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Function Notation and Mathematical Models

Figure 3.13.1 shows a 38 cm × 32 cm rectangular sheet of cardboard. You can fold the sheet into a box by cutting a square with side lengths *x* cm from each of the four corners. The animation in Figure 3.13.1 shows how to fold the sides to make a box.

**Figure 3.13.1. **Make a box by cutting square corners from a rectangular sheet of paper and folding the sides.

We want to define a function that gives the box’s volume in relation to the side length of the square cutouts.

- The volume of a rectangular box is
*area of base*times*height*. - The area of the box’s base is the box’s
*width*(the cardboard’s width) times the box’s*not**length*(the cardboard’s length).*not* - We can state each of these relationships as functions of the cutouts’ length conceptually before we worry about defining them computationally.

Let *V* be the function that gives the box’s volume in relation to a value of *x*.

Let *h* be the function that gives the box’s height in relation to a value of *x*.

Let *A* be the function that gives the area of the box’s base in relation to a value of *x*.

Let *w* be the function that gives the box’s width in relation to a value of *x*.

Let *l* be the function that gives the box’s length in relation to a value of * x*.

Now we can use Figure 3.13.1 to guide us in defining the functions *V*,
*h*, *A*, *w*, and *l*.

$V(x) = h(x) \cdot A(x)$ | The box’s volume in relation to a
value of x is its height in relation to x times the area of its
base in relation to the value of x. |

$h(x) = x$ | The box’s height in relation to a value
of x is equal to the length of the cutouts’ sides, which is x. |

$A(x) = l(x) \cdot w(x)$ | The area of the box’s base in relation to a value of x is the
base’s width in relation to the value of x times its length in
relation to the value of x. |

$w(x) = 32 - 2x$ | The box’s width in relation to a value of x is 32 cm minus
twice the cutouts’ width. |

$l(x) = 38 - 2x$ | The box’s length in relation to a value of x is 38 cm minus
twice the value of x. |

Notice that when defining *V* in relation to *x*, we started by referring to the functions *h* and *A*—even though we had not yet defined the functions *h* and *A* explicitly! We could do this because we had already defined *h* and *A* conceptually, and we knew that we would define them explicitly later in the list. When writing $V(x) = h(x) \cdot A(x)$, we *anticipated* that the definitions of *h* and *A* would produce the values we needed when given a value of *x*.

To envision the graph of $y = V(x)$, imagine that the cutouts’ side lengths varies. The smallest that the cutouts can be is 0 cm on each side. Notice that $V(0) = 0$, because $h(0) = 0$. That is, the box’s height will be 0 when the cutout’s sides are 0 cm.

The largest that the cutouts can be is 16 cm on each side. With a side length of 16 cm the four cutouts will use all the cardboard’s width, so $w(16)$, the box’s width when $x = 16$, would be 0. Therefore $A(16) = 0$, and hence $V(16) = 0$.

As the value of *x* varies, the dimensions of the box vary and therefore the box’s volume varies. Figure 3.13.2 shows the box’s height varying as the cutout sides’ length varies from 0 to 16. The box’s volume starts at 0, then increases to some maximum value, then decreases to 0.

Figure 3.13.3 shows a GC window containing definitions of *V*, *h*,
*A*, *w*, and *l*. It also shows a display of the graph of
*V* in relation to the value of *x* as it varies from 0 to 16.

- Here are the function definitions from Figure 3.13.3. Explain how GC computes $V(2.3)$. In your explanation, say what a value of each function represents about the box.

$V(x)=h(x)A(x)$

$h(x)=x$

$A(x)=l(x)w(x)$

$w(x)=32-2x$

$l(x)=38-2x$

$y=V(x),0<x<n$

- Explain how GC generates the animation in Figure 3.13.3 as the value of
*n*varies. - Enter the function definitions below into GC. Click "n" on the
*n*slider and change the domain of*n*to vary from 0 to 16. Then click the play button. Explain in terms of values of*x*and*n*how GC displays the same graph as in Figure 3.13.3. - In Exercise 3.13.3, why is it essential that we use the variable
*x*in the line $y = V(x), 0 < x < n$ for GC to display the graph that it does? (Your answer should refer to the conventions built into GC.) - The animation below shows two baseball players, one on 1st base (Player 1) and one on 2nd base (Player 2). Player 1 runs at 18 ft/sec. Player 2 runs at 26 ft/sec. The distance between bases is 90 ft. The runners leave their bases simultaneously. Assume that they accelerate immediately to their running speeds. Runner 2 stops at 3rd base the moment he reaches it. Runner 1 stops at 2nd base the moment he reaches it.
- Define a function
*B*that gives the distance between Player 1 and 2nd base as a function of the number of seconds since they began running. You will need to define your function in two parts (ctrl-shift-A), one part for Player 1’s distance from 2nd base while he is running and another part for the distance after he reaches 2nd base. - Define a function
*A*that gives the distance between Player 2 and 2nd base as a function of the number of seconds since they began running. You will need to define your function in two parts (ctrl-shift-A), one part for Player 2’s distance from 2nd base while he is running and another part for the distance after he reaches 3rd base. - Define a function
*D*that gives the distance between runners as a function of the number of seconds since they began running. - Graph your function
*D*in GC. (You will need to rescale your axes to see GC's displayed graph.) Use GC’s tracing feature (click and drag on the graph) to estimate the maximum distance and the minimum distance between runners, and the number of seconds since starting at which these distances happen. If you define your functions correctly, the minimum distance will be 73.9973 feet and the maximum distance will be 94.16403 feet. - This is a challenge problem. Assume that each player accelerates to top speed at the rate of 30 (ft/sec)/sec. Rework all of a-d with this assumption.
- Enter this function into GC: $$h(x,y) = \frac{x+y}{2} - \frac{\left| x-y \right|}{2}$$
- Modify the definition of
*h*in Exercise 6 to define a function*g*that gives the maximum of*x*and*y*.

$V(u)=h(u)A(u)$

$h(s)=s$

$A(r)=l(r)w(r)$

$w(h)=32-2h$

$l(k)=38-2k$

$y=V(x),0<x<n$

$h(s)=s$

$A(r)=l(r)w(r)$

$w(h)=32-2h$

$l(k)=38-2k$

$y=V(x),0<x<n$

What does it do? How does it do what it does?

*Hint*: $\dfrac {x + y}{2}$ gives the number half-way between the value of *x* and the value of *y*; $\left| x ‑ y\right|$ gives the distance between values of *x* and *y*.

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