Chapter 7.2, Problem 68PS

Solution

a.

To calculate: Let n be the number of times the paper towel is wrapped around the dowel, let d,, be the diameter of the roll just before the nth wrap, and let 1, be the length of paper added in the nth wrap. Copy and complete the table.

A paper towel manufacturer sells paper towels rolled onto cardboard dowels. The thickness of the paper is 0.004 inch. The diameter of a dowel is 2 inches, and the total diameter of a roll is 5 inches.

  $\begin{array}{ccc}n& d_n& l_n\\ 1& 2& 2\pi \\ 2& ?& ?\\ 3& ?& ?\\ 4& ?& ?\end{array}$

  

The required table.

  $\begin{array}{ccc}n& d_n& l_n\\ 1& 2& 2\pi \\ 2& 2.0008& 2.0008\pi \\ 3& 2.0016& 2.0016\pi \\ 4& 2.0024& 2.0024\pi \end{array}$

Given information:

Dowel is 2 inches.

  

Explanation:

Consider the given sequence.

At the beginning the diameter of a dowel is 2 inches. When we roll the paper, after the first wrap around the dowel the diameter is equal to

  $\begin{array}{l}\text{dowel}=20.0004\\ \text{paper}=2\cdot 0.0004\\ 2+20.0004=2.0008\end{array}$

Note that you observe both sides of paper, left with thickness $0.0004$ and right, too.

After the second wrap it is equal to

  $\begin{array}{l}\text{after}\text{first}=2.0008\\ \text{paper=}2\cdot 0.0004\\ 2.0008+20.0004=2.0016\end{array}$

And after the third it is

  $2.0016+2\cdot 0.0004=2.0024$

Therefore, for the diameter $d_2$ , length $l_2$ will be circle.

Calculate circumference $=2.0008\cdot \pi$

Further,

Circle, calculate circumference $=2.0016\cdot \pi$

Also, circle, calculate circumference $=2.0024\cdot \pi$

Table is

  $\begin{array}{ccc}n& d_n& l_n\\ 1& 2& 2\pi \\ 2& 2.0008& 2.0008\pi \\ 3& 2.0016& 2.0016\pi \\ 4& 2.0024& 2.0024\pi \end{array}$

b

To identify: What kind of sequence is $l_{1,}{l}_{2,}{l}_{3,}{l}_{4,}\mathrm{.....}$ Write a rule for the nth term of the sequence.

The rule for the sequence $l_n=0.0008\pi n+1.9992\pi$

Given information:

The given table is:

  $\begin{array}{ccc}n& d_n& l_n\\ 1& 2& 2\pi \\ 2& 2.0008& 2.0008\pi \\ 3& 2.0016& 2.0016\pi \\ 4& 2.0024& 2.0024\pi \end{array}$

Explanation:

Consider the given sequence.

Let's observe the differences

  $\begin{array}{c}{l}_2-l_1=2.0008\pi –2\pi \\ 2.0008\pi –2\pi =0.0008\pi \\ l_3-l_2=2.0016\pi –2.0008\pi \\ 2.0016\pi –2.0008\pi =0.0016\pi \\ l_4-l_3=2.024\pi –2.0016\pi \\ 2.024\pi –2.0016\pi =0.0024\pi \end{array}$

The differences are constant; hence this sequence has common differences and the is an arithmetic sequence.

Its first term is $l_1= 2\pi$ , the common difference is $d\text{ =}0.0008\pi$ therefore the general term is $l_n=a_1+\left(n-1\right)d$

substitute the value of for $l_n\text{and}d$

  $\begin{array}{c}{l}_n=a_1+\left(n-1\right)d\\ l_n=2\pi +\left(n-1\right)0.0008\pi \\ l_n=2\pi +0.0008n\pi -0.0008\pi \\ l_n=0.0008n\pi +1.9992\pi \end{array}$

Therefore, the rule for the sequence $l_n=0.0008\pi n+1.9992\pi$

c.

To identify: The number of times the paper must be wrapped around the dowel to create a roll with a 5 inch diameter. Use your answer and the rule from part (b) to find the length of paper in a roll with a 5 inch diameter.

  

Explanation:

If $d_n=5$ inches, then

  $\begin{array}{l}{l}_n=d_n\cdot \pi \\ l_n=5\pi \end{array}$

Now, need to determine n such that

  $l_n=5\pi$

Therefore, the length with the diameter 5 inches is 5p inches. Apply the result from part (b)

Put the value of $l_n=5\pi$ into $l_n=0.0008\pi n+1.9992\pi$

  $\begin{array}{l}5\pi =0.0008\pi n+1.9992\pi \\ 5\pi -1.9992\pi =0.0008\pi n+1.9992\pi -1.9992\pi \\ 0.0008\pi n=3.0008\pi \\ n=\frac{3.0008\pi }{0.0008\pi }\\ n=3750\end{array}$

Therefore, you need to wrap 3751 times to create a roll with 5 inch diameter.

d.

To identify: How much would you expect to pay for a roll with a 7 inch diameter whose dowel also has a diameter of 2 inches. Interpret Suppose a roll with a 5 inch diameter costs $1.50.

  $\$3.10$ if the dowel costs $\$0.10$

Explanation:

Let's assume the cardboard dowel costs 10 cents. Therefore, the paper towel costs $1.40 for the paper part of the roll when the diameter is 5 inches. Now let $x$ be the cost for the paper part of the roll when the diameter is 7 inches. We can set up a proportion to solve for $x$ as follows:

  $\begin{array}{c}\frac{1.4}{{2.5}^2\pi -1^2\pi }=\frac{x}{{3.5}^2\pi -1^2\pi }\\ \frac{1.4}{5.25\pi }=\frac{x}{11.25\pi }\\ 5.25\pi x=15.75\pi \\ x=\frac{15.75\pi }{5.25\pi }\\ x=3\end{array}$

Notice that in the denominators of the above equation are the cross-sectional area of the paper part of the 5-inch diameter roll (the fraction on the left-hand side) and the cross sectional area of the paper part of the 7-inch diameter roll (the fraction on the right hand side). In both cases, so need to subtract $1^2\pi$ since that is the area of cross-sectional area of dowel (which is empty). Also notice that use 2.5, 3.5 and 1 as the radius of the 5-inch diameter roll, 7-inch diameter roll, and the 2-inch diameter dowel, respectively.

That $x$ is $3, but that is the paper part of the 7-inch diameter roll, plus 10 cents of the dowel. The total cost will be $3.10