Chapter 6.5, Problem 52AYU

A Rotating Beacon Suppose that a fire truck is parked in front of a building as shown in the figure.

The beacon light on top of the fire truck is located 10 feet from the wall and has a light on each side. If the beacon light rotates 1 revolution every 2 seconds, then a model for determining the distance $d$, in feet, that the beacon of light is from point A on the wall after $t$ seconds is given by

$d\left(t\right)\text{ = }\left|10\tan \left(\pi t\right)\right|$

(a) Graph $d\left(t\right)\text{ = }\left|10\tan \left(\pi t\right)\right|$ for $0\le t\le \text{ 2}$.

(b) For what values of $t$ is the function undefined? Explain what this means in terms of the beam of light on the wall.

(c) Fill in the following table.

(d) Compute $\frac{d\left(0.1\right)-d\left(0\right)}{0.1-0},\frac{d\left(0.2\right)-d\left(0.1\right)}{0.2-0.1}$, and so on, for each consecutive value of $t$. These are called first differences.

(e) Interpret the first differences found in part $\left(d\right)$. What is happening to the speed of the beam of light as $d$ increases?

To determine

To find:

a. Graph $d\left(t\right)=\left|10\tan \left(\pi t\right)\right|$ for $0\le t\le 2$.

Answer to Problem 52AYU

Solution:

a.

Explanation of Solution

Given:

The function $d\left(t\right)=\left|10\tan \left(\pi t\right)\right|$.

Calculation:

a. Graphing the given expression in either a graphing calculator, or using the method outlined in the solutions and using the vertical reflection operation needed to reflect the negative halves of the tangent into the positive side of the y-axis for the absolute value operation. We obtain the following graph.

To determine

To find:

b. For what values of t is the function undefined? Explain what this means in terms of the beam of light o the wall.

Answer to Problem 52AYU

Solution:

b. $t=\frac{\left(2k+1\right)}{2}$

Explanation of Solution

Given:

The function $d\left(t\right)=\left|10\tan \left(\pi t\right)\right|$.

Calculation:

b. since $\left|10\tan \left(\pi t\right)\right|=\left|10\frac{sin\left(\pi t\right)}{cos\left(\pi t\right)}\right|$, this function is undefined when $\cos \left(\pi t\right)=0\Rightarrow \pi t=\frac{\left(2k+1\right)\pi }{2}\Rightarrow t=\frac{\left(2k+1\right)}{2}$ for integer $k$. in the interval provided $0\le t\le 2$ the undefined values are going to be:

$t_0=\frac{\left(2.0+1\right)}{2}=\frac{1}{2}$

$t_1=\frac{\left(2.1+1\right)}{2}=\frac{3}{2}$

$t_2=\frac{\left(2.2+1\right)}{2}=\frac{5}{2}\ge 2$, from this list we choose the first two, since the third is outside the range given.

In terms of the light these times $t_0=\frac{1}{2}s=0.5s\text{, }{t}_1=\frac{3}{2}s=1.5s$ correspond to the times when the light beam is parallel to the wall, where it intersect the wall at infinity, which makes $d\approx \infty$ for either time point. Assuming the light is rotating counterclockwise in the figure given, and noting that for $t_A=0\Rightarrow d\left(t\right)=\left|10\tan \left(\pi .0\right)\right|=\left|10.0\right|=0$, which corresponds to time when the light is pointing straight at the wall at point $A$, then $t_0=0.5s$ corresponding to the time when the light is pointing directly in the front of truck parallel to the wall. Then one second after that (the time it takes that same beam to traverse the other half of the rotation)the beam is pointing directly behind the truck, parallel to the wall.

To determine

To find:

c. Fill in the following table.

$\text{T}$	$0$	$0.1$	$0.2$	$0.3$	$0.4$
$d\left(t\right)=10\tan \left(\pi t\right)$

Answer to Problem 52AYU

Solution:

c.

$\text{T}$	$0$	$0.1$	$0.2$	$0.3$	$0.4$
$d\left(t\right)=10\tan \left(\pi t\right)$	$0$	$3.249196962$	$7.2654252820$	$13.76381920$	$30.77683537$

Explanation of Solution

Given:

The function $d\left(t\right)=\left|10\tan \left(\pi t\right)\right|$.

Calculation:

c.

$\text{T}$	$0$	$0.1$	$0.2$	$0.3$	$0.4$
$d\left(t\right)=10\tan \left(\pi t\right)$	$0$	$3.249196962$	$7.2654252820$	$13.76381920$	$30.77683537$

To determine

To find:

d. Compute $\frac{d\left(0.1\right)-d\left(0\right)}{0.1-0}\text{, }\frac{d\left(0.2\right)-d\left(0.1\right)}{0.2-0.1}$ and so on, for each consecutive value of $t$.

Answer to Problem 52AYU

Solution:

d.

$\frac{d\left(0.1\right)-d\left(0\right)}{0.1-0}=\frac{10 \tan \left(\pi \mathrm{.0.1}\right)-10 \tan \left(\pi .0\right)}{0.1}\approx 32.49196962$

$\frac{d\left(0.2\right)-d\left(0.1\right)}{0.2-0.1}=\frac{10\tan \left(\pi \mathrm{.0.2}\right)-10 \tan \left(\pi \mathrm{.0.1}\right)}{0.1}\approx 40.16228317$

$\frac{d\left(0.3\right)-d\left(0.2\right)}{0.3-0.2}=\frac{10\tan \left(\pi \mathrm{.0.3}\right)-10 \tan \left(\pi \mathrm{.0.2}\right)}{0.1}\approx 64.98393924$

$\frac{d\left(0.4\right)-d\left(0.3\right)}{0.4-0.3}=\frac{10\tan \left(\pi \mathrm{.0.4}\right)-10\tan \left(\pi \mathrm{.0.3}\right)}{0.1}\approx 170.1301616$

Explanation of Solution

Given:

The function $d\left(t\right)=\left|10\tan \left(\pi t\right)\right|$.

Calculation:

d.

$\frac{d\left(0.1\right)-d\left(0\right)}{0.1-0}=\frac{10 \tan \left(\pi \mathrm{.0.1}\right)-10 \tan \left(\pi .0\right)}{0.1}\approx 32.49196962$

$\frac{d\left(0.2\right)-d\left(0.1\right)}{0.2-0.1}=\frac{10\tan \left(\pi \mathrm{.0.2}\right)-10 \tan \left(\pi \mathrm{.0.1}\right)}{0.1}\approx 40.16228317$

$\frac{d\left(0.3\right)-d\left(0.2\right)}{0.3-0.2}=\frac{10\tan \left(\pi \mathrm{.0.3}\right)-10 \tan \left(\pi \mathrm{.0.2}\right)}{0.1}\approx 64.98393924$

$\frac{d\left(0.4\right)-d\left(0.3\right)}{0.4-0.3}=\frac{10\tan \left(\pi \mathrm{.0.4}\right)-10\tan \left(\pi \mathrm{.0.3}\right)}{0.1}\approx 170.1301616$

To determine

To find:

e. Interpret the first differences found in part d. What is happening to the speed of the beam of light as $d$ increases?

Answer to Problem 52AYU

Solution:

e. The results in part d are like an approximation to the velocity of the light beam along width of the wall. We see that as the beam travels further away from point $A$, the of the beam increases, and is in fact nonlinear in its variation (since for equal increments in time $t=0.1\sec$ the differences in these first differences is not uniform). In fact as we approach the point $t=0.5s$ where the light detaches from the wall at infinity the velocity increases all the way to infinity, as told by the converging upon the vertical line of the asymptote.

Explanation of Solution

Given:

The function $d\left(t\right)=\left|10\tan \left(\pi t\right)\right|$.

Calculation:

e. The results in part d are like an approximation to the velocity of the light beam along width of the wall. We see that as the beam travels further away from point $A$, the of the beam increases, and is in fact nonlinear in its variation (since for equal increments in time $t=0.1\sec$ the differences in these first differences is not uniform). In fact as we approach the point $t=0.5s$ where the light detaches from the wall at infinity the velocity increases all the way to infinity, as told by the converging upon the vertical line of the asymptote.