Question

Calculus BC problem... Can anyone please help? Image to problem included.
Edit: I got to find the equation for S part, but Idk how to find t.

Answer 1

I'd appreciate if you can look over my proof so far...? Please.

Answer 2

t is the independant variable variable

i assume this is modeling with diffy qs?

Answer 3

From the picture above, I wrote the equation for arc length
\[S=\int\limits_{0}^{\theta _{f}} \sqrt{(\frac{ dx }{ d \theta })^2+(\frac{ dy }{ d \theta })^2 + (\frac{ dz }{ d \theta })^2} d \theta\]

Answer 4

Yes, it is independent! :D

Answer 5

Given: We also know that
\[r^2 \theta= constant\]

Where r^2 theta is with respect to d theta. Constant with respect to t.
That was the relationship.

Answer 6

im not aware of the physical formulas for this, but i do know that the angular velocity of rotation increases as a mass is pulled closer to its center of gravity.

do we know a formula for this process?

Answer 7

the time it takes L to wrap depends on the angular velocity is what i would think but im bad at physics

Answer 8

\[S=\int\limits_{0}^{\theta _{f}} \sqrt{(\frac{ dx }{ d \theta })^2+(\frac{ dy }{ d \theta })^2 + (\frac{ dz }{ d \theta })^2} d \theta\]

I took the derivative of X, Y, and Z.

and plug in

\[S=\int\limits_{0}^{\theta _{f}} \sqrt{(-R sin \theta)^2+(\frac{-d}{2 \pi})^2 + (Rcos\theta)^2} d \theta\]

Answer 9

i dont have a clear idea in my head on how to approach a solution just yet, so these are some of my initial concerns :)

Answer 10

Then, I use trig identities

cos^2 + sin^2 =1

multiply by r^2

Rcos^2 + Rsin^2 = R^2

If I plug this in the equation I'd just get:

R^2 + (d/2pi)^2 inside the parenthesis.
Sorry I messed up on the latex coding so I didn't put it in the equation...

Answer 11

latex is just eyecandy .... as long sa its readable its fine

Answer 12

should be R^2cos^2 + R^2sin^2 = R^2. sorry

Answer 13

is arclength your own idea, or was it a hint of some sort? at the moment i cant really see if thats a good approach or not.

Answer 14

So I found S, but the problem is to fnd t... Idk how to set up the next step >.<

Answer 15

Oh my teach said to assume that the ball is thrown at 90 degrees angle.
Observation:
-no tangential speed, only gravity pulling down

Answer 16

t is determined by the physics formulas that govern the situation .... length of rope, angular velocity, mass, etc ...

Answer 17

and the distance d is constant.
We can also assume that
L = total lenth
S= arc length
Then each time the tetherball wrap around the pole, the remaining lenth would be

m= small length
m= L-S

idk if this is right?

Answer 18

Oh no, I think he only wants to use Calc on this. No velocity or anything.

Said the final answer should not have theta in it...

Answer 19

When I integrate to get S, I found that S is equal to

\[S= \theta _{F}\sqrt{R^2 + \frac{ d^2 }{ 4\pi^2 }}\]

Where theta F is the final angle where r ends and touch the pole.

Answer 20

my thoughts, and i do tend to overthink these things.

angular momentum is conserved in a system. this is something that kepler oberved in planetary motion. the area of the sweep remains constant as the planets move in their orbits. this means that the planets actually speed up and slow down as they travel in their orbits in order to maintain keplers law.

the ball around the pole would have to obey the same principle, which is why if youve ever played tetherball the angular velocity speeds up as the ball gets closer to the pole.

also why an iceskater can do those amazing spins ... they pull in thier arms and their angular momentum increases in order to maintain this conservation of mometum.

Answer 21

needless to say im not going to be much use in finding a suitable solution .... modeling with diffy qs is not a strong point.

Answer 22

I see.. thank you for your help! >.< Ik that physics would be so much easier to solve this. Idk why but my teacher wants it in calc with xyz haha.

Answer 23

good luck :)

Answer 24

Thank you :D You're amazing for helping me out!

Answer 25

The key is what ur teach told u:
"no tangential speed, only gravity pulling down"

I will change the naming quite a bit so please look closely at the pic below as the variables may not be defined the same as urs:



l is the length of string ALREADY wrapped on the cylinder at time t
so l=0 at t=0; when l=L, t=the answer n lets call it T.

if the string is unwrapped from the cylinder at time t, we get the triangle above.

as it only gravity pulling down, from physics we know y=1/2*g*t^2 where g is gravity.

from the angular velocity w, x=2*pi*R*w*t

it is a right-angled triangle, so x^2 + y^2 = l^2
So at time T, l=L, x=2*pi*R*w*T, y=1/2*g*T^2

(2*pi*R*w*T)^2 + (1/2*g*T^2)^2 = L

That would the equation between T and L.