Question

Help with calculus particle parabolic distance question.

Answer 1

So I have a question: A particle moves along the path defined by the equation y = x^2. Find the distance the particle travels from x = 0 to x = 6. I'm stuck on a certain step.

Answer 2

\[L^b_a = \int\limits_{a}^{b} \sqrt{(1+(f'(x))^2} dx\]
\[\int\limits_{0}^{6}\sqrt{1 + (2x)^2} dx\]
\[a^2 + x^2 > x= a \tan \theta\]
\[x = \frac{ 1 }{ 2 } \tan \theta\]
\[dx = \frac{ 1 }{ 2 } \sec^2 \theta d \theta\]
\[\int\limits_{0}^{6}\sqrt{1+4(\frac{ 1 }{2} \tan \theta)^2}(\frac{ 1 }{ 2 }\sec^2\theta d \theta)\]
\[\frac{ 1 }{ 2 } \int\limits_{0}^{6} \sqrt{1+\tan^2\theta} \sec^2 \theta d \theta\]
\[\frac{ 1 }{ 2 } \int\limits_{0}^{6} \sec^3 \theta d \theta\]
\[\int\limits \sec \theta d \theta + \int\limits \sec \theta \tan^2 \theta d \theta\]

Here's where I'm stuck. What do I do next?

Answer 3

@ganeshie8

Answer 4

\[\sec \theta \tan \theta + \int\limits \sec \theta \tan^2 \theta d \theta\] I can't figure out how to integrate the second part lol.

Answer 5

@ParthKohli

Answer 6

There's a general technique to evaluate integrals of the form \(\sin^m x \cos ^n x\). I always forget all the cases, but you can look it up somewhere.

Answer 7

\[\sec \theta \tan \theta - \int\limits \sec \theta (1-\sec^2\theta) d \theta\]

Answer 8

oops. flipped. \[\sec \theta \tan \theta - \int\limits (\sec^3 \theta - \sec \theta) d \theta\]

Answer 9

Maybe pull a sec out
then you'd have
(sec^2-1 )*sec
and replace that with your identity and do a u sub?

Answer 10

Oi but watch out, you just undid what was done hehe

Answer 11

Halt. you did a lot of steps that you shouldn't have

Answer 12

\[\int\limits_{}^{}\sec(\theta)\tan(\theta)*\sec(\theta)d(\theta)\]\[u=\sec(\theta)\]\[du=\sec(\theta)\tan(\theta)d(\theta)\]

Answer 13

How did \(\large\rm \int \sec \theta ~d\theta\) turn into \(\large\rm \sec\theta\tan\theta\) ?

Oh I see, you accidentally differentiated, ya?

Answer 14

oh shuu sorry, it was sec(theta)tan^2(theta)
I thought it was sec^2(theta)tan(theta)
:>
that was from way earlier xD

Answer 15

There is a nice reduction formula for secant :)

Answer 16

The only reason why I haven't said what the reduction formula is, is because I haven't figured out how to rederive it yet haha.

Answer 17

I remember how to derive the sine and cosine formulas...
trying to remember secant... it's probably not too bad >.<

Answer 18

derivative of sec u is uhh. sec u tan u du/dx

Answer 19

integral of sec u was I think ln |sec u + tan i| + C I think.

Answer 20

tan u*

Answer 21

Looks good!

Answer 22

\[\large\rm \int\limits \sec^3x~dx=\int\limits \sec x\cdot \left(\sec^2x~dx\right)\]Hmm, integration by parts, ya? :)\[\large\rm u=\sec x\qquad\qquad\qquad dv=\sec^2x~dx\]I'll bet this works out.... probably.... maybe...

Answer 23

Ok ok ok, yah, this is one of those weird circular things.\[\large\rm \int\limits \sec^3\theta~d \theta=\sec \theta \tan \theta-\int\limits \sec \theta \tan^2\theta~d \theta\]That second integral, applying Pythagorean Identity,\[\large\rm \int\limits \sec^2\theta~d \theta=\sec \theta \tan \theta-\int\limits \sec \theta(\sec^2\theta-1)d \theta\]Oh oh, let's call our original integral like capital i or something,\[\large\rm \mathcal I=\sec \theta \tan \theta-\int\limits \sec^3\theta~d \theta+\int\limits \sec \theta~d \theta\]Which is really,\[\large\rm \mathcal I=\sec \theta \tan \theta-\mathcal I+\int\limits\limits \sec \theta~d \theta\]

Answer 24

ya? fun stuff? lil bit? :o

Answer 25

ooh! we did these in class! :D

Answer 26

Add \(\rm \mathcal I\) to each side, divide by 2,\[\large\rm \mathcal I=\frac{1}{2}\left[\sec \theta \tan \theta+\int\limits\limits \sec \theta~d \theta\right]\]

Answer 27

*cries* my internet is so slow it can't process the equations. sorry. loading.

Answer 28

ah :3

Answer 29

Typesetting math: 0%...

Answer 30

Do pictures show up ok?