Question

find the most probable velocity in this function...confiused

Answer 1

what function?

Answer 2

What is this equation, i.e., what is it that it represents?

Answer 3

function probabilty of speed...

Answer 4

so the probability of having speed v is f(v) ?

Answer 5

yes my friend, how i solved this problem?

Answer 6

Well, that means v is anything from 0 to infinity. Use calculus to find the max.

It should be fairly obvious actually that this function is decreasing, so the max is when v = 0.

Answer 7

Now, to find C, the sum of all the probabilities is 1. Hence
\[ \int_0^\infty f(v) \ dv = 1 \]

Use that relationship to find what the value of C must be.

Answer 8

You will need to know this integral:
\[ \int_0^\infty e^{-ax^2} \ dx = \frac{\sqrt{\pi}}{2\sqrt{a}}\]

Answer 9

One last thing. Double check the definition of the function f(v). Does v go from 0 to infinity, or from -infinity to +infinity?

If it is the second, then modify the integral above.

The max of f(v) remains the same for either domain of velocities v.

Answer 10

i think from -infinity to +infinity

Answer 11

Ok. In that case, you need
\[ \int_{-\infty}^\infty f(v) \ dv = 1 \] Notice the function f(v) is an even function (f(-v) = f(v)), so you will still need and can use the other integral identity I wrote above.

Answer 12

after we get value of C, and then put v=0 in this equation?

Answer 13

This kind of function is a fundamental one in probability, and you should learn this bell shape.
http://www.wolframalpha.com/input/?i=e%5E%28-x%5E2%29

Given your function f(v) = Cexp(-av^2) for some constant a, isn't it obviously it has a max at v = 0? That's because

f'(v) = -2av.C.exp(-av^2) = 0 when v = 0.

Answer 14

yes,this funct have a max when v=0