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Help - tangent to y = 9e^(2x)cos(pi*x)

  1. Dec 13, 2007 #1
    Q: find the tangent to y = 9e^(2x)cos(pi*x) at (0,9)

    I know how to do the tangent for a polar coordinate at an angle, but how find it at (0,9)?
  2. jcsd
  3. Dec 13, 2007 #2
    Also, how do I evaluate lim n --> infinity of [ln(10+x) - ln(5+x)]?
  4. Dec 13, 2007 #3


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    Homework Helper

    For the first question...
    to find the equation of a tangent, you need the point and the gradient of the tangent at that point. So you already have the point (0,9).

    Sub x=0 into the gradient function,[itex]\frac{dy}{dx}[/itex] and you will get the gradient of the tangent at (0,9).

    For your second problem:
    [tex]lim_{n\rightarrow \infty} [ln(10+x) - ln(5+x)][/tex]

    is there any way you can simplify ln(10+x) - ln(5+x)?

    and here is a hint:
    [tex]lim_{n\rightarrow \infty} log_a [f(x)] = log_a [lim_{n\rightarrow \infty} f(x)][/tex]
  5. Dec 13, 2007 #4


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    I'm not even sure what a "tangent for a polar coordinate at an angle" is, but finding the tangent to a curve at a given point is one of the first things I learned in calculus!

    Or are you talking about the tangent of an angle? Surely not!

    To find the tangent line to a curve at a given point, first find the derivative there: [itex]f'(x_0)[/itex]. The equation of the tangent line at [itex](x_0, f(x_0)[/itex] is [itex]y= f'(x_0)(x- x_0)[/itex]. The only "hard part" here is finding the derivative- and it's not all that hard. Use the product law along with the chain rule.

    Well, you certainly know, I hope, that [itex]ln(10+x)- ln(5+ x)= ln((10+x)/(5+x))[/itex] and, since ln(x) is a continuous function, that is the ln of the limit of (10+x)/(5+x). Can you do that?
  6. Dec 13, 2007 #5
    Last time I checked, yes!

    And for the first question, I have to get the derivatie and evaluate it at [itex]y= fsingle-quote(x_0)(x- x_0)[/itex], correct?

    How would I find the y-intercept?
    Last edited: Dec 13, 2007
  7. Dec 13, 2007 #6
    One more thing, how do i show that (-5)^n over n*5^n

    converges? By the alternating series test? I am not sure how to manupulate the (-5)^n on the top.

    Sorry about the stupid questions, I am trying to prepare for the test.

    EDIT: i guess It's just (-1)^n(5)^n...
    Last edited: Dec 13, 2007
  8. Dec 14, 2007 #7


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    No, you have to evaluate the derivative at x= 0. I can see no reason why you would want to find the y-intercept.

    You mean

    What can you say about convergence of

    That's obviously what your series reduces to.
  9. Dec 14, 2007 #8
    it converges conditionally.
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