Jamastiene
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Tue Oct-03-06 06:17 PM
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Help. I need some calculus help? Please? |
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Edited on Tue Oct-03-06 06:49 PM by Jamastiene
Does anyone know how to find the answer to this problem?
A television camera at ground level is filming the lift-off of a space shuttle that is rising according to the formula s=60t^2, where s is measured in feet and t is measured in seconds. The camera is 2000 feet from the launchpad. Find the rate of change in the angle of elevation of the television camera (d angle/dt) 10 seconds after liftoff.
I'd like to see a step by step answer to this one. I'll admit it. I'm lost.
Any calculus gurus around now?
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Jamastiene
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Tue Oct-03-06 06:29 PM
Response to Original message |
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Edited on Tue Oct-03-06 06:30 PM by Jamastiene
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Jamastiene
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Tue Oct-03-06 06:40 PM
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2. Pretty please with sugar on top? |
Mike03
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Tue Oct-03-06 06:45 PM
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3. You know, I would repost this in |
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the science forum! There are some very brilliant people there who might be able to help.
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ZombieNixon
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Tue Oct-03-06 06:47 PM
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4. OK...now it's been a hell of a long time since I've done related rates, |
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but here...I hope it's right. The answer does make sense.
Step 1: write out the equations you have:
s = 60t^2 therefore ds/dt = 120t
s(10) = 6000
Step 2: identify what you're solving for: dQ/dt @ t=10 (I used "Q" in place of "theta")
You need an equation for dQ/dt.
Step 3: since dQ must be in terms of t and s is the only equation we have with t in it, we need to solve for dQ in terms of s, first (dQ/ds):
Q = tan(s)/2000 (draw a right triangle to figure this out with 2000 as the base and s as the altitude)
2000Q = tan(s)
2000(dQ/ds) = (sec(s))^2 = 1/(cos(s))^2 // secant identity
dQ/ds = 1/2000(cos(s))^2
Step 4: isolate dQ/dt by multiplying dQ/ds by ds/dt dQ/dt = dQ/ds * ds/dt
substitute equations...
dQ/dt = 1/2000(cos(s))^2 * 120t = 120t/2000(cos(s))^2
Step 5: solve dQ/dt at t = 10
120(10)/2000(cos(6000))^2
(cos(6000)) = -.5
(cos(6000))^2 = .25
2000*.25 = 500
1200/500 = 2.4
Answer: dQ/dt @ t=10 = 2.4 degrees per second
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Zavulon
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Tue Oct-03-06 07:41 PM
Response to Reply #4 |
10. Yeah, that was my first guess too ;)))))) (NT) |
wain
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Tue Oct-03-06 07:42 PM
Response to Reply #4 |
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You exhibit a keen mind breaking down problems into logical, workable pieces.
Even if the answer is nor correct, it's easy to go back and correct.
Breaking down problems into manageable pieces is a very special, marketable skill.
:)
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ZombieNixon
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Tue Oct-03-06 07:52 PM
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14. After doing C programming for the past 8 years, I would hope so. |
Fox Mulder
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Tue Oct-03-06 07:45 PM
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Where were you when I was in calculus? :P
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ZombieNixon
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Tue Oct-03-06 07:51 PM
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Probably in calculus getting stuff like this wrong. :shrug: :P
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Telly Savalas
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Tue Oct-03-06 07:59 PM
Response to Reply #4 |
15. In step 3, I think you want to start with tan(Q)=s/2000 |
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That is, tan(Q) = tangent of the angle = opposite side divided by adjacent side = s/2000.
My brain is mush, so please let me know if I'm full of shit. I'd review this more closely but I've gotta run. By the time I get back later tonight my brain will be even mushier. Sorry.
By the way, the absence of alarcojon in this thread is making me sad.
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ZombieNixon
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Tue Oct-03-06 08:01 PM
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16. Oh, yes, thank you, you're right. |
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I just checked my notes and saw that I typed it in wrong. Unfortunatly, it's too late to edit. :(
And...alarcojon...I know...:cry:
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Mike03
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Tue Oct-03-06 06:47 PM
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I might change the thread subject to something about Calculus or Equations, because it's not totally clear from the subject what the issue is.
Good luck in getting your question answered.
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Jamastiene
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Tue Oct-03-06 07:30 PM
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ZombieNixon
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Tue Oct-03-06 07:32 PM
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7. Did you see my response (#4) |
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or is that totally wrong? :(
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Jamastiene
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Tue Oct-03-06 07:36 PM
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I didn't see it before. I'm so confused by this problem. Thanks. I'll look at it now. I'm sorry. :hug:
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ZombieNixon
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Tue Oct-03-06 07:37 PM
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If you need anything clarified, tell me. :) :hug:
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Jamastiene
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Tue Oct-03-06 08:36 PM
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17. I'm that much of a math dunce. |
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:blush:
Yes, pretty much everything after the s=120t part is a blur to me. I'm trying to take it all in. Believe it or not, I have 3 more problems to do regarding related rates. I got 2 of them on my own though, so I am improving somewhat, I think.
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Jamastiene
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Tue Oct-03-06 10:30 PM
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18. Thanks to everyone who responded. |
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I'm going to have to go to bed. I don't know what it is about word problems, but they have always gotten me so addled and confused that I don't quite know what to do with myself. For me, that's rare.
Thanks, again. I'm off to bed. Hopefully, I'll be able to come back tomorrow and learn some more. Too bad, I can't stay up later and post more questions now. It is somewhat clearer to me now though.
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