# A satellite following the equation y = 1/2x^2 - 4, where x and y are in millions of kilometres, is surveying a far away planet, located at the origin (0,0). How close to the planet does the satellite get?

Mar 8, 2017

$2.65$ million kilometers.

#### Explanation:

Essentially, the problem is this:

What is the point on $y = \frac{1}{2} {x}^{2} - 4$ that is closest to $\left(0 , 0\right)$?

By the distance formula, we have:

$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

$d = \sqrt{{\left(0 - {x}_{1}\right)}^{2} + {\left(0 - {y}_{1}\right)}^{2}}$

$d = \sqrt{{\left(- x\right)}^{2} + {\left(- y\right)}^{2}}$

$d = \sqrt{{x}^{2} + {y}^{2}}$

$d = \sqrt{{x}^{2} + {\left(\frac{1}{2} {x}^{2} - 4\right)}^{2}}$

$d = \sqrt{{x}^{2} + \frac{1}{4} {x}^{4} - 4 {x}^{2} + 16}$

$d = \sqrt{\frac{1}{4} {x}^{4} - 3 {x}^{2} + 16}$

${d}^{2} = \frac{1}{4} {x}^{4} - 3 {x}^{2} + 16$

We now differentiate with respect to $x$.

$2 d \left(\frac{\mathrm{dd}}{\mathrm{dx}}\right) = {x}^{3} - 6 x$

$\frac{\mathrm{dd}}{\mathrm{dx}} = \frac{{x}^{3} - 6 x}{2 d}$

$\frac{\mathrm{dd}}{\mathrm{dx}} = \frac{{x}^{3} - 6 x}{2 \sqrt{\frac{1}{4} {x}^{4} - 3 {x}^{2} + 16}}$

We're now going to find the critical points by

a) Finding where the derivative is undefined
b) Finding where the derivative is $0$

First of all, the derivative is undefined whenever the denominator is equivalent to $0$.

$2 \sqrt{\frac{1}{4} {x}^{4} - 3 {x}^{2} + 16} = 0$

$\frac{1}{4} {x}^{4} - 3 {x}^{2} + 16 = 0$

${x}^{4} - 12 {x}^{2} + 64 = 0$

Let $u = {x}^{2}$. Then the equation becomes.

${u}^{2} - 12 u + 64 = 0$

$u = \frac{- \left(- 12\right) \pm \sqrt{{\left(- 12\right)}^{2} - \left(4 \cdot 1 \cdot 64\right)}}{2 \cdot 1}$

We can see very quickly that there exists no real solution to this equation.

Now for the other set of critical points.

$0 = \frac{{x}^{3} - 6 x}{2 \sqrt{\frac{1}{4} {x}^{4} - 3 {x}^{2} + 16}}$

$0 = {x}^{3} - 2 x$

$0 = x \left({x}^{2} - 6\right)$

$x = 0 \mathmr{and} \pm \sqrt{6}$

Next, we must check whether these are maximums or minimums. We want to minimize distance, so we will have a minimum. Check on both sides of each critical point. If the derivative is inferior to $0$ on the left-hand side and greater than $0$ on the right-hand side, we have a minimum.

We don't even have to check the entire derivative. The sign of the derivative, positive or negative will only be influenced by the numerator because a sqrt is always positive.

Test Point $x = - 1$

${\left(- 1\right)}^{3} - 2 \left(- 1\right) = - 1 + 2 = 1$

We can instantly eliminate $x = 0$ as the minimum because the derivative is increasing to the left of the point.

Test Point: $x = - 3$

${\left(- 3\right)}^{3} - 2 \left(- 3\right) = - 27 + 6 = - 21$

Since $- 3 \le - \sqrt{6} \le - 1$, we have that $- \sqrt{6}$ is a minimum.

Since quadratic functions are symmetric, $\sqrt{6}$ will also be a minimum. Therefore, the points $\left(\sqrt{6} , - 1\right)$ and $\left(- \sqrt{6} , - 1\right)$ are closest to $\left(0 , 0\right)$.

However, we must find the distance, therefore:

$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

$d = \sqrt{{\left(\sqrt{6} - 0\right)}^{2} + {\left(- 1 - 0\right)}^{2}}$

$d = \sqrt{6 + 1}$

$d = \sqrt{7}$

Since distance is a scalar quantity, it will be this in any direction on the plane. Finally, we're talking millions of kilometers, so the closest distance is $\sqrt{7}$ million kilometers, which is $2.65$ million kilometers, nearly.

Hopefully this helps!

Mar 8, 2017

$\sqrt{7} \times {10}^{6}$km

#### Explanation:

Alternative approach, if you are happy with the Lagrange Multiplier.

In simple units, actual distance units quoted above...

We wish to optimise:

$f \left(x , y\right) = \sqrt{{x}^{2} + {y}^{2}}$, the distance formula applied between any point on trajectory and the Origin

• and

$g \left(x , y\right) = y - {x}^{2} / 2 + 4$, the constraint, which the equation of the trajectory

Basic idea;

$\nabla f = \lambda \nabla g$

$\implies \left(\begin{matrix}\frac{x}{\sqrt{{x}^{2} + {y}^{2}}} \\ \frac{y}{\sqrt{{x}^{2} + {y}^{2}}}\end{matrix}\right) = \lambda \left(\begin{matrix}- x \\ 1\end{matrix}\right)$

$\lambda = \frac{- 1}{\sqrt{{x}^{2} + {y}^{2}}} = \frac{y}{\sqrt{{x}^{2} + {y}^{2}}}$

$\implies y = - 1$

$\implies x = \pm \sqrt{6}$

The optimised distance is therefore: $f \left(\sqrt{6} , 1\right) = \sqrt{7}$

In terms of showing this to be a min, note from geometry that:

$f \left(0 , - 4\right) = 4$

$f \left(\pm 2 \sqrt{2} , 0\right) = 2 \sqrt{2}$

As $\sqrt{7} < 4$ and $\sqrt{7} < 2 \sqrt{2}$, and the function is continuous throughout, this is a minimum distance.

graph{1/2x^2 - 4 [-10, 10, -5, 5]}