How can I solve sin(t^2)-(t^2)=0 without a numerical solver?

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Homework Help Overview

The discussion revolves around solving the equation sin(t^2) - t^2 = 0 for the variable t, which arises from a dynamics problem related to summing forces. Participants are exploring methods to find solutions without relying on numerical solvers.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the known solution t = 0 and the additional solution t = 1.04, as indicated by an external tool. There is a question about the validity of the latter solution, with some participants suggesting that sin(t^2) cannot exceed t^2 for t = 1.04. Others propose using small angle approximations for sin(x) and consider the implications of graphing the function.

Discussion Status

The conversation is ongoing, with participants clarifying the problem statement and addressing the discrepancies in graphing results. Some guidance has been offered regarding the use of Newton's method and ensuring the calculator is set to the correct mode. There is an acknowledgment of common mistakes related to angle measurement.

Contextual Notes

Participants note that the original problem was misstated, and there is a discussion about the implications of using degrees versus radians in calculations. The context of the problem being derived from a dynamics scenario is also highlighted.

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Homework Statement



Solve
sin(t2) - t2 =0. for t

Homework 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.

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 TI-89, 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?
 
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wutang42 said:

Homework Statement



Solve
sin(t2) - t2 =0. for t

Homework 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.

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 TI-89, 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?

t=1.04 can't possibly be a solution because

\sin(t^2)-t^2 < 0

for that value of t. How can you tell? Because the max sin can be is 1, and 1.042>1.
 
for small radian values sin(x) ≈ x hence
and that's probably the best you can do.
 
Shoot, I realized that I misstated the problem. It's actually 1.225*sin(t2) - t2 =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
 

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wutang42 said:
Shoot, I realized that I misstated the problem. It's actually 1.225*sin(t2) - t2 =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

Why doesn't it make much sense? Your graph clearly shows zeroes near ##\pm 1##. You could work it by hand with your calculator using Newton's method with a starting value of ##x=1##. Make sure your calculator is in radian mode.
 
LCKurtz said:
Why doesn't it make much sense? Your graph clearly shows zeroes near ##\pm 1##. You could work it by hand with your calculator using Newton's method with a starting value of ##x=1##.Make sure your calculator is in radian mode.

Aaaaand there's my problem. I was doing it in degrees the whole time on my calculator, and couldn't figure out why my graph looked nothing like the wolfram graph. Derp. Thanks!
 
wutang42 said:
Aaaaand there's my problem. I was doing it in degrees the whole time on my calculator, and couldn't figure out why my graph looked nothing like the wolfram graph. Derp. Thanks!

Common mistake :wink: I've done it enough times myself that it's now on my checklist of things that could have gone wrong.
 

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