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Understanding Algebra for Calculus

Erich McAlister

Section 2.1 Examples of Algebraic Expressions

An algebraic expression is a mathematical expression formed using the normal operations of arithmetic (addition, subtraction, multiplication, division, and exponents for now) and one or more variables (unknown quantities for which we may substitute numbers). Let’s look at some basic algebraic expressions, describe them verbally, and evaluate them for some values of the variables.

Example 2.1.1.

Consider the algebraic expression
\begin{equation*} (3x+1)^2-2. \end{equation*}
This expression only contains one variable, \(x\text{.}\) We can substitute values for \(x\) and get values of the algebraic expression. Exaluate this expression when \(x=2\) and \(x=-2\text{.}\) Then describe what the expression does to \(x\) in the correct order.
Solution.
If we evaluate this expression at \(x=2\) we get
\begin{equation*} (3\cdot 2+1)^2 -2 = 7^2-2 = 47. \end{equation*}
If we evaluate this expression at \(x=-2\) we have
\begin{equation*} (3\cdot(-2) +1)^2 -2 = (-5)^2 -2 = 23, \end{equation*}
Evaluating algebraic expressions isn’t that hard provided you keep track of your order of operations. An important related skill is to read an algebraic expression and verbally express what is being done to the variable(s) in the correct order. In this expression we start with the value of \(x\) and
Step 1:
multiply \(x\) by three, (We’ll label this S1 for Step One and continue labeling steps this way.)
S2:
add one to the result of S1
S3:
square the result of S2, and finally
S4:
subtract two from the result of S3.
This idea of verbally describing algebraic expressions and understanding the steps that they represent is important for seeing algebraic expressions as something other than just a bunch of random symbols on a page. This task will become very powerful when solving equations and continue to be useful well into the study of Calculus (and beyond).

Question 2.1.3.

Write an algebraic expression that represents the following process applied to \(x\text{:}\)
S1:
Add one to \(x\text{,}\)
S2:
square the result of S1
S3:
subtract two from the result of S2, and finally
S4:
multiply the result of S3 by \(5\text{.}\)
Evaluate the expression you get at \(x=2\text{,}\) \(x=-2\text{,}\) and \(x=0\text{.}\)
We will often abbreviate the result of a step Sn by just writing Sn. For example, we could just write "multiply S3 by \(5\)" for the last step above.

Remark 2.1.4. (Try to) Make the Steps Undoable.

Eventually, our goal will be to solve equations invovling algebraic expressions. We will see that this often involves taking steps and undoing them in the correct (reverse) order. Sometimes this will mean we need to take some extra special care with out language for steps.
Take as an example the expression
\begin{equation*} \frac{1}{x}\text{.} \end{equation*}
This seems like a one-step expression described as
Step: Divide \(1\) by \(x\text{.}\)
This is an adequate description until we want to solve for \(x\) in an equation (we’ll do this in the next chapter). The problem with this description is that the action is being applied to \(1\) and not \(x\text{.}\) Thus, we cannot undo the action and get back to \(x\text{.}\) Instead, we introduce the phrase "take the reciprocal of" to mean divide \(1\) by some variable quantity. In this case, we would say
Step: Take the reciprocal of \(x\text{.}\)
Undoing taking the reciprocal of a quantity is simple; you just take the reciprocal again (this is the "flip" operation with fractions, if you want another word).

Checkpoint 2.1.5.

Verbally describe the expression
\begin{equation*} \frac{2}{x+1} \end{equation*}
by placing the steps in the correct order.
Hint.
Think of \(\frac{2}{x+1}\) as \(2\cdot\frac{1}{x+1}\text{.}\)
Sometimes a variable can appear more than once in an expression, which may make describing it as a single procedure more difficult. This difficulty is dealt with by thinking of a combination of procedures using order of operations.

Example 2.1.6.

Verbally describe the expression
\begin{equation*} 4s^2 + 2\sqrt{s+1}. \end{equation*}
Solution.
If we were evaluating this at a specific value of \(s\text{,}\) we would perform the addition of \(4s^2\) with last \(2\sqrt{s+1}\text{.}\) For this reason, we take each part that will be added together and describe them separately first. For the first term (a term is a part of an algebraic expression being added to other parts):
S1:
Square \(s\text{,}\) then
S2:
multiply S1 by four.
Then we pick up with the second term:
S3:
Add one to \(s\text{,}\)
S4:
take the square root of S3, and
S5:
multiply S4 by two.
The last step is the last thing according to our order of operations:
S6:
Add the result of S2 to the result of steps S5.

Question 2.1.7.

List the steps being performed on \(x\text{,}\) in order, for each of the following algebraic expressions:
  1. \(\frac{6}{2+x}\)
  2. \(2(x+1)^2 -5\)
  3. \(3(x+2)-5(x-7)\)
  4. \(\frac{3}{x} - 4x^2 -1\)
  5. \(\frac{1}{x-a}+\frac{3}{\sqrt{x+b}}\)
Let’s finish this section with examples of algebraic expressions that you might come across in another class or the real world.

Example 2.1.8.

An algebraic expression commonly encountered in a first year physics course gives the height \(h\) of a projectile, in meters, above the ground \(t\) seconds after it has been launched. This expression has parameters for initial height and initial upward velocity. The expression is
\begin{equation*} h = -4.9t^2+v_0 t + h_0, \end{equation*}
where \(v_0\) is the initial upward velocity in m/s and \(h_0\) is the initial height in meters. For instance, if we knew the initial height was \(40\) meters and the initial velocity was \(0\) m/s (if the object was dropped, not thrown), then we would have
\begin{equation*} h = -4.9t^2 + 40. \end{equation*}

Question 2.1.9.

Using the expression above, with \(h_0 = 40\) and \(v_0 = 0\text{,}\) for the height, answer the following questions to the best of your ability (you may need a calculator).
  1. How high is the object after \(2\) seconds? \(3\) seconds?
  2. When does the object hit the ground?
  3. Is the object ever \(50\) meters above the ground? Justify your answer in practical terms and algebraically, if possible.

Example 2.1.10.

The following table gives the men’s world record times for running events as of Summer 2017, as well as the average speeds in m/s.
Table 2.1.11.
\(d=\) Distance (m) \(T=\) time (s) \(S=\) speed (m/s)
100 9.58 10.44
200 19.19 10.42
400 43.03 9.3
800 100.91 7.93
1500 206 7.28
5000 757.35 6.6
Using techniques from statistics or linear algebra (data fitting), it is possible to give an algebraic expression involving the distance \(d\) that approximately tells you the record time \(T\text{.}\) The expression one finds this way is
\begin{equation*} T = .06640993d^{1.09689}. \end{equation*}
Since speed is distance divided by time, we can then formulate an expression for the speed \(S\) in terms of the distance \(d\text{:}\)
\begin{equation*} S = \frac{d}{T} = \frac{d}{.06640993 d^{1.09689}} = \frac{15.05798907}{d^{.09689}}. \end{equation*}

Question 2.1.12.

Use the expressions in the last example to answer the following questions.
  1. The expressions given above only give approximations. After all, if we could find record times with a formula we wouldn’t need to have races. Use a calculator to evaluate the expression given for \(T\text{,}\) for each distance, and see how well it matches the data.
  2. What does the expression for \(T\) say the record time for \(10000\) meters should be? Is it accurate? (Use the internet to check.)
  3. Two versions of the expression for \(S\) are shown. Use the rules for fractions and exponents from Chapter 1 to explain why these two versions are really the same.
  4. In the expression for \(S\text{,}\) what happens to \(S\) as \(d\) gets very large (recall 1.2.10)? Does this make practical sense?
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