Calculus AB

Could someone please help me with these tangent line problems?

1) Find the equation of the line tangent to the given curve at the indicated point: 3y^3 + 2x^2 = 5 at a point in the first quadrant where y=1.

2) Show that there is no point on the graph of x^2 - 3xy + y^2 = 1 where the line tangent to the curve is horizontal.

3) Find all points on the curve xy^2 - x^3y = 6 where the tangent line is vertical. Explain your reasoning.

  1. 1
asked by Annie
  1. 3y^3 + 2x^2 = 5
    9y^2 dy/dx + 4x = 0
    dy/dx = -4x/9y^2

    when y = 1 , 3 + 2x^2 = 5
    2x^2 = 2
    x = ± 1 , but in quadrant 1, x = 1

    dy/dx = -4(1)/9 = -4/9
    equation of tangent:
    y-1 = (-4/9)(x-1)
    9y-9 = -4x + 4
    4x + 9y = 13

    check:
    http://www.wolframalpha.com/input/?i=plot+3y%5E3+%2B+2x%5E2+%3D+5,+4x+%2B+9y+%3D+13
    notice the tangent at (1,1)

    2.
    x^2 - 3xy + y^2 = 1
    2x -3x(dy/dx) - 3y + 2y dy/dx = 0
    dy/dx(-3x + 2y) = 3y - 2x
    dy/dx = (3y - 2x)/(2y - 3x)
    to have a horizontal slope, dy/dx = 0 , that is,
    3y - 2x = 0
    y = 2x/3
    sub that back into the original
    x^2 - 3x(2x/3) + 4x^2/9 = 1
    times 9
    9x^2 - 18x^2 + 4x^2 = 1
    -5x^2 = 1
    There is no real solution to this, so there is no such tangent.

    3.
    find the derivative like I showed you in the first two.
    Since the tangent is vertical, the denominator of the slope must be zero.
    Show that this is not possible.

    posted by Reiny
  2. Thank you so much! And I will

    posted by Annie

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