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10082012, 10:30 PM  #8881 
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Join Date: Mar 2007
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Yeah, first you go 2(ln(3t)). Then you times that by the derivative of ln(3t). The derivative of lnx = 1/x, so the derivative of ln(3t) is 1/3t because you just put whatever's in the argument of ln into the denominator. But, since 3t is it's own little function, you have to use the chain rule and times 1/3t by the derivative of 3t with respect to t, which is 3. So the derivative of ln(3t) is then (1/3t)*3 = 1/t. So you go and times THAT by 2(ln(3t)) and get (2/t)ln(3t).

10082012, 10:48 PM  #8882 
UG Senior Member
Join Date: Feb 2007
Location: Richmond, Va

I'm still getting it wrong. I think its something that is too complex for me to learn over the internet. Thanks for the help though. I really do appreciate it.
one more quick question: how to find the derivative of ln(4x)? it wouldn't just be 1/(4x) would it?
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10082012, 10:50 PM  #8883 
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Join Date: Mar 2007
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(1/4x)*4 = 1/x

10082012, 10:52 PM  #8884  
yoloswag420
Join Date: Nov 2007
Location: Vancouver, BC

Quote:
you missed the chain rule portion 

10082012, 10:55 PM  #8885  
UG Senior Member
Join Date: Feb 2007
Location: Richmond, Va

Quote:
how are you getting that and what rules are you using?
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10082012, 11:06 PM  #8886 
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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) 
10082012, 11:14 PM  #8887  
UG Senior Member
Join Date: Feb 2007
Location: Richmond, Va

Quote:
oh...it was that simple. I don't know why, but i always have trouble with the chain rule.
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10082012, 11:28 PM  #8888 
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Join Date: Mar 2007
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It takes some getting used to is all

10092012, 07:53 PM  #8889 
Tab Contributor
Join Date: Jul 2008

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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10092012, 08:31 PM  #8890 
Registered User
Join Date: Mar 2009

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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10092012, 08:35 PM  #8891  
Tab Contributor
Join Date: Jul 2008

Quote:
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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10092012, 08:39 PM  #8892 
Registered User
Join Date: Dec 2007

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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10092012, 08:49 PM  #8893  
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Join Date: Mar 2009

Quote:
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 yaxis. Then, find the volume of each solid separately and add them together.
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10092012, 08:51 PM  #8894  
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Join Date: Dec 2007

Quote:
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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Last edited by guitarplaya322 : 10092012 at 09:03 PM. 

10092012, 09:06 PM  #8895 
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Join Date: Mar 2009

What's the problem exactly?
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10092012, 09:18 PM  #8896  
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Join Date: Dec 2007

Quote:
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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10092012, 10:13 PM  #8897 
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Join Date: Mar 2009

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 yaxis, 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 yaxis. I believe that for sqrt(2)≤y≤2 its pretty straightforward, just g(y)f(y).
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10092012, 10:29 PM  #8898 
yoloswag420
Join Date: Nov 2007
Location: Vancouver, BC

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 yaxis, 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. Last edited by Avedas : 10092012 at 10:57 PM. 
10092012, 11:08 PM  #8899  
Thou mayest
Join Date: Jun 2008
Location: Kansas City, MO

Quote:
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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10102012, 10:22 AM  #8900  
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Join Date: Mar 2009

Quote:
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 yaxis. 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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