# 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?

##### 2 Answers

#### Explanation:

Essentially, the problem is this:

What is the point on

By the distance formula, we have:

#d = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)#

#d = sqrt((0 - x_1)^2 + (0 - y_1)^2)#

#d = sqrt((-x)^2 + (-y)^2)#

#d = sqrt(x^2 + y^2)#

#d = sqrt(x^2 + (1/2x^2 -4)^2)#

#d = sqrt(x^2 + 1/4x^ 4 - 4x^2 + 16)#

#d = sqrt(1/4x^4 - 3x^2 + 16)#

#d^2 = 1/4x^4 - 3x^2 + 16#

We now differentiate with respect to

#2d((dd)/dx) = x^3 - 6x#

#(dd)/dx = (x^3 - 6x)/(2d)#

#(dd)/dx = (x^3 - 6x)/(2sqrt(1/4x^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

#2sqrt(1/4x^4 - 3x^2 + 16) = 0#

#1/4x^4 - 3x^2 + 16 = 0#

#x^4 - 12x^2 + 64 = 0#

Let

#u^2 - 12u + 64 = 0#

#u = (-(-12) +- sqrt((-12)^2 - (4 * 1 * 64)))/(2 * 1)#

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

Now for the other set of critical points.

#0 = (x^3 - 6x)/(2sqrt(1/4x^4 -3 x^2 + 16))#

#0 = x^3 - 2x#

#0 = x(x^2 - 6)#

#x = 0 and +- 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

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

**Test Point #x = -1#**

#(-1)^3 - 2(-1) = -1 + 2 = 1#

We can instantly eliminate

**Test Point: #x = -3#**

#(-3)^3 - 2(-3) = -27 + 6 = -21#

Since

Since quadratic functions are symmetric,

However, we must find the distance, therefore:

#d = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)#

#d = sqrt((sqrt(6) - 0)^2 + (-1 - 0)^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

Hopefully this helps!

#### Explanation:

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

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

We wish to optimise:

- and

Basic idea;

The optimised distance is therefore:

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

As

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