investigate the family of curves defined by the parametric equations x=t^2 , y=t

Question

investigate the family of curves defined by the parametric equations x=t^2 , y=t^3-ct , where c is a positive constant

a) Graph the curves for c = 1/4 , c = 1, c = 4. What features do all the curves have in common? How does the shape change as c increases? Find the x- and y- coordinates of all the points where the tangent line is horizontal or vertical.

b) Verify that point (c,0) is on the curve for any c >0. How many tangent lines does the curve have at the point (c,0) What are their slopes? Check answers numerically (for c = 1/4 , c = 1, c = 4) by drawing the tangent lines on graphing calc.

c) Consider the curve corresponding to c = 1/3. Part of this curve is a loop. Find the length of that loop.

** Need in paragraph form to explain**

Explanation / Answer

x = t^2, y = t^3 - ct = t(t^2 - c)

ie y = +/- sqrt(x)[x - c]


As c increases the curves further apart

dy/dx = (3/2)x^(1/2) - (c/2)x^(-1/2) = (3x - c)/(2sqrt(x))

When tangent is horizontal dy/dx = (3x - c)/(2sqrt(x)) = 0
ie x = c/3 Therefore tangent is horizontal at
(c/3, -2c^(3/2)/(3sqrt(3))) and (c/3, 2c^(3/2)/(3sqrt(3)))

Tangent is vertical when dy/dx = (3x - c)/(2sqrt(x)) = infinity

ie when 2sqrt(x) = 0 or x=0
Point is the origen (0, 0)

b) when x = c, y = 0 ie y is independant of c for x=c. Point proved.
when x=c, dy/dx = c/sqrt(c) = sqrt(c)
So there are 2 tangent lines at (c, 0) with slopes sqrt(c) and -sqrt(c).

c) For c = 1, y = +/- sqrt(x)(x-1)
ie curve intercepts x axis at 0 and 1

Length of loop = 2*INT[sqrt((dx/dt)^2 + (dy/dt)^2)]dt between 0 and 1
=2*INT[sqrt(9t^4 - 2t^2 + 1)]dt between t=0 and t=1
=2*IN

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