Confusion with a parametric equation problem

Question

Need some help figuring this problem out, would really appreciate if someone could provide the steps to their solution as well, as I feel like I'm missing one key step in solving it.

Consider the curve defined by parametric equations x=t^2 and y=t^3 - 3t

a) Find all points on the curve where the tangent line is horizontal

and

b) Find all points on the curve where the tangent line is vertical

Thank you!

Answer 1

dy/dx = dy/dt / dx/dt = (3t² - 3) / 2t.

The graph will have a horizontal tangent line when the numerator equals zero, i.e. +/- 1. Substitute those values into (t², t³ - 3t). to get coordinates of points with horizontal tangent lines.

The graph will have vertical tangent line when dy/dx is undefined, i.e. when the denominator is zero in this case. Again substitute 0 for t in the parametric definition to get (x,y).

Answer 2

It might be easier to do these on the Cartesian (XY) plane than parametrically. To convert from parametric to rectangular, solve one equation for T and plug that T (substitute) into the other equation. Here the easiest equation would be x=T^2 or T=√x. Putting the two equations together, y=(√x)^3-3(√x) or y=x^3/2-3x^1/2.

The places where a tangent would be horizontal are any places where the first derivative (dy/dx) is 0 meaning that the slope of the tangent line is 0. The slope of a vertical tangent line would be infinite so any points with a undefined slope (vertical lines) would qualify.

To get the first derivative for this equation, we can use the Power Rule which states that the derivative of x^n is nx^(n-1) so the derivative of this is 3/2x^1/2-3/2x^-1/2. Setting that equal to 0 would give the horizontal asymptote.

Answer 3

a) Equation of the horizontal tangent line:

Step 1. Find the derivative for f(x)

x=t^2

dx/dt = 2t

Step 2. Solve for dx/dt = 0

dx/dt = 2t = 0

t= 0

Step 3. Solve for x and y at t found in step 2

x(t=0) = t^2 = 0^2

x = 0

y(t=0) = t^3 - 3t =0^3 - 3*0

y = 0

Step 4. Points on the curve where tanget line is horizontal is x=0, y=0

b) Equation of the vertical tangent line:

Step 1. Find the derivative for f(y)

y = t^3 - 3t

dy/dt = 3t^2 - 3

Step 2. Solve for dy/dt = 0

dy/dt = 3t^2 - 3 = 0

3t^2 = 3

t^2 = 1

t= 1 and -1

Step 3. Solve for x and y at t found in step 2

Solve for t =1:

x(t=1) = t^2 = 1^2

x = 1

y(t=1) = t^3 - 3t =1^3 - 3*1 = 1 - 3

y = -2

Solve for t = -1

x(t=-1) = t^2 = -1^2

x = 1

y(t=-1) = t^3 - 3t = -1^3 - 3*-1 = -1 + 3

y = 2

Step 4. Points on the curve where tanget line is vertical is x = 1, y = -2 and x = 1, y = 2

Answer 4

Hi Sha K.,

since this question is posed under calculus I will assume that you have some facility with that.

1) Perhaps easiest is to solve as y=f(x). This is easy to do; t=x^(1/2) and thus y = x^(3/2) - 3x^(1/2) .

2) Check to make sure that you have adequately set up to trace for both t>=0 and t<=0 cases.

3) Differentiate the equation above to obtain y'(x). That would be: y' = (3/2)x^(1/2) - (3/2)x^(-1/2) .

4) Now, obviously, at least one term there will blow up at x=0, and it would appear that there is at least one x value for which (3/2)x^(1/2) = (3/2)x^(-1/2) , looks like x=1 to me.

5) Also check for additional points on the function, in light of (2) above.

That should about do it?

-- Cheers, -- Mr. d.