calculusmultivariable-calculusoptimization
I have 2 graphs $y = x$ and $y = x^2 + 1$. I need to find the minimum distance between the 2 of them.
I've determined the parametric equations for them:
$r(s) = (s, s)$
and
$r(t) = (t, t^2 + 1)$
I've determined the distance formula to be:
$d = \sqrt{(t-s)^2 +(t^2+1-s)^2}$
However, from this point I'm stuck. I know I need to do some form of partial differentiation.
I don't know what I need to do with partial derivatives with respect to s and t for me to determine the distance.
In the first part, we are taking $s$ to be some fixed (but unspecified) value. The distance from a point on the parametric line to the given point is indeed $$\sqrt{(1+t)^2+\left(2-\frac12t\right)^2+(s-2t-1)^2}.\tag{$\star$}$$ To find the distance from the parametric line, itself to the given point, we must minimize this. It suffices to minimize its square. To do this, we will take the partial derivative of its square with respect to $t,$ set it equal to $0,$ solve for $t$, and plug it back into $(\star),$ to give us a function in terms of $s.$
Then, to find $s$ such that the distance is at a minimum, we will square the resulting function and minimize, by taking the derivative with respect to $s,$ setting equal to $0,$ and solving for $s$.
The minima , in the interior of the elliptical region occurs at (-1,0) where z takes the value -1. On the boundary of the ellipse, the maxima occurs at $(2\sqrt(6,0)$ where z takes the value $4(6 + \sqrt(6)$. Notice there are three more critical points here, one is a local maxima, two are local minima .
I've updated this drawing since I discovered that I wrote down the parameters of the elliptical cylinder wrong, sorry !
Again, just for fun, assume the y coordinate is zero, then $x^2 = 24$ , so x is $2\sqrt{6}$ , yes the negative root should also be checked. And, $2x + x^2$ takes the value $24 + 4\sqrt{6}$ a local maxima.
Best Answer
If you must use calculus instead of geometry, you would be much better off minimising the square of the distance. The partial derivatives are then pretty straight-forward. A point suspected of being a minimum will be a point where both partial derivatives are equal to zero. One partial derivative of $d^2$ is $$-2 (-s + t) - 2 (1 - s + t^2)$$ and the other is $$2 (-s + t) + 4 t (1 - s + t^2)$$ You can now find a point where both are equal to zero.