Finding the maximum point of a function

[AStaunton]

trying to show that the maximum of a certain function is \sqrt{\frac{2kT}{m}}.

The function in question is:

f(v)=4\pi(\frac{m}{2\pi kT})^{\frac{3}{2}}v^{2}e^{-\frac{mv^{2}}{2kT}}

so to do this find df/dv and set that = 0 and the value for v will be the v_max:

\frac{df}{dv}=2ve^{-\frac{mv^{2}}{2kT}}-\frac{2kT}{m}e^{-\frac{mv^{2}}{2kT}}=e^{-\frac{mv^{2}}{2kT}}(2v-\frac{2kT}{m})=0

now divide out the (2v-2kT/m) term leaves:

e^{-\frac{mv^{2}}{2kT}}=0

take natural log of both sides:

-\frac{mv^{2}}{2kT}=1\implies v=\sqrt{-\frac{2kT}{m}}

The answer I found was wrong by a minus sign, can anyone tell me where along the line I somehow created this extra minus? Also, I'm not sure if it is completely kosher to divide out the (2v-2kT/m) term as I did, so also please give clarification on this.

And final question; if the dividing out the (2v-2kT/m) is a correct method to solve this problem, should it also be possible to divide out the exponential term instead...that is when I'm at this point:

e^{-\frac{mv^{2}}{2kT}}(2v-\frac{2kT}{m})=0

should I be able to divide out the exponential term and then also expect to find an identical value for v_max, ie. solve the following for v and this will be my v_max:

2v-\frac{2kT}{m}=0\implies v=\frac{m}{kT}

clearly this is not the value for v_max that was stated in the start of the problem...so clarity on this is much appreciated

 

[tiny-tim]

Hi AStaunton!

Sorry, but this is completely wrong …

your product rule derivative is wrong,

e^x is never 0 (unless x = -∞),

and yes of course you can't discard that (wrong) bracketed term.

Start again, carefully. ​

 

[AStaunton]

OK...I think this is the correct derivitive now, (noting that I did not include the constant multiplying term as this is going to divide out anyway):

\frac{df}{dv}=2ve^{-\frac{mv^{2}}{2kT}}-\frac{mv^{4}}{2kT}e^{-\frac{mv^{2}}{2kT}}=e^{-\frac{mv^{2}}{2kT}}(2v-\frac{mv^{4}}{2kT})=0

at this point, my idea of dividing out one or other of the terms and solving for v was the only trick I had...and you say this is incorrect. any advice on how to proceed is appreciated

 

[tiny-tim]

Hi AStaunton!

nope, still wrong!

you needed to use the chain rule as well as the product rule …

you can't just multiply by the whole of that exponent.

(And when your bracket is correct, you know that it has to equal zero.)

 

[AStaunton]

thanks.

Is it just a peculiarity of the exponential function that I had to focus on the bracketed term..
ie. in general it is OK to divide out either of the terms to get a valid answer for v?
of course, really what this is doing is finding the roots of the equation df/dv..

 

[tiny-tim]

… in general it is OK to divide out either of the terms to get a valid answer for v?

In general, if AB = 0, then both A = 0 and B = 0 will be solutions.

The only difference in this case is that A can't be 0.