# Related Rates

## Homework Statement

Let g(x) = x^2 + 12x + 36, a parabola with one zero at x = -6. The parabola moves downward at a rate of 2 units/sec. How fast is the distance between the zeroes changing when they are 10 units apart?

## Homework Equations

g(x) = x^2 + 12x + 36

## The Attempt at a Solution

I figured since this is a parabola, the zeroes would simply be equidistant from -6, so at 10 units apart they would be at x=-1 and x=-11.

Next I took the derivative of y = x^2 + 12x + 36 to obtain
dy/dt = (2x + 12)(dx/dt) and I substituted -2 for dy/dt since this parabola is moving down and solved for dx/dt to get
dx/dt = -2/(2x+12). Next I substituted in -11 and -1 for x and obtained, respectively, dx/dt = 1/5 and -1/5. But this doesn't seem to make sense: doesn't this mean they're moving closer to each other, when they should be going further away? The correct answer is 2/5, so if the signs were what I expected I would've had it correct, but I can't figure out what's wrong.

It's worth mentioning that I was given a solution that was quite distinct from the manner in which I tried to solve it, but I didn't understand it either. This is the provided solution:
y = ax^2 + bx +c, therefore dc/dt = -2. Distance between roots = (-b+(b^2-4ac)^(1/2))/(2a) - (-b-(b^2-4ac)^(1/2))/(2a) = ((b^2-4ac)^(1/2))/a = (144-4c)^(1/2). Distance = 10 when c = 11, thus the rate at which they are moving apart is 2/5.

I understand it up until the last step, where I don't see how the jump was made. Thanks in advance for the help!

We can simplify that a bit by noting that x2+ 12x+ 36= (x+ 6)2 so we are really talking about y= (x+6)2- 2t. The zeros are given by (x+6)2= 2t so $x= -6\pm\sqrt{2t}$. The distance between those zeros is $D= 2\sqrt{2t}$. You could differentiate that directly but I think it is easier to write D2= 8t and then differentiate: 2D dD/dt= 8 so dD/dt= 4/D. When the zeros are 10 units apart, D= 10 so dD/dt= 4/10= 2/5 unit per second.