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Math Help - One Derivitive Problem & One Optimization Problem

  1. #1
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    One Derivitive Problem & One Optimization Problem

    1. Find the derivitive of 3x^2tan(x)/(sec(x))


    I put ((6xtan(x))+sec^2(x)3x^2(sec(x))-(sec(x)tan(x))(3x^2tan(x)))/(sec^2(x))
    Not sure where I messed up...


    2. (1 pt) If a projectile is fired from ground level with initial velocity of 3775 ft/sec and an inclination angle , and if air resistance can be ignored, then its range---the horizontal distance it travels---is about R=(1/16)02sincos
    feet. What value of (in radians) maximizes the range?

    I do not know how to do this problem..
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  2. #2
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    Quote Originally Posted by lmao View Post
    1. Find the derivitive of 3x^2tan(x)/(sec(x))


    I put ((6xtan(x))+sec^2(x)3x^2(sec(x))-(sec(x)tan(x))(3x^2tan(x)))/(sec^2(x))
    Not sure where I messed up...


    2. (1 pt) If a projectile is fired from ground level with initial velocity of 3775 ft/sec and an inclination angle , and if air resistance can be ignored, then its range---the horizontal distance it travels---is about R=(1/16)02sincos
    feet. What value of (in radians) maximizes the range?

    I do not know how to do this problem..


    Here is the answer:



    This whole problem surrounds the quotient rule; however for phase one you must use the product rule, then you can carry on.
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  3. #3
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    Anyone get the second one? I only have 45 minutes to turn this in!
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  4. #4
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    Re:

    I know that #1 is 100% correct (Just as teachers like it written -unsimplified), however we haven't hit the chapter on optimization yet...
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  5. #5
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    Hello, lmao!


    . . . . . . . . . . . . . . . . .3x▓Ětan(x)
    (1) Differentiate: .y .= .------------
    . . . . . . . . . . . . . . . . . . sec(x)

    . . . . . .sec(x)Ě[6xĚtan(x) + 3x▓Ěsec▓(x)] - 3x▓Ětan(x)Ěsec(x)Ětan(x)
    y' . = . ---------------------------------------------------------------------
    . . . . . . . . . . . . . . . . . . . . sec▓(x)

    . . . . . .6xĚsec(x)Ětan(x) + 3x▓Ěsec│(x) - 3x▓Ěsec(x)Ětan▓(x)
    y' . = . ------------------------------------------------------------
    . . . . . . . . . . . . . . . . . . . sec▓(x)

    . . . . . . . . . . . 3xĚsec(x)Ě[2Ětan(x) + xĚsec▓(x) - xĚtan▓(x)]
    Factor: .y' . = . ---------------------------------------------------
    . . . . . . . . . . . . . . . . . . . . .sec▓(x)

    . . . . . . . . . . . . 3xĚ
    [2Ětan(x) + x[sec▓(x) - tan▓(x)]]
    Reduce: .y' . = . -------------------------------------------
    . . . . . . . . . . . . . . . . . . . sec(x)

    . . . . . . . . . . . . . .3xĚ[2Ětan(x) + x]
    Therefore: .y' . = . ---------------------
    . . . . . . . . . . . . . . . . . sec(x)

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  6. #6
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    Re:

    Soroban is on his way...
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  7. #7
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    Quote Originally Posted by lmao View Post
    2. (1 pt) If a projectile is fired from ground level with initial velocity of 3775 ft/sec and an inclination angle , and if air resistance can be ignored, then its range---the horizontal distance it travels---is about R=(1/16)02sincos
    feet. What value of (in radians) maximizes the range?

    I do not know how to do this problem..
    I'm presuming the formula is R=(1/16)*2sincos ?

    Let's call alpha = q.

    Note that 2sin(q)cos(q) = sin(2q) so
    R = (1/16)*sin(2q)

    The non-Calculus way:
    I'm afraid you are reduced to graphing R as a function of q. However, note that R is merely a sine function, so we can easily tell where the maximum is. Note that the sine function reaches its maximum value at an angle of pi/2 rad (90 degrees). Thus the maximum R value will be for
    2q = pi/2 rad

    q = pi/4 rad (45 degrees)

    The Calculus way:
    What value of q maximizes R? Take the first derivative of R with respect to q and set it equal to 0:
    dR/dq = (1/16)*2*cos(2q) = 0

    Thus
    cos(2q) = 0

    2q = pi/2 rad

    q = pi/4 rad.

    -Dan
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  8. #8
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    Hello, lmao!

    Your post has some glitches; I have to guess what's missing.
    And what can be seen is confusing . . .


    2. If a projectile is fired from ground level with initial velocity of 3775 ft/sec
    and an inclination angle θ, then its range is about: R = [???]
    What value of θ (in radians) maximizes the range?

    The horizontal displacement is: .x .= .(3775Ěcosθ)t

    The vertical displacement is: .y .= .(3775Ěsinθ)t - 16t▓


    The projectile is on the ground when y = 0.
    . . (3775Ěsinθ)t - 16t▓ .= .0 . . t[3775Ěsinθ - 16t] .= .0

    . . Hence: .t = 0 .and .t = 3775Ěsinθ/16


    The range is: .x .= .3775ĚcosθĚ(3775Ěsinθ/16) .= .(3775▓/16)ĚsinθĚcosθ

    . . x .= .(3775▓/32)Ě(2ĚsinθĚcosθ) .= .(3775▓/32)Ěsin(2θ)


    The range is a maximum when: .sin(2θ) = 1 . . 2θ = π/2 . . θ = π/4

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