Question

An object must maintain a radius of 20 around an origin. It applies thrust along the x and y axes with a total thrust that is the square of the difference of the distance. How much thrust is applied along x?

Answer 1

Hmm... Lemme see if I get it right. You have the x,y of the object and it's 'd' far away from the middle of the circle (in radius of 20). you want to express that thrust for x using y and the parameters?

Answer 2

If y = 0, x = 22\[thrust = (x-20)^{2} = 4\]easy. But if the distance remains the same and is moved around the circle. Say x = 14 and d is still 22. How do I solve for the x component of the thrust?

Answer 3

sounds like you're building a game or something hehe
k thinking. hopefully i could help

Answer 4

I'm guessing I could use the slope, but my brain hurts from thinking about this for so long.

Answer 5

Well, the way I see it, if x and y are the sides of that triangle then.
\[
\angle K = arctan(\frac{y}{x})
\]
Now in the inner triangle, the angle is the same, because 'a' is parallel to the x axis.
so we can say
\[
d = \sqrt{x^2 + y^2} \\
a = cos(\angle K) \cdot (d - 20) \\
b = sin(\angle K) \cdot (d - 20) \\
a^2 + b^2 = (d-20)^2
\]
Is that ok?

Answer 6

That's assuming (0,0) is the middle of the circle

Answer 7

hmm.. I looked at them like lengths.. I should have used absolute values =|

Answer 8

Origin is (0, 0)

Answer 9

if x = 14 and d = 22 then y = 20, Pythagorean

Answer 10

lemme fix it sec..

Well, the way I see it, if x and y are the sides of that triangle then.
\[
\angle K = arctan \left( \begin{array}{l l} \frac{|y|}{|x|} \end{array} \right)
\]

Answer 11

Maybe if you'll be more specific about the desired behavior of this thing I could help better..

Answer 12

No you're right! The value of thrust is in the direction at angle k from x.

Answer 13

I'm confused, I don't really understand what are you trying to do... What I gave you felt like the most... trivial solution... but you didn't define any connection between x and y.. Since I already have to decide about this connection by myself, I could do just the same:
\[
\text{thrust_x}^2 + \text{thrust_y}^2 = (d - 20)^2 \\
\text{thrust_x} = \text{thrust_y} \\
2 \cdot \text{thrust_x}^2 = (d - 20)^2 \\
\text{thrust_x} = \frac{ \sqrt{d - 20} }{ \sqrt{ 2 } }\\
\]
That's obviously not what you want... but it shows how important it is so you explain what are you trying to do. perhaps the solution I gave satisfies the condition you set up, but isn't good for your purpose..