Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: True/False : 2 Questions

  1. Jun 4, 2010 #1
    1. The problem statement, all variables and given/known data
    1- The following function is one-to-one



    f(x)= -x if x belongs to [-1,0]

    f(x)=3x+2 if x belongs to [0,1]



    2- The following parametric curve represents a line segment from (0,3) to (2,0) :

    x(t)=2sin^2(t) and y(t)=3cos^2(t) where t belongs to [0,pi/2]

    2. Relevant equations



    3. The attempt at a solution

    For (1): I think its false, if we draw the function on the interval [-1,1] we will get a graph which is similar to the graph of f(x)=|x| (the v-shape). clearly there are horizontal lines intersect the curve of f in more than one point, so its not 1-1 function, so the statement is false.



    For (2): I do not have any idea!
     
  2. jcsd
  3. Jun 4, 2010 #2

    Mark44

    Staff: Mentor

    That works for me.
    You know that sin2(t) + cos2(t) = 1, right? Can you work this idea into your parametric equations to get one equation that involves only x and y (no t)?
     
  4. Jun 4, 2010 #3
    well,
    I would say:
    (x/2)+(y/3)=sin^2(t)+cos^2(t)=1
    so
    (x/2)+(y/3)=1
    which is ellipse
    ok
    then?
     
  5. Jun 4, 2010 #4

    Mark44

    Staff: Mentor

    No, x/2 + y/3 = 1 is NOT an ellipse. Also, keep in mind that 0 <= t <= pi/2.
     
  6. Jun 4, 2010 #5
    ohhh
    Sorry
    its a line
    so when i substitute the smallest value for t i will get the start of the line
    and when i substitute the biggest for t i will get the end of the line
    so its TRUE
     
  7. Jun 4, 2010 #6
    by the way.
    the first statement is true.
    its one-to-one function.
     
  8. Jun 4, 2010 #7

    Mark44

    Staff: Mentor

    Right. I stand corrected. I should have drawn the graph, because then it's obvious.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook