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Horrible Calculus - Help!

  1. Oct 30, 2004 #1
    Hello everyone!

    I have a calculus midterm next week and I need some help solving old midterm questions. I'm a first year calculus student so they're not hard, but I can't figure them out!

    I am told to solve without using the tan addition formula.
     

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    Last edited: Oct 30, 2004
  2. jcsd
  3. Oct 30, 2004 #2
    problem number 2
     

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  4. Oct 30, 2004 #3

    arildno

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    Since:
    [tex]tan(\frac{\pi}{6})=\frac{1}{\sqrt{3}}[/tex]
    the question merely asks you to calculate the derivative of tan(x) at [tex]x=\frac{\pi}{6}[/tex]
     
  5. Oct 30, 2004 #4

    arildno

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    For your other questions, show some of your own work on these.
     
  6. Oct 30, 2004 #5
    Sure, but how do I use the code thing? Or is there an easy way to type out the work I have?
     
    Last edited: Oct 30, 2004
  7. Oct 30, 2004 #6

    arildno

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    Click on a given "LATEX" code to see how you generate it. There's a sticky in General Physics which tells you a bit about it.
     
  8. Oct 30, 2004 #7
    [tex]\lim_{h\rightarrow0} \frac{1}{h}} (\frac{sin(h+(\pi/6))}{cos(h+(\pi/6))}} - \frac{1}{\sqrt3}})[/tex]

    [tex]\lim_{h\rightarrow0} \frac{1}{h}} (\frac{sin(x)cos(\pi/6) + cos(x)sin(\pi/6)}{cos(x)cos(\pi/6) + sin(x)sin(\pi/6)}} - \frac{1}{\sqrt3}})[/tex]

    [tex]\lim_{h\rightarrow0} \frac{1}{h}} (\frac{sin(x)(\sqrt3/2) + cos(x)(1/2)}{cos(x)(\sqrt3/2) + sin(x)(1/2)}} - \frac{1}{\sqrt3}})[/tex]

    [tex]\lim_{h\rightarrow0} \frac{1}{h}} ({\frac{\sqrt3sin(x) + cos(x)}{{\sqrt3cos(x) +sin(x)}} - \frac{1}{\sqrt3}}) [/tex]

    [tex]\lim_{h\rightarrow0} \frac{1}{h}} (\frac{2sin(x)}{3cos(x) + \sqrt3sin(x)}) [/tex]

    Wow, that latex is hard to code... I think I have it now though.

    This is asfar as I can go with this problem, from here, I am confused.
     
  9. Oct 30, 2004 #8
    you're making it way to diffcult, lim(h->0) 1/h * (tan(Pi/6 + h)-1/sqrt(3)) is also lim(h->0) 1/h * (tan(Pi/6 + h)-tan(Pi/6)), lim(h->0) 1/h * f(a+h)-f(a) is the differential of f in a, so you really just need to calculate tan's differential in Pi/6
     
    Last edited: Oct 30, 2004
  10. Oct 30, 2004 #9
    Okay, that makes sense and therefore:

    [tex]\lim_{h\rightarrow0^+} \frac{1}{h}} (\tan(h+(\pi/6))}- \frac{1}{\sqrt3}})[/tex]

    equals

    [tex]\lim_{h\rightarrow0^+} \frac{1}{h}} (\tan(h+(\pi/6))}- \tan(\frac{\pi}{6}}))[/tex]

    [tex]\frac{d}{dx}}\tan(x) = \sec^2(x)[/tex]

    [tex]\sec^2(\frac{\pi}{6}}) = \frac{4}{3}} [/tex]

    and


    [tex]\lim_{h\rightarrow0^+} \frac{1}{h}} (\tan(h+(\pi/6))}- \frac{1}{\sqrt3}}) = \frac{4}{3}} [/tex]

    Is this true?
     
  11. Oct 31, 2004 #10
    yeah, that's what i meant
     
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