# Trouble with finding shortest distance between $y=x^2$ and $y=x-1$ with Lagrangian multipliers

I've found another thread with a similar question, but none of the answers help with the specific part I'm stuck on.

Just to make things simpler, I've used the square of the distance $f(x_1,x_2,y_1,y_2)=(x_1-x_2)^2+(y_1-y_2)^2$, and I have constraints $G_1=x_1^2-y_1$, and $G_2=x_2-y_2-1$.

Taking the gradients of $f$, $G_1$, and $G_2$, I have the system of equations

$2(x_1-x_2)=2\lambda_1x_1$

$2(y_1-y_2)=-\lambda_1$

$-2(x_1-x_2)=\lambda_2$

$-2(y_1-y_2)=-\lambda_2$

What I've done so far is, I've first observed that the immediate implication of the 2nd and 4th equations is that $\lambda_1=-\lambda_2$. Using this, I cancelled out the 1st and 3rd equations with the substituted value for $\lambda_2$, and got that $2\lambda_1x_1=\lambda_1\implies x_1=\frac{1}{2}$, and by constraint 1 that $y_1=\frac{1}{4}$. From here though, I'm not sure how to pin down the value of $\lambda$, and therefore determine $x_2$ and $y_2$.

• From equation 1 you obtain $x_2 =(1-\lambda_1)/2$, from equation 2 you obtain $y_2= (1+2 \lambda_1)/4$. Using the constraint $G_2$, you obtain $\lambda_1 = -3/4$ and with that the remaining values. – Fabian Jan 20 '17 at 9:18

The shortest distance between the parabola $y = x^2$ and straight line $y = x - 1$ occurs when the tangent to the parabola is parallel to the line, i.e. when their slopes are equal. Now, slope of tangent of parabola $y = x^2$ is

$$m = \frac{dy}{dx} = 2x$$

and the slope of the straight line $= 1$. Hence the shortest distance occurs at $x = 1/2$. The point on the parabola corresponding to $x = 1/2$ is $(1/2, 1/4)$. Hence, perpendicular distance of this point from the given straight line is:

$$d = \frac{|0.5 - 0.25 - 1|}{\sqrt{2}} = \frac{3}{4\sqrt{2}}$$

which is the shortest distance between the curves.

EDIT: This, I guess, isn't the answer you're looking for. But I find this approach much easier and shorter than using Lagrange multipliers. Hence just shared this technique.

• Yeah, normally this would've been the technique I would've used, given how intuitive its approach is. I'm specifically preparing for a test involving Lagrange multipliers though, which is why I'm using an otherwise overkill method. – George Jan 20 '17 at 8:00

the distance between the curves is given by $$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$ with $$y_1=x_1^2,y_2=x_2-1$$ with these equations we obtain $$\sqrt{(x_2-x_1)^2+(x_2-1-x_1^2)^2}$$ now you must differentiate this with respect to $$x_1,x_2$$

• What do I use as the constraints for the Lagrange multiplier in this case? – George Jan 20 '17 at 7:55

Perhaps one could use Lagrange multipliers in more elegant way, but this will work.

Define $$D(x,y,x',y') = (x-x')^2+(y-y')^2$$ and $$g(x,y,x',y') = (y-x^2)^2 + (y'-x'+1)^2.$$

Then you want to optimize $D$ with respect to $g(x,y,x',y') = 0$. Calculate gradients:

$$\nabla D = 2(x-x',y-y',-(x-x'),-(y-y'))\\ \nabla g = 2((y-x^2)(-2x),y-x^2,-(y'-x'+1),y'-x'+1)$$

Set $\nabla D = \lambda \nabla g$ to get system

\begin{align} x-x' &=-2\lambda x(y-x^2)\\ y-y' &=\lambda (y-x^2)\\ x-x' &=\lambda (y'-x'+1)\\ y-y' &=-\lambda (y'-x'+1)\\ \end{align}

From equations 1 and 2 we get $$x-x' = -2x(y-y')$$ and from equations 3 and 4 we get $$x-x' = -(y-y')$$ which gives us $$(2x-1)(y-y') = 0$$ You can quickly check that $y-y' = 0$ is impossible since it would imply that $x-x'= 0$ as well, but our parabola and line do not intersect. Thus, $x = \frac 12$. Now you know $y = \frac 14$ and from the system \begin{align}x-x' &= y'-y\\ y' &= x'-1\end{align} you can get $x'$ and $y'$. This will give you your distance.