# Arch Length :)

• Sep 23rd 2010, 06:04 AM
CHARLIEDANCE
Arch Length :)
Attachment 19021

Hey guys I don't know if this would work but lets hope for the best. Okay i was given the follow properties of a function. The challenge was to find the formula of 2 continuous functions satisfy all three of the properties . any idea?
• Sep 23rd 2010, 06:16 AM
yeKciM
Quote:

Originally Posted by CHARLIEDANCE
Attachment 19021

Hey guys I don't know if this would work but lets hope for the best. Okay i was given the follow properties of a function. The challenge was to find the formula of 2 continuous functions satisfy all three of the properties . any idea?

first property is that your function is even meaning that $\displaystyle f(-x) = f(x)$
second (as seen on the image to) means that is above x-axis
third just say that surface under it is 1

does this helps you in any way ? what kind of functions can be even ?

Hint: for the first ....$\displaystyle f(x) = -x^2 + ?$ will give you that and second ?$\displaystyle f(x) = -x^? + ???$
• Sep 23rd 2010, 06:41 AM
Traveller
Quote:

Originally Posted by yeKciM
first property is that your function is even meaning that $\displaystyle f(-x) = f(x)$
second (as seen on the image to) means that is above x-axis
third just say that surface under it is 1

does this helps you in any way ? what kind of functions can be even ?

Hint: for the first ....$\displaystyle f(x) = -x^2 + ?$ will give you that and second ?$\displaystyle f(x) = -x^? + ???$

Does the function have to be even ? I think that several functions can be appropriately scaled and translated to satisfy the given properties, even though they might not be even.
• Sep 23rd 2010, 06:58 AM
yeKciM
Quote:

Originally Posted by Traveller
Does the function have to be even ? I think that several functions can be appropriately scaled and translated to satisfy the given properties, even though they might not be even.

try doing integration from -1 to 1 of the odd function :D

and if you look at the image attached you see what kind of functions are we talking about :D
• Sep 23rd 2010, 07:36 AM
Traveller
Quote:

Originally Posted by yeKciM
try doing integration from -1 to 1 of the odd function :D

and if you look at the image attached you see what kind of functions are we talking about :D

The images are only examples of some functions that have the given properties. That does not mean that every function that satisfies those properties will be even. And a non-even function might not be odd.

Here is a non-even function that probably satisfies the given properties, please check to be sure :D :

$\displaystyle \frac{4\pi}{16 - \pi } ( sin( \frac {\pi x}{4} + \frac {\pi}{4}) - \frac {x}{2} - \frac{1}{2} )$
• Sep 23rd 2010, 08:00 AM
yeKciM
Quote:

Originally Posted by Traveller
The images are only examples of some functions that have the given properties. That does not mean that every function that satisfies those properties will be even. And a non-even function might not be odd.

Here is a non-even function that probably satisfies the given properties, please check to be sure :D :

$\displaystyle \frac{4\pi}{16 - \pi } ( sin( \frac {\pi x}{4} + \frac {\pi}{4}) - \frac {x}{2} - \frac{1}{2} )$

check integral of that :D (not 1)
• Sep 23rd 2010, 08:12 AM
Soroban
Hello, CHARLIEDANCE!

Quote:

Attachment 19021

i was given the above properties of a function.
Find the formulas of 2 continuous functions that satisfy all three of the properties.

Assuming that the first graph is a down-opening parabola,
. . symmetric to the $\displaystyle y$-axis, we have: .$\displaystyle y \:=\:c - ax^2$

The $\displaystyle x$-intercepts are $\displaystyle (\pm1,0)$
. . So we have: .$\displaystyle c - a(\pm1)^2 \:=\:0 \quad\Rightarrow\quad c \:=\:a$

The function (so far) is: .$\displaystyle f(x) \:=\:a(1-x^2)$

The area on $\displaystyle [\text{-}1,1]$ is 1:

. . $\displaystyle \displaystyle A \;=\;2a\int^1_0(1-x^2)\,dx \;=\;1$

. . . . . . . $\displaystyle 2a\left(x - \frac{1}{3}x^3\right)\,\bigg]^1_0 \;=\;1$

. . . . . . . . . . . . . . $\displaystyle 2a\cdot\frac{2}{3} \;=\;1$

. . . . . . . . . . . . . . . . . $\displaystyle a \;=\;\frac{3}{4}$

Therefore: .$\displaystyle f(x) \;=\;\frac{3}{4}\left(1-x^2\right)$

Assuming that the second graph is a down-opening quartic,
. . symmetric to the $\displaystyle y$-axis, we have: .$\displaystyle y \:=\:c - ax^4$

The $\displaystyle x$-intercepts are $\displaystyle (\pm1,0)$
. . So we have: .$\displaystyle c - a(\pm1)^4 \:=\:0 \quad\Rightarrow\quad c \:=\:a$

The function (so far) is: .$\displaystyle g(x) \:=\:a(1-x^4)$

The area on $\displaystyle [\text{-}1,1]$ is 1:

. . $\displaystyle \displaystyle A \;=\;2a\int^1_0(1-x^4)\,dx \;=\;1$

. . . . . . . $\displaystyle 2a\left(x - \frac{1}{5}x^5\right)\,\bigg]^1_0 \;=\;1$

. . . . . . . . . . . . . . $\displaystyle 2a\cdot\frac{4}{5} \;=\;1$

. . . . . . . . . . . . . . . . . $\displaystyle a \;=\;\frac{5}{8}$

Therefore: .$\displaystyle g(x) \;=\;\frac{5}{8}\left(1-x^4\right)$

I'll let you show that the arc lengths are different . . .

• Sep 23rd 2010, 08:15 AM
Traveller
Quote:

Originally Posted by yeKciM
check integral of that :D (not 1)

Sorry, my mistake, but that is hardly a problem. :D

But the function can be multiplied with some number to adjust the area. Here is the corrected function :

$\displaystyle \frac{\pi}{4 - \pi } ( sin( \frac {\pi x}{4} + \frac {\pi}{4}) - \frac {x}{2} - \frac{1}{2} )$
• Sep 25th 2010, 05:42 AM
CHARLIEDANCE
Quote:

Originally Posted by Soroban
Hello, CHARLIEDANCE!

Assuming that the first graph is a down-opening parabola,
. . symmetric to the $\displaystyle y$-axis, we have: .$\displaystyle y \:=\:c - ax^2$

The $\displaystyle x$-intercepts are $\displaystyle (\pm1,0)$
. . So we have: .$\displaystyle c - a(\pm1)^2 \:=\:0 \quad\Rightarrow\quad c \:=\:a$

The function (so far) is: .$\displaystyle f(x) \:=\:a(1-x^2)$

The area on $\displaystyle [\text{-}1,1]$ is 1:

. . $\displaystyle \displaystyle A \;=\;2a\int^1_0(1-x^2)\,dx \;=\;1$

. . . . . . . $\displaystyle 2a\left(x - \frac{1}{3}x^3\right)\,\bigg]^1_0 \;=\;1$

. . . . . . . . . . . . . . $\displaystyle 2a\cdot\frac{2}{3} \;=\;1$

. . . . . . . . . . . . . . . . . $\displaystyle a \;=\;\frac{3}{4}$

Therefore: .$\displaystyle f(x) \;=\;\frac{3}{4}\left(1-x^2\right)$

Assuming that the second graph is a down-opening quartic,
. . symmetric to the $\displaystyle y$-axis, we have: .$\displaystyle y \:=\:c - ax^4$

The $\displaystyle x$-intercepts are $\displaystyle (\pm1,0)$
. . So we have: .$\displaystyle c - a(\pm1)^4 \:=\:0 \quad\Rightarrow\quad c \:=\:a$

The function (so far) is: .$\displaystyle g(x) \:=\:a(1-x^4)$

The area on $\displaystyle [\text{-}1,1]$ is 1:

. . $\displaystyle \displaystyle A \;=\;2a\int^1_0(1-x^4)\,dx \;=\;1$

. . . . . . . $\displaystyle 2a\left(x - \frac{1}{5}x^5\right)\,\bigg]^1_0 \;=\;1$

. . . . . . . . . . . . . . $\displaystyle 2a\cdot\frac{4}{5} \;=\;1$

. . . . . . . . . . . . . . . . . $\displaystyle a \;=\;\frac{5}{8}$

Therefore: .$\displaystyle g(x) \;=\;\frac{5}{8}\left(1-x^4\right)$

I'll let you show that the arc lengths are different . . .

Hey Guys thanks you for the help. thank you yeKciM and sorobaN for giving step by step it helped me really under stood it!

Soroban i have drawn the graphs on excel and they look good to me.(Rofl)

I have tried to show that the arc lengths are different to 4 decimal places by using :

Attachment 19058

By one of my other class mates said that i could also use the Trapezoidal rule of the Simpson's Rule and that i should try to get the smallest possible integral.

Is this true ? How would i go about that using the above formula? Be cause with what i have im not sure if i i can go even further or have the need to use the Simpson's Rule or Trapezoidal Rule.
• Sep 25th 2010, 12:26 PM
CHARLIEDANCE
my friend also said the the arc length formula should also be used anywhere possible and integration by parts could be is my friend being helpful or is he just confusing me more ><
• Sep 27th 2010, 04:17 AM
CHARLIEDANCE
When i try to intergrate the equations to the equations it doesnt work. any ideas anyone?