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Homework Help: A little help with integration and trigonometric functions.

  1. Jun 18, 2007 #1
    I was doing my exam today and ran into a couple problems.
    First one: how do you differentiate [tex]\tan^2[/tex]?
    I converted it into [tex]\sec^2 - 1[/tex] and used the u/v = (u`v - v`u)/v^2 method, but I would like somebody clever to do it for me, just to be sure, please.
    1. The problem statement, all variables and given/known data
    Another problem.
    Rate of change of population P equals [tex]\lambda P \cos(\lambda t)[/tex]
    Find the formula for population P in terms of [tex] P_0[/tex],[tex] \lambda[/tex] and t.

    Then, find t. When P = [tex]2P_0[/tex]

    3. The attempt at a solution

    I had problems with finding t.

    t came out to be arcsin of something. The problem is, they never said anything about degrees or radians and so t could vary quite a bit, depending on that. Did I do it wrong or am I missing something out?

    I don't have the exact question. It was in the exam I did two hours ago. All help is appreciated.

    Thank you.
    Last edited: Jun 18, 2007
  2. jcsd
  3. Jun 18, 2007 #2


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    differentiate tan^2 x? tan x = sinx/cosx , chain rule or remember the result.. d(tanx)/dx = sec^2 x, chain rule simply gives d(tan^2x)/dx = 2 tan x sec^2 x

    dP(t)/dt = L P(t) cos (Lt)
    => dP(t)/P(t) = L cos(Lt) dt
    => integrate both sides
    => Log_e P(t) = sin (Lt) +C
    => P(t) = A e^(sin Lt)
    where A is some constant
    I suppose P0 means initial value so A = P0
    and so when P(t) = 2 P0
    => 2 P0 = P0 e^(sin L t)
    => Log_e 2 = sin Lt
    => t = 1/L arcsin (Log_e 2)

    ok.... I am expecting you to be able to do it next time
  4. Jun 18, 2007 #3


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    1. To differentiate tan^2x, one would use the chain rule. [tex]\frac{d}{dx}\tan^2x=2\tan x\frac{d}{dx}(\tan x)=2\tan x \sec^2 x[/tex] where the last step is done either using the quotient rule for tangent, or from memory.

    2. For this question you need to solve the differential equation [tex]\frac{dP}{dt}=\lambda P\cos(\lambda t) [/tex], using the initial conditions given. Is that what you did?
  5. Jun 18, 2007 #4
    1)Chain rule, of course!
    I got [tex]\frac{\cos^4(x)}{\sin(2x)} for that one.

    2) Yes, I did that, it's easy. But then you have to make t the subject of the formula and substitute some values to find when the population doubles. That's where I had the problem, as I wasn't sure what to use: degrees or radians. Try it yourself and see.
  6. Jun 18, 2007 #5
    For the second one you get [tex] P = P_0e^\sin(\lambda t)[/tex] <= I think.

    When [tex] P = 2P_0[/tex], [tex] t = \frac{\arcsin(ln 2)}{\lambda}[/tex]. <= did it in my head, so could be wrong, but the principle is there. Arcsin can give you different values, depending on what system you use: degrees, radians, grads. Or am I getting something wrong here?
  7. Jun 18, 2007 #6


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    Looks like you've got the reciprocal of the correct answer (or just made a typo here!)

    Ok, I see your point now. It doesn't really matter which you use, but I would always use radians. They'll both give the same answer, it's just the units of time will be different.
  8. Jun 18, 2007 #7
    The units will be different, eh? How do I even figure out the units? They said to give your answer in minutes. And t is in days, I think.
  9. Jun 19, 2007 #8

    Gib Z

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    When dealing with logs, its usually best to use radians as angle measures that require units give workable but difficult results, such as 2 log seconds.
  10. Jun 19, 2007 #9
    I used degrees, doesn't matter. I'll probably lose one mark for that or something.

    And I'll know next time. I actually thought about using radians, but couldn't be bothered to change the mode on my calculator.

    Thanks to everyone for their help.
  11. Jun 20, 2007 #10
    Could someone integrate his for me, please?

    [tex]2\int\sec^4x dx[/tex]

    I had it in my exam today and I want to know if I did it correctly. Thank you.
  12. Jun 20, 2007 #11


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    The result is

    [tex]\frac{2\,\left( 2 + {\sec (x)}^2 \right) \,\tan (x)}{3}[/tex]

    Did you have that?
  13. Jun 20, 2007 #12

    [tex]\tan x + \tan^2 x + \sec^2 x[/tex]

    If I remember correctly. The sad thing is, I didn't really have to do that integration. I missed out the minus in ln, which would've given me [tex]\sec^2 x[/tex], which is considerably easier to integrate.
  14. Jun 20, 2007 #13
    Woops, I meant tan and sec to the power of 3, not too. Is that more correct in any way?
  15. Jun 21, 2007 #14


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    OK, my Mathematica gives something else (the expression above).
    Then perhaps it's time to show your calculation :)
  16. Jun 21, 2007 #15
    Forget about it, I've gotta revise for a different exam tomorrow. Thanks for your help anyway.
  17. Jun 21, 2007 #16


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    OK, good luck with the exam.
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