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AP Cal AB MC Question

  1. May 6, 2008 #1
    1. The problem statement, all variables and given/known data
    The Shortest distance from the curve xy=4 to the origin is:

    A. 2
    B. 4
    C. rad (2)
    D. 2* rad (2)
    E. .5 rad(2)


    2. Relevant equations
    ... distance equation?


    3. The attempt at a solution

    i used the pythagorean theorem to find the distances using x^2 and y^2 as a^2 and b^2 but couldnt get a right answer.. i went through each answer choice and got A but that's not the correct answer
     
  2. jcsd
  3. May 6, 2008 #2

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    Parametrize it then put it in the equation for distance in place of x and y. Minimize the distance like you would any other function of one variable.
     
  4. May 6, 2008 #3
    Parametrize?

    sorry but what does Parametrize mean?
     
  5. May 6, 2008 #4

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    Rewrite both variables as a function of t.
    x = f(t) = ?
    y = g(t) = ?

    Plug the (f(t), g(t)) and (0,0) into the distance forumula to get distance as a function of t. Then, find the value of t that minimized d.
     
  6. May 6, 2008 #5
    I got D, don't you think the shortest distance is when y=x intersects your curve? - It's at (2,2) so distance = 2*rad(2)
    Edit: above answer simply y guessing..

    or find function for Distance {D(x)} .. you know D = rad(x^2+y^2) subsitute y val from your org function
     
  7. May 6, 2008 #6
    For the AB exam, it seems parametric equations are not required. I remember this question when prepping for the BC a few weeks ago. In essence, you would use the distance formula. You might need this for the test tomorrow, so I will just post how I thought about the question.

    The simplest way I found to think about this question and get the answer is:

    The graph xy = 4 is a hyperbola, which you can see by writing it as y = 4/x. It's also symmetric about the line y = x because its inverse is the same as itself. We draw a rough graph and notice that for x close to 0 and |x| close to infinity, the distance from the graph to the origin is huge. As we move closer to the symmetry line, the distances seem to get smaller. Intuition tells us the shortest distance is at the symmetry point when the graph seems to dip toward the origin [and it looks like a sort of question they would put on the test ;)]. If this is true, then [itex]x = y = \pm 2[/itex], and the point (2,2) or (-2,-2) is nearest the origin. This is a distance of [itex]2\sqrt{2}[/itex]. This explanation is long, but on the actual test you can do this very quickly.

    The rigorous way to do it involves the distance formula. The question mentions distance from the origin to a point on the graph xy = 4. We have two points, (0,0) and the point on the graph, so we have an idea that we need to use the distance formula. Points on the graph are (x, 4/x). The distance between (0,0) and (x,4/x) as a function of x is:

    [tex]\text{distance}=d(x)=\sqrt{x^2 + (4/x)^2}[/tex]

    We want to minimize this, so take the derivative and set it equal to 0:

    [tex]d\,'(x) = \frac{2x + 2(4/x)(-4/x^2)}{2\sqrt{x^2 + (4/x)^2}} = 0[/tex]

    This is 0 when the numerator is 0, so

    [tex]x - 16/x^3 = 0 \implies x^4 = 16 \implies x = \pm 2[/tex]

    At this point, we should technically find the second derivative to confirm it's a minimum, but this is a multiple choice test, and we're pretty sure this is a minimum, since there are no other critical numbers. Use the distance formula to find the minimal distance.
     
    Last edited: May 6, 2008
  8. May 6, 2008 #7
    One shortcut :rolleyes:
    d(x)=\sqrt{x^2 + (4/x)^2} is max or min when d(x)=x^2 + (4/x)^2 is max or min .. so lot less work
     
  9. May 6, 2008 #8
    Good point :) (is that a pun? don't think so lol)
     
  10. May 6, 2008 #9
    thank you everyone :]

    i got it
     
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