calculus

posted by .

Find all points on the curve x^2y^2+xy=2 where the slope of tangent line is -1.

This is what I got.

x^2y^2+xy=2

x^2*2ydy/dx + 2x*y^2 + x*dy/dx + y =0

dy/dx(2yx^2+x)= -2xy^2-y

dy/dx = (-2xy^2-y)/(x+2yx^2)

how do I finish from here? I need to know where the slop of the tangent line equals -1.

  • calculus -

    You can insert dy/dx = -1 at this point:

    x^2*2ydy/dx + 2x*y^2 + x*dy/dx + y =0

    You then get an equation relating x and y. You also know that x and y are on te curve, so they satisfy the equation of the curve:

    x^2y^2+xy = 2

    So, you have two equations for the two unknowns x and y.

  • calculus -

    So I put -1 in for dy/dx and got this equation: -2x^2y+2xy-x+y=0. Then would I solve for either x or y and substitute back in to the original equation to get a value for x and y?

  • calculus -

    Yes.

  • calculus -

    I got this for y.

    2x^2y-2xy+x-y=0

    2x^2y-2xy+x=y

    2x^2-2x+x=1 (divide through by y)

    2x^2-x-1=0

    what do I from here? It doesn't factor evenly. I'm stuck!

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Calculus

    The line that is normal to the curve x^2=2xy-3y^2=0 at(1,1) intersects the curve at what other point?
  2. Ap Calc AB

    Determine (dy/dx) using implicit differentiation. cos(X^2Y^2) = x I'm really confused what to do now..i think the next steps are: d/dx [cos(X^2*Y^2)] = d/dx [x] = -sin(X^2*Y^2)* ((X^2*2Y dy/dx) + (Y^2*2X)) = 1 = -2YX^2 sin(X^2*Y^2) …
  3. Calculus

    Find all points on the curve x^2y^2+2xy=3 where the slope of the tangent is -1. If there is more than one point, use a comma to separate each pair of coordinates
  4. Calculus

    the curve: (x)(y^2)-(x^3)(y)=6 (dy/dx)=(3(x^2)y-(y^2))/(2xy-(x^3)) a) find all points on the curve whose x-coordinate is 1 and write an equation for the tangent line of each of these points b)find the x-coordinate of each point on …
  5. Calculus

    Use implicit differentiation to find the slope of the tangent line to the curve 2(x^2) + 2xy + 2(y^3)= -18
  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 [finding slope of tangent line]

    Find the slope of the tangent line to the curve 0x^2−2xy+2y^3=10 at point (-4,1) Help please?
  8. calculus

    Find the slope of the tangent line to the curve (2x+4y)^1/2+(2xy)^1/2=13.4 at the point (6,5).
  9. Calculus

    Find y" by implicit differentiation. x^3+y^3 = 1 (x^3)'+(y^3)' = (1)' 3x^2+3y^2(y') = 0 3y^2(y') = -3x^2 y' = -3x^2/3y^2 y' = -x^2/y^2 y" = [(y^2)(-x^2)'-(-x^2)(y^2)']/y^4 y" = [(y^2)(-2x)-(-x^2)(2y)]/y^4 y" = [(-2xy^2)-(-2yx^2)]/y^4 …
  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