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Old 10-08-2012, 10:55 PM   #8921
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Quote:
Originally Posted by MakinLattes
(1/4x)*4 = 1/x


how are you getting that and what rules are you using?
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Old 10-08-2012, 11:06 PM   #8922
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It is the chain rule. ln(4x) is like f(g(x)) where f(x) = ln(x) and g(x) = 4x. So, I took the derivative of ln(x), which is 1/x, but instead of x I put 4x in the denominator because I'm treating 4x as just x for the first part of the chain rule. Then I took the derivative of 4x, 4, and multiplied 1/4x by it.

All I did was f'(g(x))*g'(x)
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Old 10-08-2012, 11:14 PM   #8923
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Quote:
Originally Posted by MakinLattes
It is the chain rule. ln(4x) is like f(g(x)) where f(x) = ln(x) and g(x) = 4x. So, I took the derivative of ln(x), which is 1/x, but instead of x I put 4x in the denominator because I'm treating 4x as just x for the first part of the chain rule. Then I took the derivative of 4x, 4, and multiplied 1/4x by it.

All I did was f'(g(x))*g'(x)


oh...it was that simple. I don't know why, but i always have trouble with the chain rule.
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Old 10-08-2012, 11:28 PM   #8924
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It takes some getting used to is all
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Old 10-09-2012, 07:53 PM   #8925
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Ok, this one problem I have is giving me a ton of trouble. I've been trying to get the correct answer (back of the book says it's 931m), but I honestly can't get it.

"A rocket moves upward, starting from rest with an acceleration of +29.4m/s^2 for 3.98s. It runs out of fuel at the end of the 3.98s but does not stop. How high does it rise above the ground?"

Any thoughts on this one?
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Old 10-09-2012, 08:31 PM   #8926
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The tricky part is that the acceleration changes. Try dividing the flight of the rocket into two parts at the point where the acceleration changes, find how high the rocket rises in each part, and then add these numbers.

The first part will be from liftoff (t=0) until the fuel runs out (t=3.98). The second part will be from that point until the rocket hits a maximum height where velocity=0. To find that time, you'll need to use the first acceleration to find the initial velocity at t=3.98.

Try using these ideas, and if you can't figure it out I'll post a full solution.
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Old 10-09-2012, 08:35 PM   #8927
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Originally Posted by In Vein
The tricky part is that the acceleration changes. Try dividing the flight of the rocket into two parts at the point where the acceleration changes, find how high the rocket rises in each part, and then add these numbers.

The first part will be from liftoff (t=0) until the fuel runs out (t=3.98). The second part will be from that point until the rocket hits a maximum height where velocity=0. To find that time, you'll need to use the first acceleration to find the initial velocity at t=3.98.

Try using these ideas, and if you can't figure it out I'll post a full solution.


Ok, I'll try and figure it out from here. I knew that the maximum height would include velocity being 0, but I wasn't sure how to go from there.

Edit: I tried a few different ways of approaching it, yet I can't get anywhere close to the 931m that I'm supposed to get.
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Old 10-09-2012, 08:39 PM   #8928
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Calculus question guys. I have a region that I'm rotating across the y axis. Normally I know how to solve this kind of problem, but the region crosses the Y axis, so I'm confused what would happen to it.
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Old 10-09-2012, 08:49 PM   #8929
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Originally Posted by guitarplaya322
Calculus question guys. I have a region that I'm rotating across the y axis. Normally I know how to solve this kind of problem, but the region crosses the Y axis, so I'm confused what would happen to it.


Interesting, I've never actually encountered a problem like this. As such, I'm not entirely sure, but just visualizing it, it would make sense to separate the figure into two objects at the point where the curve crosses the y-axis. Then, find the volume of each solid separately and add them together.
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Old 10-09-2012, 08:51 PM   #8930
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Quote:
Originally Posted by In Vein
Interesting, I've never actually encountered a problem like this. As such, I'm not entirely sure, but just visualizing it, it would make sense to separate the figure into two objects at the point where the curve crosses the y-axis. Then, find the volume of each solid separately and add them together.

That's what I was thinking too. Just wanted to see if there any known tricks or shortcuts because this will be a pain in the ass.

EDIT: Nope I don't think that will work because some regions on either side cover each other when they're rotated around the axis.
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Old 10-09-2012, 09:06 PM   #8931
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What's the problem exactly?
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Old 10-09-2012, 09:18 PM   #8932
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Quote:
Originally Posted by In Vein
What's the problem exactly?

Rotate the region enclosed by x=(y^4)/4-(y^2)/2 and x=(y^2)/2 about the y axis.
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Old 10-09-2012, 10:13 PM   #8933
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Wow that is a doozy. Here's what I'm thinking though.
First, for simplicity, f(y)=x=(y^4)/4-(y^2)/2 and g(y)=x=(y^2)/2
Our problem with this solid is on the interval 0≤y≤sqrt(2), since here is where the regions would overlap when rotated.
We know that on 0≤y≤2 |g(y)|≥|f(y)|. So, the curve furthest from x=0 on the interval 0≤y≤sqrt(2) is g(y).
So, when we rotate the region about the y-axis, this curve will determine the outer edge of the figure. The only thing I give pause to is what happens to the region between f(y) and x=0 as it gets rotated. It does add volume, but it adds volume that has already been filled. So, I think, but am not sure, that we can just discount it as redundant.

For 0≤y≤sqrt(2), find the volume by rotating g(y) around the y-axis.
I believe that for sqrt(2)≤y≤2 its pretty straightforward, just g(y)-f(y).
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Last edited by In Vein : 10-09-2012 at 10:14 PM.
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Old 10-09-2012, 10:29 PM   #8934
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check out this plot linky (refresh the page if it doesn't work for whatever reason)

if you solve both equations you find the graphs intercept at (0,0) and (2,2)

when you rotate about the y-axis, the inner curve (red) becomes your inner radius and the outer curve (blue) becomes your outer radius. if you're calculating volumes using single variable calculus, you can use the cylindrical shell method to calculate this quite easily.

actually if you know multivariable methods (double and/or triple integrals, specifically) you can do this even more easily with polar coordinates or cylindrical coordinates.

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Old 10-09-2012, 11:08 PM   #8935
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Quote:
Originally Posted by aerosmithfan95
Ok, this one problem I have is giving me a ton of trouble. I've been trying to get the correct answer (back of the book says it's 931m), but I honestly can't get it.

"A rocket moves upward, starting from rest with an acceleration of +29.4m/s^2 for 3.98s. It runs out of fuel at the end of the 3.98s but does not stop. How high does it rise above the ground?"

Any thoughts on this one?

If you're still here 2 hours later, I can do this!
If not, well I've been cooped up in the house all day and physics is fun so I'm figuring this out anyway

So you split the problem into two parts, in the first you have these variables, assuming you are using the upwards direction as your reference frame
initial velocity = 0 m/s
acceleration = 29.4 m/s^2
time = 3.98s
and if you have any 3 variables, you can find the other two. What you need is distance up, so you can use the following equation

distance = (initial velocity)(time) + 1/2(accleration)(time^2)

So for the second part we have the variables
acceleration = -9.8 m/s^2 (maybe you forgot to factor gravity in when you were running it before?)
final velocity = 0 m/s (that'll be when the rocket hits its peak and then starts coming down)

What you need then is the initial velocity, which is the final velocity of the first, and of course the up

so

acceleration = [(final velocity) - (initial velocity)] / time
final velocity = (acceleration)(time) - (initial velocity)

and then

(final velocity)^2 = (initial)^2 + 2(a)(distance)
distance = (final^2 - initial^2) / 2a


I got 931 and some change. hope that helps. Soothed me anyway
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Old 10-10-2012, 10:22 AM   #8936
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Quote:
Originally Posted by Avedas
check out this plot linky (refresh the page if it doesn't work for whatever reason)

if you solve both equations you find the graphs intercept at (0,0) and (2,2)

when you rotate about the y-axis, the inner curve (red) becomes your inner radius and the outer curve (blue) becomes your outer radius. if you're calculating volumes using single variable calculus, you can use the cylindrical shell method to calculate this quite easily.

actually if you know multivariable methods (double and/or triple integrals, specifically) you can do this even more easily with polar coordinates or cylindrical coordinates.


The way it is graphed on Wolfram is misleading. Instead of graphing them on a normal axis orientation, Wolfram just renamed the axes. In the link you provided, you would have to rotate the region around the horizontal axis, since that is the y-axis. The confusing part of this problem is that the region crosses through the axis about which it is being rotated, so some of the parts would overlap. Check out my previous post and see if you agree with the logic.
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Old 10-10-2012, 12:22 PM   #8937
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Originally Posted by In Vein
The way it is graphed on Wolfram is misleading. Instead of graphing them on a normal axis orientation, Wolfram just renamed the axes. In the link you provided, you would have to rotate the region around the horizontal axis, since that is the y-axis. The confusing part of this problem is that the region crosses through the axis about which it is being rotated, so some of the parts would overlap. Check out my previous post and see if you agree with the logic.


Oh you're right. It doesn't really change how the problem is solved though. The outer radius alone would be suitable for volume calculations. Overlapping volumes mean nothing without a density function anyway.
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Old 10-10-2012, 01:14 PM   #8938
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to those studying mathematics at uni, i've a simple question for you:

are you enjoying it?
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Old 10-10-2012, 11:52 PM   #8939
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Originally Posted by laid-to-waste
to those studying mathematics at uni, i've a simple question for you:

are you enjoying it?


I like the idea of learning it and applying it, but pure theoretical math doesn't really do it for me. Also I'd infinitely rather focus on my engineering courses than the math courses. I still have so many more math courses to take
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Old 10-11-2012, 12:11 AM   #8940
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Is there anyone here who knows C++ pretty well? I could really use some help cause i'm struggling right now with some programming assignments
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