth and its position. To measure its position, an arbitrary axis is agreed upon and the satellite’s position is its angle 𝜃 measured from this axis. In orbital dynamics, the angle 𝜃 is called the true anomaly. I don’t know why it is called that, but a definition of anomaly is “a deviation from the normal”. If we consider the normal being on the reference axis, this term makes a bit more sense.

Now I chose a circular orbit because the equations describing it are simpler than for other shapes. For example, the time it takes for a body to complete one orbit, either circular or elliptical is

\[T=\frac{2\pi}{\mu^2}\left(\frac{h}{\sqrt{1-e^2}}\right)^3\ \ \ \ \ \ \ \ \ (1)\]

where

T = the period, the time it takes to complete one orbit

μ = gravitational parameter. It is a combination of the gravitation constant (a constant in the universe) and the masses of the bodies

h = angular momentum per unit mass. Since the body has mass and is rotating around the other body, it has angular momentum

e = the eccentricity of the orbit. A measure of how elliptical the orbit is.

The eccentricity of a circular orbit is 0 and this simplifies equation (1).

So now the word problem.

The Word Problem

Equation (1) applies to elliptical and circular orbits. For a circular orbit, e = 0. The radius of a circular orbit is r. There is a relationship between the angular momentum and the radius of a circular orbit:

\[r=\frac{h^2}{\mu}\ \ \ \ \ \ \ \ \left(2\right)\]

If a satellite is at 𝜃 = 0 at t = 0, the time it takes to travel 𝜃 radians is

\[t=\frac{\theta}{2\pi}T\ \ \ \ \ \ \ \ \left(3\right)\]

So using equations (1), (2), and (3), develop an expression for t in terms of r and 𝜃 and an expression for 𝜃 in terms of r and t. Remember that 𝜋 and μ are constants so they can be in these expressions.

The Solution

First, we can simplify equation (1) by substituting e = 0:

\[T=\frac{2\pi}{\mu^2}\left(\frac{h}{\sqrt{1-e^2}}\right)^3\Longrightarrow\frac{2\pi}{\mu^2}h^3\ \ \ \ \ \ \ \ \ \ \left(4\right)\]

We can rearrange equation (2) to get h in terms of r:

\[r=\frac{h^2}{\mu}\Longrightarrow h=\sqrt{r\mu}\ \ \ \ \ \ \ \ \ \ \ \ (5)\]

Now substitute h from equation (5) into equation (4):

\[T=\frac{2\pi}{\mu^2}h^3=\frac{2\pi}{\mu^2}\left(({r\mu)}^\frac{1}{2}\right)^3=\frac{2\pi r^\frac{3}{2}\mu^\frac{3}{2}}{\mu^2}=\frac{2{\pi r}^\frac{3}{2}}{\sqrt\mu}\ \ \ \ \ \ \ (6)\]

To follow the development in (6), you need to remember the exponent rules and how to convert between a fractional exponent and square root notation.

Now substitute this into equation (3):

\[t=\frac{\theta}{2\pi}T=\frac{\theta}{2\pi}\frac{2{\pi r}^\frac{3}{2}}{\sqrt\mu}=\frac{{\theta r}^\frac{3}{2}}{\sqrt\mu}\ \ \ \ \ \ \ \ \ \ \ (7)\]

which is one of the answers. Rearranging this to solve for 𝜃 gives the second answer:

\[t=\frac{{\theta r}^\frac{3}{2}}{\sqrt\mu}\Longrightarrow\theta=\frac{t\sqrt\mu}{r^\frac{3}{2}}\ \ \ \ \ \ \ \ \ \ \ (8)\]

Note that only basic algebra was used to find the answers. It doesn’t matter if you have equations with numbers or a lot of letters. The steps to find a solution are the same.

Word Problems – 1

A constant complaint from my students is that they struggle with word problems. So I thought I would produce several posts showing examples of how to approach and solve these problems.

Now every word problem is unique and different approaches can apply (hence, why these problems are scary). But there are techniques to help you through this. I have found that, when confronted with a word problem, it helps to do the following:

  1. Read through the entire problem.
  2. Draw a picture(s) if the problem lends itself to this, and label things with known values and use letters to label other things that you feel are useful to know.
  3. Read through the problem again and as you read, write equations that express mathematically what the words are saying. Use letters for unknown things. I find it helps to use letters that easily refer to the unknown thing. For example, use “C” to be an unknown price of a child’s ticket and “A” for the adult ticket. You can also use subscripts to differentiate unknowns. For example, use “tA” for the time to travel on path A and “tB” for the time to travel on path B.
  4. Identify the thing(s) that the problem is asking for, and look over the pictures and equations you have, to come up with a way to solve for the requested unknown(s).

Example

The digits of a 3-digit number add up to 13. The sum of the first two digits is 6. The number formed when the digits are reversed and 1 is subtracted is 3 times the original number. What is the original number?

I will get to this in a minute, but as you progress through maths, you will find that, more and more, you need knowledge from your past. This is an example of that. You may be tempted to use xyz to represent the unknown number, but that would be incorrect. In maths, xyz means x×y×z. If x = 1, y = 2, and z = 3, 123 does not equal 1×2×3 = 6.

From primary school (grade school in the USA), you learned that 123 means 1×100 + 2×10 + 3×1 because 1 is in the hundreds place, 2 is in the tens place, and 3 is in the units place. So let’s get to the problem.

So reading through the problem, you see that we need to find a 3-digit number. This problem does not lend itself to drawing pictures, but we can generate some equations. From the first sentence, we can write:

x + y + z = 13     (1)

From the second sentence, we can write:

x + y = 6     (2)

Now you may notice at this point, that we can use equation (2) in equation (1) as equation (1) has x + y in it and equation (2) says that we can replace this with 6:

x + y + z = 6 + z = 13 ⟹ z = 7

Wow! We haven’t finished going through the problem yet, but we have 1/3 of the problem solved. Now what about the third sentence? The numerical value of the number when the digits are reversed is:

100z + 10y + x

The third sentence says that if we subtract 1 from this, we get 3 times the value of the original number. A key word to look for in word problems is “is”. This means “=” in maths:

100z + 10y + x – 1 = 3(100x + 10y + z)     (3)

Since we now know that z = 7, this can be simplified to:

700 + 10y + x – 1 = 3(100x + 10y + 7)

⟹ 10y + x + 699 = 300x + 30y +21     (4)

Now we can get all the variables on the left side and the numbers on the right side, combine like terms to get:

-299x-20y = -678 ⟹ 299x+20y = 678     (5)

Equations (2) and (5) are a set of simultaneous equations in the variables x and y. You can manually use the substitution or elimination methods or use your CAS calculator to solve for x and y. I will assume that you can do these things to get x = 2 and y = 4. So the number that solves the problem is 247.

What is important here is to follow the process. Each word problem is unique but with practice, you will learn to look for key word and phrases that will help you to convert the word problem to a standard textbook problem. I will present more examples in future posts.