Post by: luffy on April 18, 2011, 07:29:00 pm
I'm not entirely sure if this question is within the scopes of the specialist maths course because I couldn't do it :P I didn't get it from any textbook, and I just thought up the question myself. It seems like a very basic question, yet I couldn't do it... >:(
Question: "Find the minimum distance between the two graphs
I know how to do the question if a point on either graph is known. However, in this case, neither is known.
Good luck and thanks in advance :D
Post by: brightsky on April 18, 2011, 07:41:07 pm
the parabola has gradient dy/dx = 2x. we want 2x = 1, since the tangent is parallel to y = x, which means x = 1/2. so Q = (1/2, 5/4)
so the aforementioned perpendicular line has equation y - 5/4 = -(x - 1/2) --> y = -x + 7/4
we want this to intersect y = x, so x = -x + 7/4, 2x = 7/4, x = 7/8. so y = 7/8. So (7/8, 7/8) is the closest point.
hopefully there's no computational errors.
Post by: jane1234 on April 18, 2011, 07:47:36 pm
Find dy/dx of that.
Let dy/dx = 0 (to get minimum distance)
x=0.5
EDIT: Just read question again and it asked for minimum distance. Sub x=0.5 into distance equation and you get 0.75
Post by: luken93 on April 18, 2011, 07:48:17 pm
If we let f(x) = x^2 + 1 and g(x) = x
Then h(x) = f(x) - g(x)
= x^2 - x + 1
With a quick check of b^2 - 4ac, we see that it will have no solutions, hence they will never have any points of intersection.
Thus, when you find the minimum (-b/2a), the y value will be the minimum distance between the 2 graphs.
- (-1/2) = 1/2
h(1/2) = 1/4 - 1/2 + 1
= 3/4
Thus, the minimum distance will be 3/4 or 0.75, and this will occur at f(1/2) and g(1/2)
Post by: luffy on April 18, 2011, 07:55:03 pm
Let distance be y = |x^2+1 - x|
Find dy/dx of that.
Let dy/dx = 0 (to get minimum distance)
x=0.5, y=0.5
Your answer would be correct if the minimum distance occurred at a set x-value. I.e. the minimum distance will not be a simple difference in y-values.
Subtraction of ordinates will also give you the minimum/maximum distance.
If we let f(x) = x^2 + 1 and g(x) = x
Then h(x) = f(x) - g(x)
= x^2 - x + 1
With a quick check of b^2 - 4ac, we see that it will have no solutions, hence they will never have any points of intersection.
Thus, when you find the minimum (-b/2a), the y value will be the minimum distance between the 2 graphs.
- (-1/2) = 1/2
h(1/2) = 1/4 - 1/2 + 1
= 3/4
Thus, the minimum distance will be 3/4 or 0.75, and this will occur at f(1/2) and g(1/2)
Your assuming the difference occurs when the difference in y-values is the least. This may not be the case....
call the closest point on y = x to the parabola y = x^2 + 1 point P. then we imagine a tangent to the parabola that is parallel to y = x. call the point where this tangent touches the parabola Q. the condition for the tangent is that PQ is perpendicular to the tangent, which means it's perpendicular to y = x.
the parabola has gradient dy/dx = 2x. we want 2x = 1, since the tangent is parallel to y = x, which means x = 1/2. so Q = (1/2, 5/4)
so the aforementioned perpendicular line has equation y - 5/4 = -(x - 1/2) --> y = -x + 7/4
we want this to intersect y = x, so x = -x + 7/4, 2x = 7/4, x = 7/8. so y = 7/8. So (7/8, 7/8) is the closest point.
hopefully there's no computational errors.
Of all the answers, yours seems the most correct. However, how do you know the condition that the tangent will be perpendicular to the line. I actually got the same answer as you, but I don't think we are allowed to assume such things? Or maybe I'm just "over-thinking" the question. Can someone please confirm this answer? It will take a large "load" off my mind if its correct... :P
- Sorry if any of my "rebuttals" were completely incorrect. I just want an answer, which I can completely understand with full confidence.
Post by: brightsky on April 18, 2011, 08:00:20 pm
Post by: xZero on April 18, 2011, 08:02:37 pm
we have x1 = x, y1 = x (y=x) and x2 = x, y2 = x^2+1 (y=x^2+1), sub them into the equation, diff it and find the min distance
Post by: luffy on April 18, 2011, 08:03:20 pm
it's not an assumption as such, but a definition of the "shortest distance".
Actually, I just looked at your working out more clearly. It seems more logical than mine and I don't see any reason why it would be wrong.
Thanks a lot Brightsky :D -> Lifesaver :D
Why don't you use the distance between 2 point formula,
we have x1 = x, y1 = x (y=x) and x2 = x, y2 = x^2+1 (y=x^2+1), sub them into the equation, diff it and find the min distance
I tried that... However, you get two variables for the same equation and it doesn't work out (because the 'x' in the y2 equation is x2, which will be different to x1).
Post by: jane1234 on April 18, 2011, 08:04:22 pm
Let distance be y = |x^2+1 - x|
Find dy/dx of that.
Let dy/dx = 0 (to get minimum distance)
x=0.5, y=0.5
Your answer would be correct if the minimum distance occurred at a set x-value. I.e. the minimum distance will not be a simple difference in y-values.
Ah my bad. I assumed you were talking about vertical distance. I think brightsky's method is probably correct then.
Post by: luken93 on April 18, 2011, 08:30:26 pm
Hahaha also did this, soz luffy :PLet distance be y = |x^2+1 - x|
Find dy/dx of that.
Let dy/dx = 0 (to get minimum distance)
x=0.5, y=0.5
Your answer would be correct if the minimum distance occurred at a set x-value. I.e. the minimum distance will not be a simple difference in y-values.
Ah my bad. I assumed you were talking about vertical distance. I think brightsky's method is probably correct then.
Post by: enpassant on April 18, 2011, 08:41:43 pm
Hey guys,
I'm not entirely sure if this question is within the scopes of the specialist maths course because I couldn't do it :P I didn't get it from any textbook, and I just thought up the question myself. It seems like a very basic question, yet I couldn't do it... >:(
Question: "Find the minimum distance between the two graphsand
I know how to do the question if a point on either graph is known. However, in this case, neither is known.
Good luck and thanks in advance :D
Only you can get away with this
Others will be told to include the word 'challenge'
Post by: luffy on April 18, 2011, 09:09:07 pm
Hey guys,
I'm not entirely sure if this question is within the scopes of the specialist maths course because I couldn't do it :P I didn't get it from any textbook, and I just thought up the question myself. It seems like a very basic question, yet I couldn't do it... >:(
Question: "Find the minimum distance between the two graphs
I know how to do the question if a point on either graph is known. However, in this case, neither is known.
Good luck and thanks in advance :D
Only you can get away with this
Others will be told to include the word 'challenge'
Oh. Sorry, I didn't realize it was necessary - I put it in the other thread I made. Thanks for telling me. :D
Post by: Mao on April 18, 2011, 11:11:24 pm
Hey guys,
I'm not entirely sure if this question is within the scopes of the specialist maths course because I couldn't do it :P I didn't get it from any textbook, and I just thought up the question myself. It seems like a very basic question, yet I couldn't do it... >:(
Question: "Find the minimum distance between the two graphs
I know how to do the question if a point on either graph is known. However, in this case, neither is known.
Good luck and thanks in advance :D
Only you can get away with this
Others will be told to include the word 'challenge'
Heh, at least his intention is genuine.