Delta function to calculate density of probablity

[anaisabel]

Homework Statement

Find integral of delta function in problem

Relevant Equations

equation 1 from solution

 

[vela]

In the future, please type your post rather than posting images of your work. Your writing (to me at least) isn't the easiest to read.

In any case, what do you mean by "my teacher breaks the integral"? Also, where does the length of the crocodile enter into the solution?

 

[anaisabel]

Sorry , my handwriting is a bit messy, but basically he integrates fron 0 to pi/4 and then from pi/4 to pi/2.
The length of the crocodile isn't really relevant, i just wrote the problem for some context. What i really want to know is how to solve that integral. I know you have to change variables because inside the dirac there is a function depending on theta, i just don't know how to do it. It is the calculus part that i can't figure out.

 

[vela]

I'd say it has to do with the fact that the range is maximum when $\theta = \pi/4$.

 

[anaisabel]

I'd say it has to do with the fact that the range is maximum when $\theta = \pi/4$.

Can you elaborate a bit more, i am not understanding very well.

 

[vela]

Try calculating the projectile's range for $\theta=\pi/6$ and $\theta=\pi/3$. You'll get the same distance. Do you see how that would complicate the evaluation of the original integral?

 

[anaisabel]

I have forgotten a bit of my calculus, so I know the distance is the same, but if i don't know why it would complicate?.

 

[vela]

The length of the crocodile isn't really relevant.

I don't see how it couldn't be. If the crocodile were infinitely long and lying on the +x axis, for instance, the probability of hitting it would be 1.

I have forgotten a bit of my calculus, so I know the distance is the same, but if i don't know why it would complicate?.

What have you tried to do in evaluating the integral?

 

[anaisabel]

I don't see how it couldn't be. If the crocodile were infinitely long and lying on the +x axis, for instance, the probability of hitting it would be 1.What have you tried to do in evaluating the integral?

In when comes to length of the cocrodile, I haven't gotten that far because I can't solve the integral. I have tried to solve it, and I understand now why you have to separate in two intervals, from 0 to pi/4 and pi/4 to pi/2, because when you perform tha change of variable and change the limits it would go from 0 to 0 if you didnt break into intervals. I understood that, but what I don't understand after. After I separate into intervals and perform the change of variable y like it is in my attempt, I don't know how to evaluate the integral, using the equation 1. The density is 2/pi, so is a constant, so what is the meaning of what is inside of delta?

 

[vela]

The problem says $\theta$ is uniformly distributed, which is a continuous distribution. If you take the crocodile to be a point particle, you're asking for the probability that $\theta$ take on one or two particular values. When you have a continuous distribution, that probability would be 0. I'm not sure why you're using a delta function here or what your thinking is behind your integral.

Nevertheless, here's an example of how to change variables. Consider
$$\int f(x) \delta(x^2-a^2)\,dx,$$ where $a>0$. Let's call the argument of the delta function $g(x) = x^2-a^2$. The formula for the change of variables is
$$\delta(g(x)) = \sum_i \frac{\delta(x-x_i)}{\lvert g'(x_i) \rvert},$$ where the sum is over all the roots of $g$. In this example, $g(x)=0$ at $x=a$ and $x=-a$, so we get
$$\int f(x) \delta(x^2-a^2)\,dx = \int f(x)\left[\frac{\delta(x-a)}{2a} + \frac{\delta(x-(-a))}{2a}\right]\,dx = \frac{f(a)+f(-a)}{2a}.$$