Question

Find the arc length of the curve r=θ^3, 0≤θ≤π. Where are the tangent lines vertical? Horizontal?

Answer 1

you need y = f(θ) and x = f(θ), do you know how to find them?

Answer 2

I belive so. wouldn't the y function be cos(θ) - θsin(θ) and the x function be sinθ + θcos(θ)?

Answer 3

Well, let's do the arc length first. Do you know the formula?

Answer 4

\[\sqrt{dx + dy}\] ?

Answer 5

dx and dy are squared though.

Answer 6

well you could do it that way, but there is a better alternative.

L = ∫sqrt(r^2 + (dr/dθ)&2) dθ. Look familiar?

Answer 7

Not at all. Could you elaborate on that?

Answer 8

typo....
L = ∫sqrt(r^2 + (dr/dθ)^2) dθ

Answer 9

Ok, so it should look like this: \[\sqrt{r ^{2} + dr/dTheta^2}\] albeit with the integration, correct?

Answer 10

yes.
r = θ^3
so dr/dθ = 3θ^2

yes?

Answer 11

Ok, that makes sense. So, can you explain why I need to find the arc length?

Answer 12

well the question asks for arclength doesn't it?

Answer 13

Sorry, you're correct. (Long day with a decent amount of medicine makes one a bit oblivious)

Answer 14

so, when plug those in the formula what do you get?

Answer 15

I ended up getting sqrt(theta^5 + 9theta^4). Right?

Answer 16

uhm, what is (θ^3)^2 ?

Answer 17

Ohh, I see what I did. That should be to the sixth.

Answer 18

simplify further what do you get?

Answer 19

theta to the third + 3(theta)^2

Answer 20

O.O holy crap man. You can't do that!

Answer 21

D:

Answer 22

sqrt(θ^6 + 9θ^4) = sqrt[θ^4 (θ^2 + 9)] <--- do you see what I did?

Answer 23

Ok, I see what you did.

Answer 24

ok, what did i do? and how do you simplify it further

Answer 25

took a theta^4 out of both terms, and the quantity can be broken down into theta + 3 and theta -3

Answer 26

Nooooooooooooooooooooo!!!!!! D: You told me the same mistake again

\[\sqrt{a+b} \neq \sqrt{a} + \sqrt{b}\]

Answer 27

*facepalm*

Answer 28

hower, \[\sqrt{ab} = (\sqrt{a}) (\sqrt{b})\]

Answer 29

*however*,...

Answer 30

Using that formula above, can you simplify it further?

Answer 31

Take the theta^4 and the quantity, separate them into two different square root functions, and go from there?

Answer 32

yes, specifically simplified to what?

Answer 33

theta^2 * sqrt(theta^2 + 9)

Answer 34

very good
so,
arclength = ∫θ^2 sqrt(9 + θ^2) dθ, from 0 to pi,
which is approximately 39.77

Answer 35

Just got it. Thank you!

Answer 36

as for horizontal and vertical tangent, do you know how to find them?

Answer 37

Yes, I do.

Answer 38

excellent.