How to find the minimum distance between the two curves $e^x$ and $\ln
(x)$ ?
No idea how to find the common normal in this case.Help!
[Math] Minimum distance between $e^x$ and $\ln x$
functionsgraphing-functions
Related Solutions
So with the curve y=2^x and the arc between (2,4) and (10,1024)
Formula is: s = Int(a,b,sqrt(1+f'(x)^2)) when y is function of x (little more involved for more general curves like a circle or something)
f'(x) = ln(2)*2^x
s=Int(2,10,sqrt(1+ln(2)*2^x))
Answer is ~72.8. makes some intuitive sense. it must be longer than the straight line connecting the points, so that gives us a lower bound. sqrt(8^2 + 1020^2) = 31.8
The formula i gave above follows pretty closely from standard distance / pythagorean theorem
We need to sum up a bunch of infintesimal distances, should be a pretty familiar calculus concept. We are solving for s, the arc length
ds^2 = dx^2 + dy^2 <--divide thru by dx^2
(ds/dx)^2 = 1 + (dy/dx)^2 <-- take square root
ds/dx = sqrt(1 + (dy/dx)^2) <-- multiply by dx
ds = sqrt(1 + (dy/dx)^2)*dx <-- take integral. sub f'(x) for (dy/dx) since we have function
s = int(sqrt(1+f'(x)^2))*dx
this leaves us with indefinite integral. take definite integral between a and b to get specific result.
PS - sorry for lack of TeX skills
Are you sure that your expression for $d^2$ is correct? I get $(x_1)^4+4(x_2)^4-4(x_1)^2(x_2)^2-(x_1)^2+5(x_2)^2-2x_1x_2+1$, but I may have made a mistake.
Assuming your $d^2$ is correct (though I'm pretty sure it's not): consider the square of the distance formula as a function of $x_1$ and $x_2$ (which you've already got): $f(x_1, x_2) = 4(x_2)^4 + 5(x_2)^2-4(x_1)^2(x_2)^2-2x_1x_2$.
Now use $\frac{\partial f}{\partial{x_1}}$ and $\frac{\partial f}{\partial{x_2}}$ to find where the critical points of the function are (i.e., set both partial derivatives equal to $0$ and solve the system of equations). So, we have $2(8(x_2)^3-4x_2(x_1)^2+5x_2-x_1) = 0$ and $2(-x_2 - 4x_1(x_2)^2) = 0$.
Solving that system of equations will yield values of $x_1$ and $x_2$ that minimize $f(x_1, x_2)$, which minimizes $d^2$. From there you can get $d$.
Even if your expression for $d^2$ is incorrect, it's still the same idea to find the critical points.
Edit: your expression for $d^2$ is definitely incorrect. Going through this process with your expression gives the correct values ($x_1 = 0$ and $x_2 = 0$) but it also yields $d = 0$, which is obviously wrong as you can show very easily that these parabolas never meet (try to set them equal to each other and solve for $x$).
Best Answer
Hint:
Each graph forms the boundary of a convex region. The line segment of minimal distance between the two curves must therefore be unique.
The picture is symmetric about the line $y = x$, so reflecting the line segment through this line must yield another line segment of minimal length.
We can deduce that the slope of the common normal must therefore be $-1$.