U = b (a^2 over x^2 - a over x)
or, more readably,
U = b(a^2/x^2 - a/x)
We can ignore b, since it does not affect where the extrema occur. It's just a scale factor.
U = a^2/x^2 - a/x = (a^2-ax)/x^2 = a(a-x)/x^2
clearly the asymptote is at x=0
(asymptotes occur when the denominator is zero)
U=0 where x=a
dU/dx = -2a^2/x^3 + a/x^2
= (-2a^2+ax)/x^3 = a(x-2a)/x^3
clearly dU/dx=0 when x=2a
For positive a, b, the potential energy, U, of a particle is given by
U = b (a^2 over x^2 - a over x) for x is greater than 0
a) find the x-intercept
b) find the asymptotes
c)compute the local minimum
d) sketch the graph
I don't know how to start this problem
Well first I'd assume that we would have to find the derivative
so u= (a^2x^-2) -ax^-1
so u' = -2a^2x^-3 - ax^-2?
and then do we have to factor and set it equal to zero to find the x-intercepts?
Are finding the x-intercepts the same thing as finding the critical points?
How do you find the asymptotes? Do we just graph what we got for the derivative and then see? But wait we can't graph the derivative because of the many unknown variables.
I am unable to find the x-intercept, asymptotes, local minimum and graph. Please help me!
1 answer