Question

Help!!! Maybe a challenge for you

Answer 1

so,
from outer most dotted line to inner..
\[
b+\\
b\sin\theta+\\
b\sin^2\theta\\
\vdots\\
b\sin^n\theta
\]

Answer 2

yep thats the first answer...

Answer 3

i dont understand that why is it in series, just because it is approaching to zero

Answer 4

so,
tootal length=
\[L=b\sum_{n=0}^\infty \sin^n\theta\]
this is a geometric series with
\[a_0=b\\r=\sin\theta\]
so,
\[L=\frac{b}{1-\sin\theta}\]

Answer 5

nice how do we calculate sum of series

Answer 6

I just did!!!!
the last equation

Answer 7

oh so it is asking for formula...gotcha

Answer 8

you will need both of them
part (a) says to express it as infinite summation series
part (b) compute the solution
(c) apply limit as \(\theta\to\pi/2\)
(d) Justify the answer from part (c) using geometry

Answer 9

hmm. so for c is it\[L = b \sum_{n=0}^{\infty} \sin^n (\pi/2)\]

Answer 10

why apply limit to the complicated one?
apply to the evaluation

Answer 11

b/1-sin(pi/2) ???

Answer 12

how to justify answer using geometry

Answer 13

before you justify..
what was the limit you got?

Answer 14

zero

Answer 15

right?

Answer 16

@electrokid would you please help me solve the rest part

Answer 17

On putting the limit tends to pi/2 the sum becomes 1/0 or infinity.
This is bcoz if the angle becomes pi/2 the lines become parallel. So they wont intersect and there'll infinite dotted lines.

Answer 18

1/0 is 0 i think

Answer 19

No 1/0 is an indeterminate form and infinity...Think of it as u r dividing a finite non zero no. with a very very very small no.. The result will be a very big no. i.e. infinity

Answer 20

ok for parts c and d :
part c(method)
to show convergence first re-write \[sin^n(\theta) \]as a series disregarding the odd terms because ours is between 0 and pi/2
this yields:
\[sin^n(\theta)= \sum_{n=0}^{\infty}[\sum_{k=0}^{\infty} \frac{\theta^{1+2k}}{(1+2k)!} ]^n\]

now we know that the sin function between 0 and pi/2 is bounded let us call the sinx fn a_k

we know that \[0\le a_k\le1 for 0\le\theta\le\pi/2\]

it can easily be shown that the \[a_k^n \rightarrow0\] knowing this info, for theta equals pi/2 this series converges to zero.


@electrokid do you agree?

Answer 21

not that hard!!!!
The geometric series will only converge for \(|\sin \theta|<1\)
\[L=\lim_{\theta\to{\pi\over2}}\frac{b}{1-\sin\theta}\]
has a un-removable pole and is \(+\infty\).
i.e., think about what it will look like..
you will not have a triangle!!!! but a straight line segment

Answer 22

by definition, for a geometric series to converge, the common ratio must be less than "1"

Answer 23

yes sir

Answer 24

i still can't get ;ast part....i got limit as 0 for part c

Answer 25

i mea, in this case, parallel lines

Answer 26

last*

Answer 27

you will get
b/0

which is undefined

Answer 28

so once again what's b?

Answer 29

b is some random starting constant

Answer 30

@FibonacciChick666 hahaha. I do not think that what he/she meant. "part b" of the question....

Answer 31

oops lol