The potential energy of a particle on the x-axis is given by U= 8xe ^-x ^2/19 where x is 1 meter and U is in Joules. Can you explain for me how to find the point on the x-axis for which the potential is a maximum or minimum and is this answer a point of maximum potential energy or minimum potential energy? Please show me how to figure this. Thanks
You use your calculus.
dU/dx, set to zero, solve for x
I don't know where the 19 goes, but take the derivative, set to zero
remember d/dx (uv)= vdu/dx + u dv/dx
To find the point on the x-axis where the potential energy is a maximum or minimum, we need to determine the critical points of the potential energy function.
To find the critical points, we need to take the derivative of the potential energy function with respect to x, and then find the values of x where the derivative is equal to zero.
First, let's find the derivative of the potential energy function, U(x):
U(x) = 8xe^(-x^2/19)
To find the derivative, we can use the product rule:
dU/dx = 8e^(-x^2/19) + 8x(-2x/19)e^(-x^2/19)
Simplifying further:
dU/dx = 8e^(-x^2/19) - (16x^2/19)e^(-x^2/19)
Now, set this derivative equal to zero and solve for x:
0 = 8e^(-x^2/19) - (16x^2/19)e^(-x^2/19)
To simplify the equation, let's divide both sides by e^(-x^2/19):
0 = 8 - (16x^2/19)
Cross-multiplying gives:
16x^2 = 8 * 19
Dividing both sides by 16:
x^2 = 76/16
Taking the square root:
x = ± √(76/16)
Simplifying further:
x = ± √(19/4)
Therefore, the critical points on the x-axis where the potential energy may be a maximum or minimum are x = √(19/4) and x = -√(19/4).
Now, to determine if these points represent maximum or minimum potential energy, we need to analyze the second derivative of the potential energy function.
Taking the second derivative of U(x):
d^2U/dx^2 = - (64x/19)e^(-x^2/19) - (16/19)e^(-x^2/19) + (32x^3/19^2)e^(-x^2/19)
Simplifying further:
d^2U/dx^2 = - (64x/19 + 16/19 - (32x^3/19^2))e^(-x^2/19)
Now, substitute the critical points into the second derivative equation.
For x = √(19/4):
d^2U/dx^2 = - (64√(19/4)/19 + 16/19 - (32(√(19/4))^3/19^2))e^(-√(19/4)^2/19)
Simplifying further:
d^2U/dx^2 = - (64√19/19√4 + 16/19 - (32(19√19/4)/19^2))e^(-19/19)
d^2U/dx^2 = - (64√19/19√4 + 16/19 - (32√19/19))e^(-1)
The value of d^2U/dx^2 at x = √(19/4) can be determined.
Similarly, substitute x = -√(19/4) into the second derivative equation to find its value.
If the second derivative is positive at a critical point, it indicates a local minimum point.
If the second derivative is negative at a critical point, it indicates a local maximum point.
Therefore, by determining the signs of the second derivatives at x = √(19/4) and x = -√(19/4), we can determine if these points represent maximum or minimum potential energy.