Calculus PLEASE check my work ,

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1.) which of the following represents dy/dx when y=e^-2x Sec(3x)?
A.3e^-2x sec(3x) tan (3x)-2e^-2x Sec(3x)<<<< my choice
B.)3e^-2x sec(3x)tan (3x)-2xe^-2x sec(3x)
C.)3e^-2x sec(3x)tan (x)-2e^-2x Sec(3x)
D.)3e^-2x sec(3x)tan (x)-2xe^-2x sec(3x)

2.)calculate dy/dx if y=Ln(2x^3+3x)
A.)1/2x^3+3x
B.)1/6x^2+3
C.)2x^3+3x/6x^2+3
D.)6x^2+3/2x^3+3x <<<< my choice

3.)Given the curve that is described by the equation r=3 cos theta, find the angle that tangent lines makes with the radius vector when theta=120.
A.) 30deg
B.) 45deg
C.) 60deg
D.) 90deg << my choice

4.)A line rotates in a horizontal plane according to the equation theta=2^t-6t,where theta is the angular position of the rotating line, in radians ,and tis the time,in seconds. Determine the angular acceleration when t=2sec.

A.) 6 radians per sec^2
B.) 12 radians per sec^2
C.) 18 radians per sec^2
D.) 24 radians per sec^2 << my choice.

  • Calculus PLEASE check my work , -

    1. Correct
    2. Correct
    3.
    dy/dx
    =(dy/dt) / (dx/dt)
    put
    x=rcos(t)=3cos²(t)=
    y=rsin(t)=3sin(t)cos(t)
    Calculate dy/dx and evaluate at t=120° (2π/3) to get
    dy/dx=-1/√3
    => θ=150°
    Angle with radius vector=150°-120°=30°.

    4.
    There may have been a typo in the question. If the question is asking for θ=2^t-6t, the answers should not be in round numbers.

  • Calculus PLEASE check my work , -

    A
    line
    rotates
    in
    a horizontal
    plane
    according
    to
    the
    equation
    9
    =
    2t3-
    6t,
    where
    (J
    is
    the
    angular
    position
    of
    the
    rotating
    line,
    in
    radians,
    and
    t
    is the
    time,
    in
    sec-
    onds.
    Determine
    the
    angular
    acceleration
    when
    t
    =
    2
    sec.

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