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?

Answer 1

#2.65# million kilometers.

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!

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

#sqrt 7 times 10^6#km

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
#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]}

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 3

The satellite gets as close as 8 million kilometers to the planet.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7