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Tspirit
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I can't understand the solution to Problem 1.4(a). The solution is the following:
What puzzles me is that ρ(θ)dθ=ρ(x)dx ? Why are they equal?
Tspirit said:[
What puzzles me is that ρ(θ)dθ=ρ(x)dx ? Why are they equal?
Yes, and it must be a one-to-one-mapping (more precisely a diffeomorphism).PeroK said:What he means is that if you take a small interval (a physical interval), then there is a definite probability that the needle lies in that interval, independent of coordinates.
If you express this probability in polar coordinates you get ##\rho (\theta) d\theta## and if you express this probability in cartesian coordinates you get ##\rho (x) dx## and, therefore, they must be equal.
What you must be careful of is that when you set up your problem, ##d \theta## and ##dx## do indeed cover the same physical interval.
It helps me to think of it this way (similar to PeroK's reply). The only way for the shadow to be between x and x + dx is for the pointer to be between the corresponding θ and θ + dθ as shown below (and vice versa). So, the probability that the shadow is between x and x + dx equals the probability that the pointer is between the corresponding values θ and θ + dθ. The values of dθ and dx that correspond to each other are given by dx = -r sinθ dθTspirit said:What puzzles me is that ρ(θ)dθ=ρ(x)dx ? Why are they equal?
Tspirit said:View attachment 106575
I can't understand the solution to Problem 1.4(a). The solution is the following:
View attachment 106576
What puzzles me is that ρ(θ)dθ=ρ(x)dx ? Why are they equal?
This explanation really helps me understanding the physical meaning of that, thank a lot!TSny said:It helps me to think of it this way (similar to PeroK's reply). The only way for the shadow to be between x and x + dx is for the pointer to be between the corresponding θ and θ + dθ as shown below (and vice versa). So, the probability that the shadow is between x and x + dx equals the probability that the pointer is between the corresponding values θ and θ + dθ. The values of dθ and dx that correspond to each other are given by dx = -r sinθ dθ
View attachment 106579
The specific problem in Griffiths' book that involves calculus is problem 3.11, which deals with finding the electric field due to a charged rod using integration.
This problem is considered challenging because it requires knowledge of both calculus and electrostatics concepts. It also involves multiple steps and requires careful manipulation of equations.
Some tips for solving this problem include drawing a diagram to visualize the problem, breaking the problem down into smaller, more manageable steps, and checking your work step by step to avoid mistakes.
Yes, there are resources available to help with this problem. You can consult your textbook or look for online tutorials and practice problems to gain a better understanding of the concepts involved.
Understanding this problem in calculus is important for building a strong foundation in mathematics and physics. It also helps develop critical thinking and problem-solving skills, which are essential for any scientific field.