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Math Help - trigonometry derivative

  1. #1
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    trigonometry derivative


    1. Use the inequalities sinx<x<tanx for a suitable value of x to show that pi lies between 3 and 2root3.

    2.By writing tanx as sinx/cosx, show that d(lntanx)/dx = 2cosec2x.

    3. A tuning fork sounding A avobe middle C oscillates 440 times a second. The displacement of the tip of the tuning fork is given by 0.02cos(2pi x 440t) millimetres, where t is the time in seconds after it is activated. Find
    a. the greatest speed, b. the greatest acceleration of the tip as it is oscillates.

    If you know the answers to any of these questions, please, please help!
    Thanks!
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by autkcat View Post

    1. Use the inequalities sinx<x<tanx for a suitable value of x to show that pi lies between 3 and 2root3.
    plug in x = pi/6
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by autkcat View Post
    2.By writing tanx as sinx/cosx, show that d(lntanx)/dx = 2cosec2x.
    what have you tried

    follow the hint (though i don't think it helps--if you know the derivative of tan(x), it's easier to do it straight): \frac d{dx} \ln \tan x = \frac d{dx} \ln \frac {\sin x}{\cos x} = \frac d{dx} ( \ln \sin x - \ln \cos x)

    now use the chain rule and simplify
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  4. #4
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    Quote Originally Posted by autkcat View Post
    3. A tuning fork sounding A avobe middle C oscillates 440 times a second. The displacement of the tip of the tuning fork is given by 0.02cos(2pi x 440t) millimetres, where t is the time in seconds after it is activated. Find
    a. the greatest speed, b. the greatest acceleration of the tip as it is oscillates.
    Speed is v = dx/dt. So what is the largest value of dx/dt?

    Acceleration is a = dv/dt. What is the greatest value of dv/dt?

    -Dan
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    2.By writing tanx as sinx/cosx, show that d(lntanx)/dx = 2cosec2x.

    I think i figured this out. the derivative of tanx is sec^2.
    So, d(lntanx)/dx = sec^2x(cosx/sinx)
    = 1/cosxsinx
    = 1/(0.5sin2x) [double angle formula]
    = 2cosec2x

    Right??
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