Posted by Chelsea on Sunday, October 31, 2010 at 9:17pm.
a)The curve with equation: 2y^3 + y^2  y^5 = x^4  2x^3 + x^2 has been linked to a bouncing wagon. Use a computer algebra system to graph this curve and discover why.
b)At how many points does this curve have horizontal tangent lines? Find the xcoordinates of these points.
I used MAPLE to graph the equation, but I do not understand what they mean. The graph has a closed odd shape at the top and two lines that extend from slightly below the origin, one going to the left and one to the right. Please help?
For part b, the curve appears to have horizontal tangent lines at 3 points??? I used implicit differentiation on the equation and then solved for the values of x, which I got (0, 1, and 0.5)?

Calculus  Damon, Sunday, October 31, 2010 at 9:38pm
what does dy/dx look like?
6 y^2 dy +2 y dy 5 y^4 dy = 4 x^3 dx  6 x^2 dx + 2 x dx
so
dy/dx = (4x^36x^2+2x) / (6y^2+2y5y^4)
the curve is horizontal where dy/dx = 0
2x^33x^2+x = 0
x(2x1)(x1)=0
so horizontal when x = 0, x = 1/2, and x = 1
when the denominator is 0, the slope is undefined (curve is vertical)
6y^2+2y5y^4 = 0
y(6y +2 5y^3) = 0
so jumps (hits a rock or pot hole or something) at y = 0 and possibly at three other points.
When you graph this, watch out for the points that are undefined (denominator = 0)

Calculus  Reiny, Sunday, October 31, 2010 at 9:41pm
I don't know which "computer algebra system" you are using, but I differentiated the relation and got
y' = (4x^3  6x^2 + 2x)/(6y^2 + 2y  5y^4) = 0
4x^3  6x^2 + 2x = 0
2x(2x^2  3x + 1) = 0
2x(2x1)(x1) = 0
x = 0 or x = 1/2 or x = 1
so you were right in part b)
(I actually volunteered to be on the evaluation group of the original MAPLE project at my old alma mater , the University of Waterloo, in Canada.
Glad to see the project worked out, was wondering what happened to it.)
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