Help with finding stationary point of y=e^x•sin(2x)?

y=e^x•sin(2x)

Find the stationary point where 0≤x≤0.5π

4 Answers

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  • 5 months ago
    Best answer

    Stationary points are where the derivative is 0.

    Take the derivative using product rule and chain rule.

    y' = eˣ cos(2x) (2) + eˣ sin(2x)

    y' = eˣ (2 cos(2x) + sin(2x))

    Set equal to 0.

    0 = eˣ (2 cos(2x) + sin(2x))

    eˣ is never 0, so:

    0 = 2 cos(2x) + sin(2x)

    0 = 2 + tan(2x)

    tan(2x) = -2

    2x = atan(-2) + kπ

    x = ½ atan(-2) + k/2 π

    Trying different integer values of k:

    k = 0, x ≈ -0.554

    k = 1, x ≈ 1.017

    k = 2, x ≈ 2.588

    Only k = 1 gives us an answer within 0≤x≤0.5π.

    Therefore, the stationary point is at x = ½ atan(-2) + π/2, or:

    (x, y) ≈ (1.017, 2.474)

    • Some Body
      Lv 7
      5 months agoReport

      Tangent has a period of π.

      tan θ = tan (θ+π) = tan (θ+2π) = tan (θ+3π) = ...

      So for any value of tangent, there are an infinite number of angles that work: θ + kπ, where k is an integer.

      The same thing is done for sine and cosine, except they have periods of 2π.

  • alex
    Lv 7
    5 months ago

    Hint:

    f '(x) = 0 <--> Stationary points

  • 5 months ago

    y = e^(x) * sin(2x) ← this is a function i.e. a curve

    You can obtain the stationary point when the derivative is zero.

    The function looks like (u.v), so the derivative looks like: (u'.v) + (v'.u) → where:

    u = e^(x) → u' = e^(x)

    v = sin(2x) → v' = 2.cos(2x)

    y' = (u'.v) + (v'.u)

    y' = [e^(x) * sin(2x)] + [2.cos(2x) * e^(x)]

    y' = e^(x) * [sin(2x) + 2.cos(2x)] → then you solve for x the equation: y' = 0

    e^(x) * [sin(2x) + 2.cos(2x)] = 0 → you know that: e^(x) ≠ 0 of course

    sin(2x) + 2.cos(2x) = 0

    sin(2x) = - 2.cos(2x)

    sin(2x)/cos(2x) = - 2

    tan(2x) = - 2 ← the corresponding angle is ≈ - 1.107148 rd → i.e.: 2.034444 rd

    2x = 2.034444 + kπ → where k is an integer

    x = 1.017222 + k.(π/2)

    First case: k = 0 → x = 1.017222 ← this is the solution where: 0 ≤ x ≤ 0.5π

    Second case: k = 1 → x = 1.017222 + (π/2) ← over 0.5π

    Third case: k = 2 → x = 1.017222 + π ← over 0.5π

  • 5 months ago

    In mathematics, particularly in calculus, a stationary point of a differentiable function of one variable is a point on the graph of the function where the function's derivative is zero.

    y = e^x•sin(2x)

    I assume that that is

    y = (e^x)•sin(2x)

    y' = (e^x)•sin(2x) + (e^x)•2cos(2x) = 0

    sin(2x) + 2cos(2x) = 0

    2cos(2x) = cos(2x+(π/2))

    solve for x

    x = about π/3, a bit less, approx 0.343π or 1.076

    product rule

    (fg)' = f'g + fg'

    (fgu)' = fgu' + fug' + guf'

    (e^x)' = e^x

    (sin ax)' = a cos ax

    Attachment image
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