Problem Solving Techniques for Physics

  • Thread starter Thread starter 01trayj
  • Start date Start date
  • Tags Tags
    Physics
AI Thread Summary
The discussion revolves around effective problem-solving techniques for physics problems, emphasizing a structured approach to understanding and solving word problems. Key steps include carefully reading the problem to grasp the given information and what is being asked, identifying relevant formulas, and visualizing the physical process involved. Participants highlight the importance of breaking down complex problems into manageable parts, using methods like vector/matrix setups for linear equations or back substitution for non-linear equations. They also suggest employing optimization strategies and experimenting with trivial values to understand relationships between variables. Additionally, the conversation touches on techniques for integrating complex expressions, such as using polynomial long division and recognizing patterns in rational functions. Overall, the emphasis is on practice and experience as essential for mastering problem-solving in physics.
01trayj
Messages
2
Reaction score
0
I was wondering if anyone could enlighten me on their own problem solving techniques generally for physics problems. I realize there is no steadfast method for tackling every problem, but a sort of detailed thought process / flow chart style approach would help me an awful lot.

Thankyou!
 
Physics news on Phys.org
For a word problem I...

1.) Read, and reread the problem carefully to make sure I understand what it is asking and what it is giving.
2.) Jot down what I think may be useful or relevant formulas. (Actually I generally know what formulas exist relating various quantities and only jot them down when I think I'll be using them.)

3.) I look at the given and asked for quantities and see if I can picture a chain of relations getting from the given set to the needed set. While doing this I try to visualize the physical process involved...

For example, if I'm given initial conditions for a projectile and am asked how high will it go, I think about a thrown object and picture it going higher, higher, higher, then lower, lower, lower... and so I identify the highest point with the point in time when it stops moving upward, translate that to d/dt height = 0. I can then if I have the general equations of motion solve for that time, then plug the time into get position and so on.

4.) I start working out possibly helpful quantities from the known formulas using the known quantities, viewing this as increasing my pool of knowns. I keep this up, paying attention to what I need to know until I have solved for it.

Variations occur when I identify that I am solving a specific type of problem... e.g. I will be solving a system of linear equations so I try to set things up in vector/matrix format.

In another example, I will be solving a system of non-linear equations, I write them down and use the simplest to back substitute an eliminate a variable.

In another example, I identify the problem as an optimization issue and try to express the quantity to be optimized as a function of free variables.

Sometimes when the "givens" are actual variable names (initial height of a) I will pick trivial values and work through to see the structure of how quantities relate. Do this a couple of times and then work it out with the abstract value.

Most of all, the more problems you work the more experience you get and the easier they become.
 
1. Write down all information in a table. (e.g x=1, y=2, z=3...).
2. Write down the unknown in there as (Q = ?)
3. Write down relevant formula's and formula's that relate to it.
4. Also, it's always good to look at certain variables and think of ways of rewriting it, especially when you get a problem like... but they didn't give me the X variable?! How am I supposed to solve it with the most direct approach. Most of the time if they omit information, either it cancels out, or you use a method like n by n matrix to solve for it and another variable, or you use some kind of substitution). Also trying to rewrite an equation into something that can be canceled is always good. Look at the equation carefully and try to recognize something that can be easily rewritten into something else and then factor things out and to your surprise.

For example: there's no direct way to take the integral of t^2/1 + t^2, u-substitution fails, by parts is pretty difficult and who the hell wants to do partial fractions?,
you think to your self, well... i realize t^2/(1 + t^2) makes me think of 1/t^2 + 1 which it's integral is tan^-1 (t), so you write that 1/t^2+1 now if you subtract - t^2+1/t^2+1 it would give you the integrand, but you know t^2+1/t^2+1 is 1, so you found a loophole in a sense, 1/t^2+1 - 1, now you can do the integral in a separate manner.
 
der.physika said:
...
For example: there's no direct way to take the integral of t^2/1 + t^2, u-substitution fails, by parts is pretty difficult and who the hell wants to do partial fractions?

What's wrong with partial fractions? A perfectly simple method but not applicable to your example.

FWIW
Whenever dealing with rational expressions, after reducing as far as possible one should always express any improper rationals to mixed form. N(x)/D(x) = Q(x) + R(x)/D(x)

N=numerator, D = denominator/divisor, Q=quotient, R=remainder.
In general one may apply long division of polynomials to find quotient and remainders.

Your example is an improper rational function and so can be reduced to mixed form:

\frac{t^2}{1+t^2} = \frac{t^2+1 - 1}{t^2 +1} = 1-\frac{1}{1+t^2}
To integrate then one gets:
I= t - \int \frac{1}{1+t^2}dt = t - \int \frac{\sec^2(\theta)d\theta}{1+\tan^2{\theta}}\quad t=\tan(\theta)
I = t - \theta + C = t - \tan^{-1}(t) + C
(Here I have used the substitution but you may recall the inverse tan's derivative and integrate directly.)
 
jambaugh said:
I= t - \int \frac{1}{1+t^2}dt = t - \int \frac{\sec^2(\theta)d\theta}{1+\tan^2{\theta}}\quad t=\tan(\theta)
I = t - \theta + C = t - \tan^{-1}(t) + C

I know this is quite a simple question but can you explain this part to me, please? =) How did the secant and tangent come in? Thanks!
 
trig triangle substitution

tangent of theta is equal to t over sq rt of 1, then it follows that secant squared of theta times d theta is equal to dt.
 
lordkelvin said:
trig triangle substitution

tangent of theta is equal to t over sq rt of 1, then it follows that secant squared of theta times d theta is equal to dt.

lol. i got it now. i went and google "trig triangle substitution" and found a simple website showing the basics. thanks for pointing out to me. i learned something new today. XD
 
Back
Top