Find the minimum distance between the curves y=e^x and y=lnx. Hint: Use the fact that e^x and lnx are inverse relationships. I have no idea where to start. Thanks!

Question

calculus - Reiny, Monday, January 18, 2016 at 9:02pm
to get you going:

let P(a , e^a) and Q(b, lnb) be the closest points on their respective graphs

since they are inverses the line PQ must be perpendicular to y = x
but y = x has a slope of 1, so the
sope of PQ = -1

See what you can do with that.

I tried to start with that, and using graphs, I got (0,1) on e^x and (1,0) on lnx, with the distance of 2sqrt2. But I don't know how to do this algebraically.

Answer 1

for y=e^x, the slope of the normal at x=a is -e^-a

for y=lnx, the slope of the normal at x=b is -b.

So, we want to find a and b such that the normals through (a,e^a) and (b,lnb) are the same line. That is,

y-e^a = -e^-a(x-a)
and
y-lnb = -b(x-b)

eliminating y, we then want

-e^-a*x + ae^-a+e^a = -bx+b^2+lnb

For these two polynomials to be identical, we need

-e^-a = -b
ae^-a+e^a = b^2+lnb

The first equation says that b = e^-a. Using that in the 2nd equation, we have

ae^-a + e^a = e^-2a - a

It is easy to see that a solution to this is a=0. That means that the two closest points on the curves are

(0,1) and (1,0)

and the minimum distance is thus √2.