Should that be U= instead of U+ ?
Is the /19 part of the exponent,
as in U = 8x e^[(-x^2)/19] ?
That equation has a maximum when x = 0, and no minimum. It asymptotically approaches 0 and x = + and - infinity.
The potential energy of a particle on the x-axis is given by U+ 8xe ^-xsquared/19 where x is 1 meter and U is in Joules. Please find the point on the x-axis for which the potential is a maximum or minimum. I've worked it three times and I get these three answers. Which one is correct?
4.359 m
3.082 m
1.202 m
I now think it is 3.082m. Is this correct?
No one has answered this question yet.
3 answers
yes you are correct that U = 8x e^[(-x^2)/19] is the way the problem reads and the choices are
a)-4.359 m b)4.359 m c) -3.082m
d)3.082 m e)1.202 m f) -1.202 m
a)-4.359 m b)4.359 m c) -3.082m
d)3.082 m e)1.202 m f) -1.202 m
Thanks for clarifying that. I was wrong in stating that the maximum is at x=0, because of the x term multiplying the exponenital.
The derivative dU/dx = 0 when
8 + 8x*(-2x/19)= 0
x^2 = 19/2
x = + or -3.082
One is a maximum and the other is a minimum. You figure out which.
The derivative dU/dx = 0 when
8 + 8x*(-2x/19)= 0
x^2 = 19/2
x = + or -3.082
One is a maximum and the other is a minimum. You figure out which.
1 answer