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 x-coordinates 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)?

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^3-6x^2+2x) / (6y^2+2y-5y^4)

the curve is horizontal where dy/dx = 0

2x^3-3x^2+x = 0
x(2x-1)(x-1)=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+2y-5y^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)

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(2x-1)(x-1) = 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.)

a) Ah, the mysterious bouncing wagon curve! I must say, this equation sounds like it was derived by a mathematician with a sense of humor.

Now, let's break it down. The reason this curve is linked to a bouncing wagon is because of its interesting shape. The closed odd shape at the top represents the wagon bouncing up and down, while the lines extending from below the origin symbolize the horizontal motion of the wagon.

b) Ah, horizontal tangent lines, the moments when the curve is momentarily flat like a pancake. Well, it seems you're on the right track. By using implicit differentiation and solving for the values of x, you got (0, 1, and 0.5) as the x-coordinates of these points. Bravo!

So, in conclusion, this bouncing wagon curve has a fun and playful nature, with its bouncing shape and flat pancake-like moments. Keep exploring, my friend, and have a wheel-y good time with your mathematical adventures!

To answer part a) of the question, you mentioned that you used MAPLE to graph the equation. That's a great start! The graph you described with a closed odd shape at the top and two lines extending from slightly below the origin sounds correct. Let me explain why this graph is linked to a bouncing wagon.

The equation you provided, 2y^3 + y^2 - y^5 = x^4 - 2x^3 + x^2, represents the motion of a bouncing wagon. The x-coordinate represents time, and the y-coordinate represents the height of the wagon at a given time. The graph shows the trajectory or path that the wagon follows as it bounces.

The closed odd shape at the top of the graph represents the highest point the wagon reaches when it bounces. After reaching this maximum height, the wagon starts descending due to gravity. The two lines extending from slightly below the origin represent the wagon bouncing back and forth.

To answer part b) of the question, you correctly mentioned that the curve has horizontal tangent lines at three points: (0, 1) and (0.5, 0). These are called critical points, where the slope (or derivative) of the curve is zero. However, you missed one more critical point.

To find the x-coordinates of these points, you mentioned that you used implicit differentiation on the equation. Implicit differentiation is indeed the right approach to find the derivative of y with respect to x. Let's go through the steps to find the critical points.

1. Differentiate the equation with respect to x using implicit differentiation. This involves differentiating each term of the equation with respect to x and treating y as a function of x. The derivative of x with respect to x is simply 1, and for y with respect to x, we use the chain rule.

2y^3 + y^2 - y^5 = x^4 - 2x^3 + x^2
Differentiating both sides with respect to x:
6y^2(dy/dx) + 2y(dy/dx) - 5y^4(dy/dx) = 4x^3 - 6x^2 + 2x

2. Set the derivative equal to zero since the question asks for horizontal tangent lines.
6y^2(dy/dx) + 2y(dy/dx) - 5y^4(dy/dx) = 4x^3 - 6x^2 + 2x
(dy/dx)(6y^2 + 2y - 5y^4) = 4x^3 - 6x^2 + 2x

3. Solve for dy/dx:
(dy/dx) = (4x^3 - 6x^2 + 2x) / (6y^2 + 2y - 5y^4)

4. To find the x-coordinates of the points where the derivative is zero, we set dy/dx equal to zero and solve for x. Let's solve the equation:
0 = (4x^3 - 6x^2 + 2x) / (6y^2 + 2y - 5y^4)

Multiplying both sides by (6y^2 + 2y - 5y^4):
0 = 4x^3 - 6x^2 + 2x

0 = 2x(x - 1)(2x + 1)

This equation has three solutions, x = 0, x = 1, and x = -1/2.

So, the x-coordinates of the points where the curve has horizontal tangent lines are 0, 1, and -1/2. You were correct in finding the values of x as 0 and 1, but you missed -1/2.

I hope this explanation helps clarify the concepts and process behind graphing the curve and finding the points of horizontal tangents. Let me know if you have any further questions!