Calculus 2 Parametric Question

Question

A curve is defined parametrically by

x = t^2 - 2 and y = t^3 - t for t ∈ R

(a) Find the point where the curve crosses itself.

(b) Find dy/dx and d^2y/dx^2

(c) Determine the points at which the tangent line is horizontal and the points at which the tangent line is vertical.

(d) Perform a sign analysis on dx/dt and dy/dt to determine the orientation of the curve.

(e) Use the information obtained in previous parts to graph the curve.

Answer 1

a) we find possible points where it crosses by seeing where multiple values of t yield the same value of x and y

x = t^2 - 2, so t and - t yield the same value of x for all t

y = t^3 - t, so, from x having the same value for t and - t, we need to find where t and -t yield the same value for y.

t^3 - t = (-t)^3 - (-t

2t^3 - 2t = 0

2t(t^2 - 1) = 0.

t = 0, +-1

t = 0 is not a solution, as 0 and -0 are the same value.

Not surprisingly, we have negatives of each other as the other solution.

At t = 1 and -1, (x,y) = (1^2 - 2, 1^3 - 1) = (-1, 0)

To show that it is a crossing, consider dx/dt and dy/dt for t = -1 and t = 1

dx/dt = 2t

dy/dt = 3t^2 - 1

Thus, at t = -1, dy/dx = (3*(-1)^2- 1)/(2 * -1)) = 2/-2 = -1

at t = 1, dy/dx = (3*(1)^2- 1)/(2 * 1)) = 2/2 = 1

Thus, this is a crossing.

b) We could build on our answer above and say that dy/dx = (3t^2 - 1)/2t and then substitute for t in terms of x; x = t^2 - 2, or t^2 = x+2 or t = +- √(x+2)

The +- is important here, as the + value yields one value of y, and the - value yields the negative of it.

We have two solutions, then, for dy/dx and d²y/dx²

From t > 0,

dy/dx = (3x+5)/(2(x+2)^1/2)

d²y/dx² = (3x+7)/(4(x+2)^3/2)

From t < 0,

dy/dx = -(3x+5)/(2(x+2)^1/2)

d²y/dx² = -(3x+7)/(4(x+2)^3/2)

c) We can build on the solutions in b or go back to dy/dt and dx/dt

building, we see that dy/dx = 0 when 3x+5 = 0, or x = -5/3; this is where the curve will be horizontal

recall that there will be 2 values of y at this point - t = +- √(-5/3 + 2) = +-1/3^1/2

for t = sqrt(1/3), y = t³ - t = 1/3^3/2 - 1/3^1/2 = -2/3^3/2

for t = -sqrt(1/3), y = t³ - t = -1/3^3/2 + 1/3^1/2 = 2/3^3/2

horizontal at (-5/3, -2/3^3/2 ) and (-5/3, 2/3^3/2)

Vertical where the denominator of dy/dx = 0. This occurs when x = -2; as t = 0 here, we have one value of y, (-0)³ - (-0) = 0; at (-2, 0), the tangent line is vertical.

d) From c), we see three turning points for dy/dx

In terms of t, we see that dy/dx is negative on (-∞,-1/3^1/2), positive on (-1/3^1/2, 0), negative on (0,1/3^1/2), and positive on (1/3^1/2, ∞)

For t on (-∞,-1/3^1/2), we see (x,y) go from (∞, -∞) to (-5/3, 2/3^3/2) where dy/dx is negative

For t on (-1/3^1/2, 0), we see (x,y) go from (-5/3, 2/3^3/2) to (-2, 0) dy/dx is positive, note that we are moving in a "negative" direction - y and x decrease as t increases.

For t on (0,1/3^1/2), (x,y) goes from (-2, 0) to (-5/3, -2/3^3/2) - x increases and y decreases, dy/dx is negative

For t on (1/3^1/2, ∞), (x,y) goes from (-5/3, -2/3^3/2) to (∞, ∞)

e) I gave the turning points and directions in d. It is easy to graph this by just having t range from -2 to 2 and calculating x and y.