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
Noah G
Mar 8, 2017

#2.65# million kilometers.

Explanation:

Essentially, the problem is this:

What is the point on #y = 1/2x^2 - 4# that is closest to #(0, 0)#?

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 #x#.

#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 #0#.

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

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

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

Let #u = x^2#. Then the equation becomes.

#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 #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#

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

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

Since #-3 <= -sqrt(6) <= -1#, we have that #-sqrt(6)# is a minimum.

Since quadratic functions are symmetric, #sqrt(6)# will also be a minimum. Therefore, the points #(sqrt(6), -1)# and #(-sqrt(6), -1)# are closest to #(0, 0)#.

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 #sqrt(7)# million kilometers, which is #2.65# million kilometers, nearly.

Hopefully this helps!

Eddie
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(x,y) = sqrt(x^2 + y^2)#, the distance formula applied between any point on trajectory and the Origin

  • and

#g(x,y) = y - x^2/2 + 4#, the constraint, which the equation of the trajectory

Basic idea;

#nabla f = lambda nabla g#

#implies ((x/sqrt(x^2 + y^2)),(y/sqrt(x^2 + y^2))) = lambda ((-x),(1))#

#lambda = (-1)/sqrt(x^2 + y^2) = y/sqrt(x^2 + y^2)#

#implies y = -1#

#implies x = pm sqrt 6#

The optimised distance is therefore: #f(sqrt 6, 1) = sqrt 7#

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

#f(0,-4) = 4#

#f(pm 2 sqrt 2, 0) = 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]}

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