I need URGENT Help!!!

**Lane** · July 9th 2006, 08:12 AM

two LORAN stations are positioned 232 miles apart along a straight shore. the stations trasnmit synchronized pulses which travel at the speed of light (186,000 miles per second) in order to find the coordinates of ships in the area. the station closest to a ship being tracked by the LORAN stations is called the master station.
A: a ship records a time difference of 0.000860 second between the LORAN signals. determine where the ship would reach shore if it were to follow the indicted byperbolic path.
B: if the ship were to enter a habor located between the two stations 32 miles from the master station, which time difference should it be looking for?
C: if the ship is 80 miles offshore when the desired time difference is obtained, what is the exact location of the ship?

You are beginning a small business, Barely Bears, that will manufacture plush toy bears. The cost of buying the intial sewing equipment is $425. additionally the material for each plush bear cost $1.75. the average cost per bear is C=1.75x+425/x where x represents the number of bears the business has produced. use a graphing utility to sketch a graph of the average cost equation and then find the the cost to manufacture 30 toy bears. What happens to the average cost if you produce 300 toy bears? what does the limiting cost per bear appear to be?

**galactus** · July 9th 2006, 02:45 PM

For #1.3

The stations are sending out signals at the speed of light or 982 $ft/{\mu}sec$

The difference in their signals is 860 $ft/{\mu}sec$

Let the stations be at some points on the x-axis: (116,0) and (-116,0) and the ship be on the line y=80.

Since it took 860 microseconds longer for the signal to arrive from one station than the other, say station A and station B, the difference in their distances( $d_{1}-d_{2}$ ) to the ship at that time is (982)(860)=844520 ft.

$d_{1}-d_{2}=844520/5280=159.95 miles$ .

Using the hyperbola equation:

$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$

In the derivation of the hyperbola equation, one can let $d_{1}-d_{2}=2a$ . Do that here.

If we let the difference in our distances equal 2a, then

a=159.95/2=79.97 miles. So, $a^{2}=6395.76$

Since the distance from the origin to either focus is 116,

$b^{2}=c^{2}-a^{2}\approx{116^{2}-6395.76^{2}}\approx{7060.24}$

Therefore, an equation for the hyperbola is:

$\frac{x^{2}}{6396}-\frac{y^{2}}{7060}=1$

Setting y=80 and solving for x, we find x=110.43

So, the coordinates of the ship are approximately (110.43,80).

I hope I didn't go off on a tangent somewhere. Get it?.Tangent?. Get it?.

**Soroban** · July 9th 2006, 03:09 PM

Hello, Lane!

Here's the second one . . . I had to read it over several times.
Some information is unnecessary and confusing.
Also your cost function needs parentheses: C = (1.75x + 425)/x

You are beginning a small business, Barely Bears, that will manufacture plush toy bears.
The average cost per bear is: $C(x) = \frac{1.75x + 425}{x}$ , where $x$ is the number of bears produced.

(a) Use a graphing utility to sketch a graph of the average cost equation.
(b) Then find the the cost to manufacture 30 toy bears.
(c) What happens to the average cost if you produce 300 toy bears?
(d) What does the limiting cost per bear appear to be?

They gave us the cost function: . $C \;= \;\frac{1.75x + 425}{x}$

$(b)\;C(30)\;=\;\frac{1.75(30) + 425}{30} \;\approx\;\$15.82$

$(c)\;C(300)\;=\;\frac{1.75(300)+425}{300}\;\approx \;\$3.17$

$(d)$ Consider $\lim_{x\to\infty} C(x)$

We have: . $\lim_{x\to\infty}\left(\frac{1.75x + 425}{x}\right) \;=\;\lim_{x\to\infty}\left(1.75 + \frac{425}{x}\right) \;=$ $1.75 + 0 \;= \;1.75$

So here's the Big Picture . . .

As we produce more bears, the cost per bear decreases.

This is great news for us.
Assuming that our production increases because of increased demand
(meaning that our bears are very popular), we can charge less.
This will probably make them more popular, increasing sales even more.
We're golden!

However, there is a limit to the decreasing cost.
It will bottom-out at $1.75 per bear (theoretically).

**Quick** · July 9th 2006, 05:08 PM

Originally Posted by Lane

A: a ship records a time difference of 0.000860 second between the LORAN signals. determine where the ship would reach shore if it were to follow the indicted byperbolic path.

What, may I ask, is a byperbolic path? is it like a line that reaches that would reach the middle if the ship was right in-between the two stations?

**ThePerfectHacker** · July 9th 2006, 05:14 PM

Originally Posted by Quick

What, may I ask, is a byperbolic path?

He meant to say, hyperbolic because the "b" is next to the "h" on the QWERTY keyboard.

**Soroban** · July 9th 2006, 05:57 PM

Hello, Quick!

What, may I ask, is a byperbolic path?

As the name implies, it is a path along a hyperbola.

Recall one of the definitions of a hyperbola: the locus of a point such that
the difference of its distances from two fixed point (foci) is a constant.

Suppose the time difference from the two signals is 0.0005 second.
Then one station is: $0.0005 \times 186,000 \,= \, 93$ miles further away.

You can set this up on a graph.
Place one station at $A(116,0)$ , the other at $B(-116,0).$

The ship is at $P(x,y)$ where: . $\overline{PA} - \overline{PB}\:=\:93$

Then: . $\sqrt{(x-116)^2+y^2} - \sqrt{(x+116)^2+y^2}\;=\;93$

Simplify . . . and you get the equation of a hyperbola.
(I'll wait in the car.)

**Quick** · July 9th 2006, 06:04 PM

Originally Posted by Lane

You are beginning a small business, Barely Bears, that will manufacture plush toy bears. The cost of buying the intial sewing equipment is $425. additionally the material for each plush bear cost $1.75. the average cost per bear is C=1.75x+425/x where x represents the number of bears the business has produced. use a graphing utility to sketch a graph of the average cost equation and then find the the cost to manufacture 30 toy bears. What happens to the average cost if you produce 300 toy bears? what does the limiting cost per bear appear to be?

just a side note to Soroban, is $\frac{x}{\infty}=0$

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