Plot always appears as a straight line

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v = [0:100];
m = 4.65*10.^-26;
y = ((m*v)/(4.14*10.^-21)).*exp((-m*v.^2)/2*4.14*10.^-21);
plot(v,y)
title('Density fucntion at t1 and t2')
Hi, I don't know how to fix this but my graph appears to be a straight line intead of an exponential curve. Thanks a lot!
  1 Commento
Roger Stafford
Roger Stafford il 5 Dic 2017
@Hieu. Here's a demonstration of why you should be careful with how you use parentheses:
x = 60/5*12
y = 60/(5*12)
z = 60/5/12
Notice that in 'y' and 'z' there is division by 12, whereas in 'x' it is multiplied. That is apparently the source of your difficulty with the quantity 4.14*10.^-21. It was multiplied rather than divided because you left out the all-important parentheses.

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Risposta accettata

Roger Stafford
Roger Stafford il 4 Dic 2017
Are you sure the second "4.14*10.^-21" doesn't belong in the denominator:
y = ((m*v)/(4.14*10.^-21)).*exp((-m*v.^2)/(2*4.14*10.^-21));
If so, and if you change the range of v to:
v = [0:1000];
you will get a nicely curved plot.
  2 Commenti
Hieu Nguyen
Hieu Nguyen il 4 Dic 2017
Let's define that quantity as A. I am trying to plot this function: f(x) = ((mx)/A) * exp((-mv^2)/2A). I still get a straight line.
Greg
Greg il 4 Dic 2017
Modificato: Greg il 4 Dic 2017
Did you put the parenthesis around 2.*A?

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Più risposte (1)

Greg
Greg il 4 Dic 2017
First, I suspect you missed parenthesis is your exp denominator. The magic of using constants. Second, most functions look like straight lines if you stay near the origin. Try plotting further out in v.
v = 0:1000;
m = 4.65*10.^-26;
denom = (4.14*10.^-21);
y = ((m*v)/denom).*exp((-m*v.^2)/(2*denom));
a = axes(figure);
plot(a,v,y);
title('Density function at t1 and t2')
  2 Commenti
Hieu Nguyen
Hieu Nguyen il 4 Dic 2017
I see. Thanks a lot! One of the reasons I am not fond of matlab sometimes.
Greg
Greg il 4 Dic 2017
Modificato: Greg il 4 Dic 2017
That has nothing to do with MATLAB. That's a mathematical given.
Also, it was unfair to accept my answer. It was identical, and 10 minutes later than, Roger Stafford's answer.
Please unaccept mine and accept his.

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