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By Wunderkind__Not (Wunderkind__Not) on Thursday, January 29, 2004 - 09:05 pm: Edit |

Okay, I need some help with this math problem I was assigned. It is hard to describe over this post but I will do it anyway.

@= spider

* = degrees

... = IGNORE them, it was the only way I could semi-draw the shapes. THEY SHOULDN"T BE THERE AND ARE NOT RELEVANT TO THE PROBLEM

This "right angled spider" can't go backwards. It shoots web only in right angles (or forming right angles).

(I will explain this since I cannot draw it) The spider starts down at the bottom and goes to the top as shown (200 ft.). At the top, the "right angeled spider" drops a web vertically down to the ground making a right angle at the lower right part of this triagnle formed.

.............@

............/

.........../

........../

........./

200ft./

......./

....../

.@ /_____________________________ (ground)

angle = 61*

..............@

............./| angle = 29*

............/.|

.........../..| (web dropped down)

........../...|

200ft../....|

......../.....|

......./...._|

...@/___|_|________________________(ground)

angle = 61*

From here, I will explain again though it may be confusing: If you can imagine, the next web the spider shoots is from the 90* angle (in pic) towards the 200ft line (hypotenus)--thus forming a 90* angle with the hypotenus and the new web line (the 90* angle box would be above the new web line if this helps you understand). From here, the spider will drop down a web vertically like she did in the beginning. This process is repeated over and over.

QUESTION: How far does the spider travel (in exact measurements or decimals (if it has one)) until it cannot go anymore?

By Transferkid (Transferkid) on Thursday, January 29, 2004 - 09:21 pm: Edit |

Ok, here goes:

The first leg, using trig, has length 200*Sin(61)

The second has length 200*sin61*cos61

The third has length 200*sin61*cos61*cos61

The fourth 200*sin61*cos61*cos61*cos61

And so on.

Basically, toobtain the length of each leg, you multiply the previous one by cos61. Therefore, each leg length is a term of a geometric sequence of common ratio cos61.

The sum is given by the formula:

Firstterm(1-(common ratio)^n)/(1-common ratio)

So total length is: 200sin61(1-(cos61)^n)/(1-cos61)

There. Now find the limit as n goes to infinity. Since 0<cos61<1, the limit of cos61^n is 0

Hence, total length is 200sin61/(1-cos61)

Done!

By Beero1000 (Beero1000) on Thursday, January 29, 2004 - 09:28 pm: Edit |

^^u forgot the original 200 ft

the first drop is 174.92ft and then every time after that to find the distance u multiply by cos61

sum=b(1-r^n)/(1-r)

b=174.92

r=cos61

n= infinity - i think since the spider wasnt given a size that it will keep going

so the sum of the series is

174.92(1-cos61^infinity)/1-cos61 =339.525

add the original 200

and my final answer is 539.525 ft

By Transferkid (Transferkid) on Thursday, January 29, 2004 - 09:34 pm: Edit |

Oh that's right!

Yeah, sorry about that

Thanks Beero

By Wunderkind__Not (Wunderkind__Not) on Thursday, January 29, 2004 - 10:04 pm: Edit |

Beero--what does "b" stand for?

By Wunderkind__Not (Wunderkind__Not) on Thursday, January 29, 2004 - 10:06 pm: Edit |

Beero--what does "b" stand for again? It has been so long, lol

By Transferkid (Transferkid) on Thursday, January 29, 2004 - 10:07 pm: Edit |

Beero used b for the first term of his geometric sequence. In this case, b=200sin61=174.92

By Beero1000 (Beero1000) on Thursday, January 29, 2004 - 10:11 pm: Edit |

yea thats the formula i learned

sum=b(1-r^n)/(1-r)

b=1rst number

r=common ratio

n=nth term

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