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Determine which points on graph of X = sqrt(Y) are closest to (0,3)

  1. Oct 30, 2008 #1
    1. The problem statement, all variables and given/known data

    Determine which points on graph of X = sqrt(Y) are closest to (0,3). What is the minimum distance between the point(s) found and the graph of x = sqrt(Y)


    2. Relevant equations

    I used the distance formula.

    3. The attempt at a solution

    I came up with (sqrt(2) , 2) but i hardly doubt its right because i have no clue what the question is asking, can somebody just tell me if my answer is right or not?
     
  2. jcsd
  3. Oct 30, 2008 #2

    Office_Shredder

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    A quick confirmation check: if the point is the closest to (0,3) then the derivative of the graph at that point must be perpendicular to the line connecting the two points (in this case it is). So your answer is correct
     
  4. Oct 30, 2008 #3

    Mark44

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    It's wrong. The point you found isn't on the curve (but (2, sqrt(2)) is, though).
    How did you use the distance formula? To use it, you need to have the coordinates of two points, one of which is the point on the curve that is the closest, and I don't see any work that shows that you found the closest point.

    This problem is tricky because it gives x as a function of y, which is the opposite of how we're usually given it.

    Here's what I did.
    I drew a graph of x = sqrt(y). The vertical axis is the x axis, and the horiz. axis is the y axis. The graph looks exactly like that of y = sqrt(x), but the axes are labeled differently and all the points are (y, x) pairs.

    Call the closest point on this graph P, with coordinates (y, sqrt(y)).

    Find the slope of the function (namely, dx/dy). This will be dx/dy = 1/[2*sqrt(y)].

    Find the slope of the line from P to the point B(0, 3) on the vertical (x) axis. The slope will be the result of delta x / delta y.

    At the point on the curve that's closest to B, the tangent line at P will be perpendicular to the line segment OP. This means that their slopes will be negative reciprocals. This will give you an equation involving radicals that you can (maybe) solve. After moving things around and getting rid of the radicals in the denominators, I ended up with a cubic polynomial that I was able to factor into a linear factor and a quadratic factor. The quadratic had no real roots.

    The point (2, sqrt(2)) (coordinates reversed from the one you reported) is about 2.55 units away from (0, 3). The point I found is about 2.236 units away, so is closer.

    Hope this helps.
     
  5. Oct 30, 2008 #4

    Mark44

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    The trouble is, the point the OP found isn't on the graph of x = sqrt(y).
     
  6. Oct 30, 2008 #5

    Office_Shredder

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    y=2. sqrt(y)=sqrt(2)=sqrt(2)=x. I assumed he was still writing his coordinates in (x,y) fashion
     
  7. Oct 30, 2008 #6
    Wait so is my answer right or is Mark's answer right?

    What i did is i solved for y, and i got Y = X^2. and then from there, i used the distance formula. (0,3) were my first points, my second points were (X,Y) then i knew that Y = X^2 so i plugged it into my distance equation, then i squared the sides and took the derivative. When i did that i got 0 = 3y-6 then i solved for Y, and it gave me 2.
     
  8. Oct 30, 2008 #7

    Dick

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    Mark44 didn't give an answer. He described one way to get an answer. Your description of the method is roughly correct as well. But y=2 is quite wrong. Can you show us how you got that in detail?
     
  9. Oct 31, 2008 #8

    Mark44

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    As I mentioned before, this problem is tricky due to x being given as a function of y, and the point (0, 3) being given. For this point, 0 is a value of y and 3 is a value of x. y = x^2 (for x >= 0) is equivalent to the first equation, and its graph will look exactly the same as that of x = sqrt(y) using the same coordinate axes as before. However, if you switch the x and y axes, you're going to also have to switch the coordinates of the point on what was the vertical axis, so that it's now on the horizontal axis, and its new coordinates will be (3, 0).
     
  10. Oct 31, 2008 #9

    HallsofIvy

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    x= sqrt(y) is the the right half of the parabola y= x^2. The (square of the) distance from the point (x, x^2) to (0, 3) is x^2+ (x^2- 3)^2. Minimize that and discard any negative solution for x. Looks lilke a pretty simple problem to me.
     
  11. Oct 31, 2008 #10

    Mark44

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    Since x is given as a function of y, I am interpreting this problem to mean that y is the independent variable and x is the dependent variable so that (0, 3) is a (y, x) pair. Certainly y = x^2 for x >= 0 is equivalent to x = sqrt(y), and their graphs are identical. If you relabel the axes, though by switching the roles of y and x, your new graph is the reflection across the line y = x of the original. In particular, the point (0, 3) is reflected to (3, 0).
     
  12. Oct 31, 2008 #11

    Dick

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    You can minimize |(sqrt(y),y)-(0,3)| and find y. Or you minimize |(x,x^2)-(0,3)| and find x. Why 'rotate'? It just makes it easier to lose track of which variable is which. The real problem here is that the OP has gotten 0=3y-6 as an equation. I would like to know how.
     
    Last edited: Oct 31, 2008
  13. Oct 31, 2008 #12

    Office_Shredder

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    You realize that if you have a pair (x,y) you can still graph x as a function of y? Considering the OP even stated that the coordinates are in the form (x,y) I think continuing to use it in (y,x) form is just being obstinate
     
  14. Oct 31, 2008 #13

    Mark44

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    I have qualified my answer in each post by saying what my assumptions are. If you were to graph a function given as b = f(p), you would probably put b on the vertical axis and p on the horizontal axis, the usual practice for dependent and independent variables.

    There was nothing in the OP's original post that stated the meaning of the order in any ordered pairs. IMO, the problem could have been stated more simply by giving the function as y = x^2 for x >= 0. Then there would not have been any uncertainty on my part about whether the distance was measured from the curve (whatever it is) to either (0, 3) or (3, 0).

    If my asssumption here is incorrect, I'll be happy to concede my argument.
     
  15. Nov 1, 2008 #14

    statdad

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    If you are speaking of the graph of the graph of [tex] y = \sqrt x [/tex], and discussing ordered pairs on that graph, and in the plane, the order for the coordinates is [tex] (x,y) [/tex], by convention. There is no requirement for the OP to specify which "specify which order is meant for ordered pairs."
    An explanation of another why another approach is valid is perfectly fine, but saying the choice is "arbitrary" is not.
     
  16. Nov 1, 2008 #15

    HallsofIvy

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    I'm a bit puzzled as to why this is still a problem. It is not too difficult to minimize the distance from [itex](\sqrt{y}, y)[/itex] to (0, 3). It is almost trivial to minimize the distance from [itex](x, x^2)[/itex] to (0, 3) and drop the negative x solution as I suggested before.
     
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