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Deathslayer7
01-31-2010, 08:46 PM
The boat moves along a path defined by

r^2 = 10,000*cos(2theta) ft^2 where theta is in rad.

If theta = .4t^2 where t is in seconds, determine the radial component of the boats velocity at the instant t=1s

from what i know (this teacher sucks really bad. can't teach. totally not worth the money but he's the only who teaches it) the radial component of v is just the derivative of r with respect to theta.

so we get this

r= sqrt(10000*cos(2theta))

so r prime = -10000*sin(2theta)*theta_prime/(sqrt(10000*cos(2theta)))

rearrange it so we get this:
r prime = -10000*sin(2theta)*theta_prime*(10000*cos(2theta))^ (-1/2)

and i believe that is product rule three times. ugh....

now i need to take r_prime_prime = ???

anyone know of an easy way to take that?

Mando Knight
01-31-2010, 09:28 PM
It should be r'=(-sin(2θ))(200θ')(cos(2θ))^(-1/2) for the first derivative.

Now, because I haven't done derivatives like this by hand for a good three-four semesters, I'd probably just jump straight to the product rule if I can't use a calculator.

The derivative of (-sin(2θ)) is (-2cos(2θ)θ'), the derivative of θ' is θ'', and the derivative of (cos(2θ))^(-1/2) is ugly, and equal to (-1/2)(r')(100cos(2θ))^-1 at first glance.

Griever
01-31-2010, 10:48 PM
r= sqrt(10000*cos(2theta))

Please just turn that into r = 100 * sqrt(cos(2theta))

DSCrankshaw
01-31-2010, 11:02 PM
If I'm understanding the question right, shouldn't v_r = dr/dt, not dr/dtheta?

Given that, don't bother solving for r. Just take the derivative for both sides.

2r(dr/dt) = 10000 (-sin 2theta) 2 dtheta/dt, and
(dr/dt) = -10000/r * (sin 2theta) dtheta/dt

Note that if you substitute sqrt (10000*cos 2theta) for r in the above equation, you get Mando Knight's (and your) solution.

Then, dtheta/dt = .8t, solve for r and theta, and you have your answer.

I'm not sure why you'd need the second derivative of r to solve the problem.

Mando Knight
01-31-2010, 11:10 PM
Note that if you substitute sqrt (10000*cos 2theta) for r in the above equation, you get Mando Knight's solution.

I actually had to go back and change the solution to the right one... I forgot to check that Deathslayer7 separated the constants before taking the derivative.

DSCrankshaw
01-31-2010, 11:14 PM
I actually had to go back and change the solution to the right one... I forgot to check that Deathslayer7 separated the constants before taking the derivative.
I revised my post when I saw that Deathslayer had that part already, anyway.

My question, though, is why is there a need for the second derivative? Am I mistaken in my belief that v_r = dr/dt?

Deathslayer7
01-31-2010, 11:48 PM
i take the derivative with respect to theta cause this is circualr motion. and the reason why i have theta prime is because im trying to follow the books example and thats what they do but they dont see how they got it.

and i can't ask the teacher how cause he couldn't answer it even if i asked <.<

at DSC: the radial component is just the derivative of r. but what you did to divide by r should theretically work since r is given. Didn't think of doing it like that. Would have made it easier to solve for the acceleration that way probably too.

and the second derivative is for acceleration which the homework problem asked me for as well.

DSCrankshaw
02-01-2010, 12:31 AM
Okay, but the derivative with respect to theta doesn't give you a velocity. dr/dtheta just gives you feet per radian--velocity needs to be feet per second. So radial velocity should be dr/dt, which equals dr/dtheta * dtheta/dt. (r' and theta' are dr/dt and dtheta/dt, respectively.)

So, dr/dtheta = -10000/r * (sin 2theta),
and dr/dt = -10000/r * (sin 2theta) dtheta/dt.

You can use this to simplify things a bit, if you apply the product rule to d2r/dt2, which gives d2r/dtheta2 * dtheta/dt + dr/dtheta*d2theta/dt2.

So,
d2r/dtheta2 = (10000/r^2) dr/dt * (sin 2theta) - 10000/r * (cos 2theta) * 2.

Now, use d2r/dt2 = d2r/dtheta2 * dtheta/dt + dr/dtheta*d2theta/dt2, to get the results--you can just plug in numbers here, or you can multiply it out, to get:

d2r/dt2 = ((10000/r^2) dr/dt * (sin 2theta) - 20000/r * (cos 2theta)) * dtheta/dt + (-10000/r * (sin 2theta))* d2theta/dt2.

You get the same result if you go directly, but sometimes it's a little bit easier to compartmentalize like this.

Okay, that's a little bit complicated, but the bottom line is this: r', in this example, is dr/dt, and theta' is dtheta/dt. If you take d/dt of both sides, you get your answer. If you just take d/dtheta of both sides, you don't get a velocity. The book does d/dt, but doesn't seem to explain that that is what it's doing.

Mando Knight
02-01-2010, 02:08 AM
Just be glad you're doing something with a defined, analytical dθ/dt. I mean, you don't want to have to solve for r=a(1-e*cos(E)) as a function of time where Δt=(E-e*sin(E))/n, n=sqrt(μ/(a^3)), and μ, e, and a are constants related to the shape of the ellipse, do you?

That stuff's rocket science. Literally.

DSCrankshaw
02-01-2010, 03:34 AM
Or Schrodinger's equation.

Enough said.