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Finals exam!

  1. Jan 12, 2004 #1
    Ok, so I have my Calculus Finals this coming Thursday, and I hope I receive some "tutoring" or review. The teacher sent out a review sheet via email, and I mirrored it www.hodeez.com/max3_finals.doc

    Just need a little/some help on 1,4,6,7,10 and the long problem.

    Of course I know the rules and I'll show you my work first, but as of right now, I'm at work and my cable @ home keep disconnecting me. So I'll breif what I can get by today hopefully :)
  2. jcsd
  3. Jan 12, 2004 #2
    For Prob 4

    f(x)= |2x+3|-|x-4|

    draw the no line
    Code (Text):

                         |           |           |
    x                  -3/2          0           4
                           |                     |
    f(x)= -(2x+3)-{-(x-4)} | 2x+3 -{-(x-4)}      | 2x+3 -(x-4)    
                           |                     |
    Code (Text):

    for   x<-3/2 u have   |2x+3|=-(2x+3)  & |x-4|= -(x-4)
    -3/2<=x<4             |2x+3|= (2x+3)  & |x-4|= -(x-4)
          x>=4            |2x+3|= (2x+3)  & |x-4|=  (x-4)
    now i believe u can recombine the above in specified interval to get f(x)
  4. Jan 12, 2004 #3
    For problem 1 [tex] \frac{df(x)}{dx}= 0 \text{ as the tangent is horizontal say at point x=t}[/tex]

    u will get Quadratic Equation find the roots to 3 decimal place
    Last edited: Jan 12, 2004
  5. Jan 12, 2004 #4
    For the problem 6 find -dx/dy which will be the slope of the normal

    For problem 7

    Apply the formula [tex]\frac{d (\frac{u}{v})}{dx}=\frac{ v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}[/tex]
    here u and v are u=cosx and v= sqrt(2x-3)
  6. Jan 12, 2004 #5
    from what I am understanding, (6) you want me to find the reciprocal negative of the derivative of the original equation? OR find the implicit differentiation and then plugging in the (2,1)

    (7) = seems like the division rule, how did i ever miss that? [zz)]

    thanks for the replies, im going to look more indepth into the answers you gave me. :smile:
  7. Jan 12, 2004 #6
    Giving Hints for long problem

    i.e f(3-h)=f(3+h) for h->0 and h>0

    and [tex] f'(x)=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}[/tex]
    u can substitute x= 4 here and so on
  8. Jan 12, 2004 #7
    DOH! another simple one over looked
  9. Jan 12, 2004 #8
    I must say all are simple

    u can do it either way
  10. Jan 12, 2004 #9
    im quite consfused on the way you are approaching it. I am thinking, if its continuous the 1st piece must match the 2nd piece @ 3 and then i set it up for k/3+2=k-3-1 and then solved for k which = -9 , then the slope must equal so im finding the deriv of both pieces and then trying to make them equal again.
  11. Jan 12, 2004 #10
    May you check this (#9):

    y'' = -3(sin^2(x))*(cos^2(x))
    and then plug in pi/6 = 180/6 = 30, yes?
  12. Jan 12, 2004 #11
    For long problem 2

    Take the first hiker along x-axis and other at an angle of 35 deg
    After time t its(1 hiker)coordinate will be(2.25t,0) and for second hiker it would be (3.8cos35 t,3.8sin35 t}.

    Now apply the distance formula and and differentiate it to get ur answer substitute for t=5.4/2.25

    Also it could have been done easily be relative velocity concept but i feel that u have to give Calculus exam right
  13. Jan 12, 2004 #12
    may i know how u reach at y'' = -3(sin^2(x))*(cos^2(x))

    it is incorrect Pls show me ur steps
    though pi/6=30 is correct
  14. Jan 12, 2004 #13
    y'= 3(sin(x))(cos(x))
    y''= 3 * product rule (sin(x))(cos(x))
    = 3 * (sin(x))*(-sin(x))+(cos(x))(cos(x))
    = -3(sin^2(x))*(cos^2(x))
  15. Jan 12, 2004 #14
    [tex]y=sin^3x \Rightarrow y'=3sin^2x (cosx)[/tex]

    u got y' wrong in ur reply
  16. Jan 12, 2004 #15
    is the answer 56.52 rounded?

    I used the Cosine Formula for triangles.

    Also, im interested in hearing your relative velocity concept. =)
  17. Jan 12, 2004 #16
    wow, that would have caused havok on the final
  18. Jan 12, 2004 #17
    Again how u reached the answer for hiker prob it is incorrect
  19. Jan 12, 2004 #18
    The method, I'm sure is correct, because that is the method we were taught to use. But my mechanics are probably incorrect. I'm using the Google calculator as I don't have my graphing calculator right now.
  20. Jan 12, 2004 #19
    Yup i know the method using the cosine formula is correct u need to recheck ur steps u might be making small errors in calulating

    [tex]cos35 = \frac{(2.25t)^2+(3.8t)^2-x^2}{2*2.25*3.8 t^2}[/tex]

    After little rearrangement

    so u will get x=2.344141643t
  21. Jan 12, 2004 #20
    My formula is C= A^2+B^2-cos(c)*A*B
    Is that the correct form?

    But 2.344 seems too way out of proportional
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