Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

What would m be?

 equation of the tangent line to the graph of y=xcos(3x) at x=π is given by y=mx+bfor 

4 Answers

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  • Philip
    Lv 6
    1 month ago

    y = xcos(3x). y' = cos(3x) + x[-sin(3x)](3) = cos(3x) -3xsin(3x). At x = pi, y' =;

    cos(3pi) - 3pi*sin(3pi) = -1. Also, at x = pi, y = pi*cos(3pi) = -pi.;

    Eqn of tangent line L to y at x = pi is y = mx + b where m is the slope of tangent line L and b is the y-intercept of L. At x = pi, m = -1.  Anonymous only wanted to

    know the value of m, not the eqn of the tangent line {y = -x}.

  • 1 month ago

    Differentiate the function

    y = xcos(3x)

    dy/dx = cos(3x) d/dx 1 + x d/dx - 3sin(3x)

    dy/dx = cos(3x) - 3xsin(3x)

    when x = π

    m = dy/dx = cos(3π) - 3πsin(3π)= - 1 

    the slope is m = - 1

    solving for y

    y = πcos(3π) = -π

    solving its y - intercept from slope intercept form

    y = mx + b

    - π = - π + b

    - π + π = b

    b = 0

    The equation of the tangent line is:

    y = - x Answer//

     

  • 1 month ago

    Need the derivative of:

    y = x cos(3x)

    Product rule:

    u = x and v = cos(3x) so we have y = uv

    y' = u'v + v'u

    so we need the derivatives of u and v.  u is easy, v needs the chain rule:

    u' = 1

    v = cos(3x)

    setting: w = 3x and v = cos(w)

    dv/dx = dv/dw * dw/dx

    Now we get the derivatives of w and v:

    w = 3x and v = cos(w)

    dw/dx = 3 and dv/dw = -sin(w)

    Substitute 3x back in for w to get:

    dw/dx = 3 and dv/dw = -sin(3x)

    Now we have what we need to complete the chain rule:

    dv/dx = dv/dw * dw/dx

    dv/dx = -sin(3x) * 3

    v' = -3sin(3x)

    Now we have what we need to complete the product rule.  Summarizing what we know:

    u = x and v = cos(3x)

    u' = 1 and v' = -3sin(3x)

    y' = u'v + v'u

    y' = 1cos(3x) + (-3)sin(3x)x

    y' = cos(3x) - 3x sin(3x)

    Now that we have this, the slope at x = π can be found by solving for y'(π):

    y' = cos(3π) - 3π sin(3π)

    y' = -1 - 3π(0)

    y' = -1 - 0

    y' = -1

    The slope is -1.

    We are given x = π, so now we need to find "y" at this point:

    y = x cos(3x)

    y = π cos(3π)

    y = π(-1)

    y = -π

    Now that we have the slope, x, and y, we can solve for the intercept:

    -π = (-1)π + b

    -π = -π + b

    0 = b

    The tangent line at x = π is:

    y = -x

  • Ash
    Lv 7
    1 month ago

    y=xcos(3x)

    y' = cos(3x)  + x (-sin(3x) * 3)

    y' = cos(3x) - 3x sin(3x)

    y' = slope of tangent

    At x = π, 

    slope of tangent =  cos(3π) - 3π sin(3π)

    slope of tangent = -1 - 3π (0)

    slope of tangent = -1

    m = -1

    If you want to find the equation of line then find the y-coordinate for point with x=π

    y=πcos(3π)

    y = -π

    tangent point is (π,-π)

    y-(-π) = -1(x-π)

    y+π = -x+π

    y = -x   ← equation of tangent

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