Question

Find the points on the curve where the tangent is horizontal or vertical. If you have a graphing device, graph the curve to check your work. (Enter your answers as a comma-separated list of ordered pairs.)x = t^3 - 3t, y = t^2 - 4

Answer 1

Answer:

There is a horizontal tangent at (0,-4)

The tangent is vertical at (-2,-3) and (2,-3).

Step-by-step explanation:

The given function is defined parametrically by the equations:

$x=t^3-3t$

and

$y=t^2-4$

The tangent function is given by:

$\frac{dy}{dx}=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }$

$\implies \frac{dy}{dx}=\frac{2t}{3t^2-3}$

The tangent is vertical at when $\frac{dx}{dt}=0$

$\implies \frac{3t^2-3}{2t}=0$

$\implies 3t^2-3=0$

$\implies 3t^2=3$

$\implies t^2=1$

$\implies t=\pm1$

When t=1,

$x=1^3-3(1)=-2$ and $y=1^2-4=-3$

When t=-1,

$x=(-1)^3-3(-1)=2$ and $y=(-1)^2-4=-3$

The tangent is vertical at (-2,-3) and (2,-3).

The tangent is horizontal, when $\frac{dy}{dx}=0$ or  $\frac{dy}{dt}=0$

$\implies 2t=0$

$\implies t=0$

When t=0,

$x=0^3-3(0)=0$ and $y=0^2-4=-4$

There is a horizontal tangent at (0,-4)