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Finding position from velocity (trig function) 
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#1
Mar211, 09:28 PM

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1. The problem statement, all variables and given/known data
S(0)=3, find S(2) position wise. 2. Relevant equations V(t)=xsin(x^2) 3. The attempt at a solution I tried to integrate with usubstitution and I got t^4/4cos(t^2). I tested it by taking the derivative and it didn't work out. 


#2
Mar211, 09:41 PM

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[tex]v(t)=\frac{dx}{dt}\quad\to\quad\frac{dx}{x\sin(x^2)}=dt[/tex]
Is the equation on the right what you integrated? 


#3
Mar211, 09:46 PM

P: 24

@SammyS,
Thanks for the reply, but I have never seen the method you used before. I understand that you manipulated the first equation to get the second, but I do not know why. If you could explain it a bit or give me a link to a website that explains it I would appreciate it. 


#4
Mar211, 10:39 PM

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Finding position from velocity (trig function)
Are you sure that your velocity function is v(t) = x*sin(x^2), where x is a distance? Is it possible that it's v(t) = t*sin(t^2) instead?



#5
Mar211, 10:50 PM

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If [tex]\frac{dx}{dt}=x\sin(x^2)\,,[/tex]
then [tex]\frac{1}{x\sin(x^2)}\ \frac{dx}{dt}\,dt=dt\,.[/tex] But, [tex]\frac{dx}{dt}\,dt=dx\,.[/tex] Therefore, [tex]\frac{dx}{x\sin(x^2 )}=dt[/tex] Now integrate both sides to find t as a function of x. 


#6
Mar211, 10:53 PM

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#7
Mar211, 11:05 PM

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Your earlier suggestion: [tex]v(t)=t\sin(t^2)[/tex] is probably correct. 


#8
Mar311, 12:17 AM

P: 24

I switched t with x in the equation. Now that I think about it I'm even unsure of why I did this. Anyway, I solved the problem by taking t out of the function before integrating. Thank you all for the quick replies.



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