# APC.3.1.2 shortest distance between curve and origin

• MHB
• karush
In summary, the x-coordinate of the point on $f(x)=\dfrac{4}{\sqrt{x}}$ that is closest to the origin is (2, 4/\sqrt{2})

#### karush

Gold Member
MHB
Find the x-coordinate of the point on $f(x)=\dfrac{4}{\sqrt{x}}$
that is closest to the origin.

a. $1$
b. $2$
c $\sqrt{2}$
d $2\sqrt{2}$
e $\sqrt[3]{2}$

not real sure but, this appears to be dx and slope problem
I thot there was an equation for shortest distance
between a point and a curve but couldn't find it offhand

$D^2 = (x-0)^2 + \left(\dfrac{4}{\sqrt{x}} - 0\right)^2 = x^2 + \dfrac{16}{x}$

minimizing $D^2$ will minimize $D$ ...

$\dfrac{d(D^2)}{dx} = 2x - \dfrac{16}{x^2} = 0$

finish and confirm the value is a minimum

Another approach would be Lagrange Multipliers (optimization with constraint. The objective function could be the square of the distance:

$$\displaystyle f(x,y)=x^2+y^2$$

Subject to the constraint:

$$\displaystyle g(x,y)=y-\frac{4}{\sqrt{x}}=0$$

Hence:

$$\displaystyle 2x=\lambda\left(2x^{-\frac{3}{2}}\right)$$

$$\displaystyle 2y=\lambda(1)$$

This implies:

$$\displaystyle y=\frac{x^{\frac{5}{2}}}{2}$$

Substituting into the constraint, there results:

$$\displaystyle \frac{x^{\frac{5}{2}}}{2}-\frac{4}{\sqrt{x}}=0$$

This leads to the same root as above, and to verify it is a miniimum we could pick another point on the constraint to verify the objective function is greater at that point than at our critical point.

interesting,,,

I've never did anything with Lagrange

x=2

My first thought would be to just try each possibility:
a) x= 1. The point is (1, 4) which has distance $\sqrt{17}$, about 4.12 from the origin.
b) x= 2. The point is (2, 4/\sqrt{2}) which has distance $\sqrt{4+ 8}= \sqrt{12}$, about 3.46, from the origin.
c) $x= \sqrt{2}$. The point is $(\sqrt{2}, 4/\sqrt[4]{2})$ which has distance $\sqrt{2+ 16/\sqrt{2}}= \sqrt{2+ 8\sqrt{2}}$, which is about 3.65, from the origin.
d) $x= 2\sqrt{2}$. The point is $(2\sqrt{2}, 4/\sqrt[4]{8})$ which has distance $\sqrt{8+ 16/\sqrt{8}}= \sqrt{8+ 8/\sqrt{2}}= \sqrt{8+ 4\sqrt{2}}$, which is about 3.70 from the origin.
e) $x= \sqrt[3]{2}$. The point is $\sqrt[3]{2}, 4/\sqrt[6]{2})$ which distance $\sqrt{\sqrt[3]{4}+ 16/\sqrt[3]{2}}$ which is about 3.78 from the origin.

Of the four distances, the smallest is 3.46 so (b) x= 2 gives the point closest to the origin!

Heavy use of calculator, light use of brain!