Chapter 6.6, Problem 27PPS

a.

To determine

To calculate: the speed of the boat at 1,2, and 3 seconds.

Answer to Problem 27PPS

The speed of the boat at 1,2, and 3 seconds are $0.26\text{ feet per second},5.76\text{ feet per second},19.86\text{ feet per second}$ respectively.

Explanation of Solution

Given information:

The speed of the boat in feet per second is given by function $f(t)=-0.04t^4+0.8t^3+0.5t^2-t$ where t is the time in seconds.

Formula used:

Synthetic Substitution :- Synthetic division is used to find the value of a function.

Remainder Theorem:- If a polynomial $P(x)$ is divided by $x-r$ , the remainder s a constant $P(r)$ , and $P(x)=Q(x)(x-r)+P(r)$ ,

Where $Q(x)$ is a polynomial with degree one less than $P(x)$ .

Calculation:

Consider ,

  $f(t)=-0.04t^4+0.8t^3+0.5t^2-t$

By using Remainder Theorem, $f(1),f(2),f(3)$ should be the remainder when the polynomial is divided by $x-1,x-2,x-3$ respectively.

Therefore, we have

  $\begin{array}{l}\overline{)1}\begin{array}{cccc}-0.04& 0.8& \begin{array}{cc}0.5& \begin{array}{cc}-1& 0\end{array}\end{array}& \end{array}\\ \begin{array}{cccc}-0.04& 0.76& \text{ 1}\text{.26}& \text{ 0}\text{.26}\end{array}\\ \overline{\begin{array}{cccc}-0.04& 0.76& 1.26& \text{ 0}\text{.26}\end{array}\hspace{.2em}|}\text{ 0}\text{.26}\end{array}$

  $\begin{array}{l}\overline{)2}\begin{array}{cccc}-0.04& 0.8& \begin{array}{cc}0.5& \begin{array}{cc}-1& 0\end{array}\end{array}& \end{array}\\ \begin{array}{cccc}-0.08& 1.44& \text{ 3}\text{.88}& \text{ 5}\text{.76}\end{array}\\ \overline{\begin{array}{cccc}-0.04& 0.72& 1.94& \text{ 2}\text{.88}\end{array}\hspace{.2em}|}\text{ 5}\text{.76}\end{array}$

  $\begin{array}{l}\overline{)3}\begin{array}{cccc}-0.04& 0.8& \begin{array}{cc}0.5& \begin{array}{cc}-1& 0\end{array}\end{array}& \end{array}\\ \begin{array}{cccc}-0.12& 2.04& \text{ 7}\text{.62}& \text{ 19}\text{.86}\end{array}\\ \overline{\begin{array}{cccc}-0.04& 0.68& 2.54& \text{ 6}\text{.62}\end{array}\hspace{.2em}|}\text{ 19}\text{.86}\end{array}$

The remainder is $0.26,5.76,19.86$ respectively.

Therefore, by synthetic substitution, $f(1)=0.26;f(2)=5.76;f(3)=19.86$

Hence, the speed of the boat at 1,2, and 3 seconds are $0.26\text{ feet per second},5.76\text{ feet per second},19.86\text{ feet per second}$ respectively.

b.

To determine

To calculate: the speed of the boat at 6 seconds.

Answer to Problem 27PPS

The boat is travelling with $\text{132}\text{.96 feet per second}$ when passing the second buoys.

Explanation of Solution

Given information:

The speed of the boat in feet per second is given by function $f(t)=-0.04t^4+0.8t^3+0.5t^2-t$ where t is the time in seconds.

Formula used:

Synthetic Substitution :- Synthetic division is used to find the value of a function.

Remainder Theorem:- If a polynomial $P(x)$ is divided by $x-r$ , the remainder s a constant $P(r)$ , and $P(x)=Q(x)(x-r)+P(r)$ ,

Where $Q(x)$ is a polynomial with degree one less than $P(x)$ .

Calculation:

Consider ,

  $f(t)=-0.04t^4+0.8t^3+0.5t^2-t$

By using Remainder Theorem, $f(6)$ should be the remainder when the polynomial is divided by $x-6$ .

Therefore, we have

  $\begin{array}{l}\overline{)6}\begin{array}{cccc}-0.04& 0.8& \begin{array}{cc}0.5& \begin{array}{cc}-1& 0\end{array}\end{array}& \end{array}\\ \begin{array}{cccc}-0.24& 3.36& \text{ 23}\text{.16}& \text{ 132}\text{.96}\end{array}\\ \overline{\begin{array}{cccc}-0.04& 0.56& 3.86& \text{ 22}\text{.16}\end{array}\hspace{.2em}|}\text{ 132}\text{.96}\end{array}$

The remainder is $132.96$ .

Therefore, by synthetic substitution, $f(6)=132.96$

Hence, the speed of the boat at 6 seconds is $\text{132}\text{.96 feet per second}$ . This means the boat is travelling with $\text{132}\text{.96 feet per second}$ when passing the second buoys.