Calculus AB

posted by Annie

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. Reiny

    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.

  2. Annie

    Thank you so much! And I will

Respond to this Question

First Name

Your Answer

Similar Questions

  1. 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, …
  2. Calculus - Damon

    Find the line which passes through the point (0, 1/4) and is tangent to the curve y=x^3 at some point. So I found the derivative which is 3x^2. Let (a, a3) be the point of tangency. 3x^2 = (a3 - 1/4)/(a-0) I'm not sure how to solve …
  3. Calculus

    I have a two part question that pertains to a curve (r(x)) and its tangent line at x=3. We are given that at x=3, r(x)=8. In order to find the slope of the tangent line, we are given another point (on the tangent line): (3.2, 8.5). …
  4. calculus

    Consider line segments which are tangent to a point on the right half (x>0) of the curve y = x^2 + 1 and connect the tangent point to the x-axis. If the tangent point is close to the y-axis, the line segment is long. If the tangent …
  5. calculus

    Consider line segments which are tangent to a point on the right half (x>0) of the curve y = x^2 + 1 and connect the tangent point to the x-axis. If the tangent point is close to the y-axis, the line segment is long. If the tangent …
  6. calculus

    Consider the curve defined by 2y^3+6X^2(y)- 12x^2 +6y=1 . a. Show that dy/dx= (4x-2xy)/(x^2+y^2+1) b. Write an equation of each horizontal tangent line to the curve. c. The line through the origin with slope -1 is tangent to the curve …
  7. Calculus

    Consider line segments which are tangent to a point on the right half (x>0) of the curve y = x^2+7 and connect the tangent point to the x-axis. If the tangent point is close to the y-axis, the line segment is long. If the tangent …
  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. AP AB Calculus

    Linear approximation: Consider the curve defined by -8x^2 + 5xy + y^3 = -149 a. find dy/dx b. write an equation for the tangent line to the curve at the point (4,-1) c. There is a number k so that the point (4.2,k) is on the curve. …
  10. 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?

More Similar Questions