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Solving sin(t^2)(t^2)=0 
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#1
Dec712, 11:31 PM

P: 3

1. The problem statement, all variables and given/known data
Solve sin(t^{2})  t^{2} =0. for t 2. Relevant equations None, besides various trig identities. This was actually a dynamics problem where I had to solve for time, and this is simply the equation you get after summing up all the forces. 3. The attempt at a solution Now obviously 0 is a solution, and when I plug it into wolfram alpha I get t=1.04, which I know is the correct answer because my professor told us that in class. However, when I plug it into my TI89, the only solution that comes out is 0, and when I graph it, it doesn't show the 1.04 solution either. So, my question is, is there any way to solve this problem without a numerical solver? And if not, is there any way to plug it into my TI 89 in order to get the correct answer? 


#2
Dec812, 12:09 AM

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P: 3,515

[tex]\sin(t^2)t^2 < 0[/tex] for that value of t. How can you tell? Because the max sin can be is 1, and 1.04^{2}>1. 


#3
Dec812, 12:47 AM

P: 2,812

for small radian values sin(x) ≈ x hence
and that's probably the best you can do. 


#4
Dec812, 12:12 PM

P: 3

Solving sin(t^2)(t^2)=0
Shoot, I realized that I misstated the problem. It's actually 1.225*sin(t^{2})  t^{2} =0
But yeah, I realize that it still doesn't make much sense, but this is the answer and the graph that wolfram alpha spits out, and I'm just trying to figure out how to replicate that either on paper or on my calculator 


#5
Dec812, 01:56 PM

HW Helper
Thanks
PF Gold
P: 7,583




#6
Dec812, 02:06 PM

P: 3




#7
Dec812, 03:27 PM

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P: 3,515




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