Word Problems – 2

I am writing posts of word problems to give students practice and underlying methods to solve these. My first post in this series provides some techniques to use to help you get to an answer. The most important thing is to not panic. Here is an example of a problem that could cause you to panic, but as you will see, calm logical thinking will get you to the answer.

A suspension bridge has been constructed over a river. The suspension cable is parabolic in nature. The distance between the two towers holding the cable is 180 metres and the minimum height of the cable above the road is 15 metres. At a point 40 metres from the vertex of the cable the height above the road is 30 metres.
How high above the road are the cables attached up the tower?

First, let me repeat from my last post, the things you should do to get started:

  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).

Hopefully, you have already done the first step. Now let’s do step 2:

This picture has all the information provided in the problem.

Looking at step 3, I do not see any equations we can write, but there is a clue in the problem. The shape of the suspension cable is a parabola and this is the shape of a quadratic equation. But before we can write down equations, we need a coordinate system.

Welcome to your first engineering lesson! When engineers are presented with a problem, they frequently must come up with a coordinate system. But they are free to place the system anywhere they want to. You can choose one that makes the equations difficult or choose one that makes the equations as simple as possible. So where do we choose ours? There is no right answer but some answers are better than others.

Here are some possibilities:

I think we can agree that the random coordinate system 1 would be a poor choice. But 2, 3, and 4 look good because the unknown value would just be the y coordinate over an appropriate value of x. Any of these three systems can be used to solve the problem, but if you review the turning point form of a quadratic equation, you should see that system 3 would generate the simplest equation of the parabola.

I will now redraw the figure using coordinate system 3 and interpreting the distances given in the problem as coordinates:

Figure 1

You may notice the two points (-40,30) and (40, 30). As a parabola is symmetric, the statement “At a point 40 metres from the vertex of the cable the height above the road is 30 metres” actually defines two points.

Now the turning point form of a quadratic equation is

y = a(xh)2+k

where (h,k) are the coordinates of the vertex. So from figure 1, we can immediately write the equation

y = ax2+15

Notice how simple this equation is because of our choice of coordinate system. We can find the value of a by substituting the point (40,30) into this equation:

30 = a(40)2+15 ⟹ a = 3/320

So the equation of the suspension cable is

y = 3x2/320 + 15

Now we can find the height of the cable support by letting x = 90:

y = 3(90)2/320 + 15 = 90.9375 m

So you see that this was not too difficult of a problem.

Transforming Quadratic (Parabolas) Graphs

Forms of Quadratic Equations

Quadratic equations come in several generic forms (or patterns) but they all have several things in common:

  1. The highest power of x (the independent variable) is 2 when all expressions are expanded in polynomial form.
  2. The other integer powers, 1 meaning just x, and 0 meaning a constant term, may or may not be present. But the squared term must be present.
  3. Other powers (negative integers, and non-integers), cannot be present.

The most common general form of a quadratic equation is \[y=ax^2 +bx+c,\] where the ab, and c are constants specific in a particular equation. For example, \[y=3x^2 -2x+7.\] Here a = 3, b = -2, and c = 7. There are other forms as well:\[\begin{array}{l}
y=k\left(x+a\right)\left(x+b\right),
y=a{\left(x-h\right)}^2 +k
\end{array},\]where the unspecified constants a, and k are specific to each form and are not the same numbers when converting between each of these forms. The most basic quadratic equation is \[y=x^2.\]Choosing various values of x, then squaring them to get the corresponding y values, and plotting these on a Cartesian coordinate grid, creates the following curve:

\[y=x^2\]

This shape is called a parabola. All quadratic equations have this shape when plotted but their position, orientation, and scale may be different. Each form of quadratic equations have their advantages. This lesson however, will concentrate on the form \[y=a{\left(x-h\right)}^2 +k.\]

The ‘a’ Factor

So we will be looking at the quadratic form \[y=a{\left(x-h\right)}^2 +k.\] This form is called the turning point form. Let’s start out simple and look at the effect of a alone by setting the h and k to zero. This leaves us with the equation \[y=ax^2.\] This coefficient in front of the x2 term scales and orients the parabola. If a is negative, all the y values are now negative. This flips (reflects) the parabola across the x-axis. If a is a large number, greater than 1, then the y values are larger for a given x than the y values in the basic y = x2 parabola. This has the effect of making the parabola sharper, that is, it is dilated along the x-axis. If a is a fraction between -1 and 1, then the y values increase more slowly. This has the effect of making the parabola flatter which is also a dilation along the x-axis. Below are several graphs of y = ax2 for various values of a. The basic parabola is shown (dashed curve) for comparison:

\[y=3x^2\]
\[y=\frac{1}{3}x^2\]
\[y=-3x^2\]
\[y=-\frac{1}{3}x^2\]

Notice how the negative sign flips the parabola across the x-axis.

The k Effect

Now let’s look at the equation \[y=ax^2 +k.\]I have just added a k to the previous equation form. If you add or subtract a constant number to an equation, it just raises or lowers the graph of the equation by k units. This is independent of the effect that a has on the curve. Below are examples for two choices of k using an a that was used above for comparison:

\[y=-\frac{1}{3}x^2+4\]
\[y=3x^2-4\]

The h Reaction

Now so far we have scaled and inverted our parabola and moved it up or down. What about moving it right or left? That’s what the h does in the form \[y=a{\left(x-h\right)}^2 +k.\] We get to this form by replacing x with x – h. Whenever this is done in any equation, as well as the quadratic equation, this moves the curve to the right h units if h  is positive or to the left if h is negative. But be careful. There is a negative sign in the form y=a(xh)2+k, so in y=a(x-3)2+k, h = 3 so this parabola is moved 3 units to the right. Whereas,  a(x+3)2+k can be thought of as a(x-(-3))2+k, so h is -3 and this moves the parabola 3 units to the left. The effect of h on the graph of a parabola is independent of the effects of a or k.

Below are two examples of the effect of h using the last example above for comparison:

\[y=3(x-1)^2-4\]
\[y=3(x+1)^2-4\]