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.

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

Respond to this Question

First Name

Your Response

Similar Questions

  1. Calculus AB

    Sorry but I've got a lot of problems that I don't understand. 1) Let f(x)= (3x-1)e^x. For which value of x is the slope of the tangent line to f positive? Negative? Zero? 2) Find an equation of the tangent line to the oven curve
  2. cal

    find the equation of a quadratic function whose graph is tangent at x=1 to the line whose slope8, tangent at x=-2 to solve the line with slope-4 and tangent to the line y=-8 find the equation of the tangent lines at x=1 and x=-2
  3. Math (equation of tangent line)

    Consider the implicit equation 2xy-1=(x+y+1)^2 a) Compute and solve for the derivative dy/dx as a function of x and y. b) Find the equation of the tangent line to the graph of the above when y=-1. For part a, I found the
  4. Calc.

    Find the area of the region bounded by the parabola y=x^2, the tangent line to this parabola at (1,1) and the x-axis. I don't really get what this question is asking. It looks like the area of right triangle to me...try the graph,
  5. Calculus - Tangent Line

    Hi, im having problems with the following problem. The main issue is actually starting the problem. Find the two points on the curve y = x^4 - 2x^2 - x that have a common tangent line. First, find the derivative of y(x) so that
  6. calculus

    find the equation of a quadratic function whose graph is tangent at x=1 to the line whose slope8, tangent at x=-2 to solve the line with slope-4 and tangent to the line y=-8
  7. Math (Calculus) (mean value theorem emergency)

    Consider the graph of the function f(x)=x^2-x-12 a) Find the equation of the secant line joining the points (-2,-6) and (4,0). I got the equation of the secant line to be y=x-4 b) Use the Mean Value Theorem to determine a point c
  8. Calculus

    If F(x)=x^3−7x+5, use the limit definition of the derivative to find FŒ(5), then find an equation of the tangent line to the curve y=x^3−7x+5 at the point (5, 95). FŒ(5)= The equation of the tangent line is y = x
  9. Circles (Conic Sections)

    I have no idea how to do these problems: Find an equation of each circle. 1. Center (3,5); tangent to the x axis 2. Center (5,-3); tangent to the y axis 3. Tangent to the x axis, y axis, and the line y=5 (two answers) I just
  10. math

    Use implicit differentiation to find the equation of the tangent line to the curve xy3+xy=14 at the point (7,1) . The equation of this tangent line can be written in the form y=mx+b i don't seem to no how to find m or b

More Similar Questions