Question

Calculus
Find the length of the parabolic segment
r=14/(1+cos(theta)), 0<=theta<=pi/2

Answer 1

I know I need to use this formula, but I'm not sure how:
\[L=\int\limits_{0}^{\pi/2}\sqrt{r^2 +(dr/d \Theta)^2}\]

Answer 2

Hmmm the algebra isn't working out for me.
Let's try to take this slow and see what's going on.

Answer 3

So at some point we'll need our r'.
Are you able to determine the derivative of your function r?

Answer 4

Even that step is giving me trouble.

Answer 5

Writing it like this might help,
\[\Large\rm r=\frac{14}{1+\cos \theta}=14(1+\cos \theta)^{-1}\]Allows us to apply the power rule, into chain rule.

Answer 6

Alternatively, you can apply the quotient rule, but that's not very fun when your numerator is just a constant.

Answer 7

Still too hard? :3

Answer 8

\[\Large\rm r'=-14(1+\cos \theta)^{-2}\color{royalblue}{\left(1+\cos \theta\right)'}\]We apply our power rule, then set up our chain rule.
We need to multiply by the derivative of the inner function.

Answer 9

\[\Large\rm r'=-14(1+\cos \theta)^{-2}\color{orangered}{\left(-\sin \theta\right)}\]

Answer 10

\[\Large\rm \frac{dr}{d\theta}=\frac{14\sin \theta}{(1+\cos \theta)^2}\]Something like that, yes? :o

Answer 11

ok. Just got there, working it out on paper

Answer 12

Now I have
\[L= \int\limits_{0}^{\pi/2}\sqrt{14/\cos(\theta) + (14\sin(\theta))/(1+\cos(\theta))^2}\]

Answer 13

Woops, we want to square our r as well, yes?

Answer 14

Lemme fix that up real quick:

Answer 15

\[\Large\rm L= \int\limits\limits_{0}^{\pi/2}\sqrt{\left[\frac{14}{(1+\cos \theta)}\right]^2 + \left[\frac{14\sin \theta}{(1+\cos \theta)^2}\right]^2}~d\theta\]Mmmm like this I think?

Answer 16

Nice! looks much cleaner and has the squares I left out

Answer 17

The algebra gets a little nasty here.... hmm

Answer 18

\[\Large\rm L= 14\int\limits_{0}^{\pi/2}\sqrt{\frac{(1+\cos \theta)^2}{(1+\cos \theta)^4} + \frac{\sin^2\theta}{(1+\cos \theta)^4}}~d\theta\]Lemme know if you're confused by this step.
I did a few different things here all at once.

Answer 19

I factored the 14 out of the root,
I applied the squares,
then I multiplied the first term by (1+cos theta)^2 on top and bottom to get a common denominator.

Answer 20

Ok, I follow.

Answer 21

Now the numerators can be on the same line and we can take the square root?

Answer 22

Yes, good good.
Combine fractions, take the square root of the bottom and pull it out,\[\Large\rm L= 14\int\limits\limits_{0}^{\pi/2}\frac{1}{(1+\cos)^2}\sqrt{(1+\cos \theta)^2+\sin^2\theta}~~d\theta\]

Answer 23

Still have some work to do with the top though,
keep in mind you can't take the root of the top,\[\Large\rm \sqrt{a^2+b^2}\ne a+b\]

Answer 24

Is there a trig equivalency for this part? \[(1 + \cos \theta )^2\]

Answer 25

No, we have to expand it out.

Answer 26

We'll have a nice simplification after expanding it though.

Answer 27

\[\large\rm L= 14\int\limits\limits\limits_{0}^{\pi/2}\frac{1}{(1+\cos)^2}\sqrt{(1+2\cos \theta+\cos^2\theta+\sin^2\theta}~~d\theta\]Somethingggggg like that, yah?

Answer 28

okay. so under the square root turns to this?\[2+2\cos \theta\]

Answer 29

Mmmm ok good.
Factoring a 2 out of each term and pulling it out of the root gives us,\[\large\rm L= 14\sqrt{2}\int\limits_{0}^{\pi/2}\frac{1}{(1+\cos)^2}\sqrt{1+\cos \theta}~~d\theta\]

Answer 30

Now a substitution for 1+cos?

Answer 31

Hmmm :(
Here's where I'm getting stuck.
We can divide out sqrt{1+cos(theta)} from top and bottom,
giving us,\[\large\rm L= 14\sqrt{2}\int\limits_{0}^{\pi/2}\frac{1}{(1+\cos)^{3/2}}~d\theta\]Buttttttt, we don't have a nice u-sub we can apply.
See how there is no Sin(theta) showing up anywhere?
We would have trouble with our du.

Answer 32

I'm gonna throw it into wolfram a sec, to try and see where I goofed up.

Answer 33

Hmm I can't figure it out ;c

@aum @dan815 @jim_thompson5910 @ganeshie8

Answer 34

Whats wrong with making u = (1 +cos theta)?

Answer 35

\[\Large\rm du=-\sin \theta~d \theta\]That sine is a little difficult to deal with.

Answer 36

\(\large 1+\cos\theta = 2\cos^2(\theta/2)\)

Answer 37

\[\large\begin{align}\rm L&= 14\sqrt{2}\int\limits_{0}^{\pi/2}\frac{1}{(1+\cos)^{3/2}}~d\theta \\~\\&= 14\sqrt{2}\int\limits_{0}^{\pi/2}\frac{1}{(2\cos^2(\theta/2))^{3/2}}~d\theta\\~\\&=7 \int\limits_{0}^{\pi/2} \sec^{3}(\theta/2)~d\theta\\~\\\end{align}\]

Answer 38

Note that in the given interval all trig functions are positive so you don't need to wry about canceling radicals

Answer 39

evaluating sec^3 should be pretty straightforward ?

Answer 40

Yes. Thank you

Answer 41

What would be the correct etiquette here? Who should get the medal from me?
zepdrix lead me through a larger portion of the work, but ganeshie8 brought the problem closer to the answer

Answer 42

We're not medal hogs, it shouldn't matter either way c:
Other people gave some medals for this question also, so that's appreciated ^^

Answer 43

^^

Answer 44

sorry, still can't get this one...