# Thread: Parametric Equations - Horizontal and Vertical Tangent

1. ## Parametric Equations - Horizontal and Vertical Tangent

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.)

$\displaystyle x = t^{3} - 3t$

$\displaystyle y = t^{2} - 3$

I'm assuming setting the parametric derivative to 0 would get the horizontal tangent. But how about finding the vertical tangent (undefined)?

2. ## Re: Parametric Equations - Horizontal and Vertical Tangent

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

A fraction is 0 when the numerator, here $\displaystyle \frac{dy}{dt}$, is 0.

It does not exist (is undefined) when the denominator, here $\displaystyle \frac{dx}{dt}$, is 0.

3. ## Re: Parametric Equations - Horizontal and Vertical Tangent

I came up with (-2,2) for the horizontal tangent and (0,-3) for the vertical one (but they are wrong on homework) ???

Horizontal

$\displaystyle y = t^{2} - 3$

$\displaystyle \dfrac{dy}{dt} = 2t$

$\displaystyle 2t = 0$

$\displaystyle t = 0$

$\displaystyle y = (0)^{2} - 3 = -3$

$\displaystyle x = (0)^{3} - 3(0) = 0$

$\displaystyle (0,-3)$

Vertical

$\displaystyle x = t^{3} -3t$

$\displaystyle \dfrac{dx}{dt} = 3t^{3} - 3t$

$\displaystyle 3t^{2} - 3 = 0$

$\displaystyle t = 1$

$\displaystyle y = (1)^{2} - 3 = -2$

$\displaystyle x = (1)^{3} - 3(1) = -2$

$\displaystyle (-2,-2)$

4. ## Re: Parametric Equations - Horizontal and Vertical Tangent

Originally Posted by Jason76
I came up with (-2,2) for the horizontal tangent and (0,-3) for the vertical one (but they are wrong on homework) ???

Horizontal

$\displaystyle y = t^{2} - 3$

$\displaystyle \dfrac{dy}{dt} = 2t$

$\displaystyle 2t = 0$

$\displaystyle t = 0$

$\displaystyle y = (0)^{2} - 3 = -3$

$\displaystyle x = (0)^{3} - 3(0) = 0$

$\displaystyle (0,-3)$

Vertical

$\displaystyle x = t^{3} -3t$

$\displaystyle \dfrac{dx}{dt} = 3t^{3} - 3t$

$\displaystyle 3t^{2} - 3 = 0$

$\displaystyle t = 1$

$\displaystyle y = (1)^{2} - 3 = -2$

$\displaystyle x = (1)^{3} - 3(1) = -2$

$\displaystyle (-2,-2)$
I get horizontal tangent at (0,-3)

Vertical tangents at (-2,-2) and (2,-2)

$(-1)^2=1$ also.

5. ## Re: Parametric Equations - Horizontal and Vertical Tangent

Originally Posted by Jason76
I came up with (-2,2) for the horizontal tangent and (0,-3) for the vertical one (but they are wrong on homework) ???

Horizontal

$\displaystyle y = t^{2} - 3$

$\displaystyle \dfrac{dy}{dt} = 2t$

$\displaystyle 2t = 0$

$\displaystyle t = 0$

$\displaystyle y = (0)^{2} - 3 = -3$

$\displaystyle x = (0)^{3} - 3(0) = 0$

$\displaystyle (0,-3)$

Vertical

$\displaystyle x = t^{3} -3t$

$\displaystyle \dfrac{dx}{dt} = 3t^{3} - 3t$
This is wrong. Possibly a typo but then you continue with it. The derivative of $\displaystyle t^3$ is $\displaystyle 3t^2$ not $\displaystyle 3t^3$.

$\displaystyle 3t^{2} - 3 = 0$

$\displaystyle t = 1$

$\displaystyle y = (1)^{2} - 3 = -2$

$\displaystyle x = (1)^{3} - 3(1) = -2$

$\displaystyle (-2,-2)$

6. ## Re: Parametric Equations - Horizontal and Vertical Tangent

Originally Posted by romsek
I get horizontal tangent at (0,-3)

Vertical tangents at (-2,-2) and (2,-2)

$(-1)^2=1$ also.

Are you sure you don't have those backwards? The horiz tangent is vertical one etc..

7. ## Re: Parametric Equations - Horizontal and Vertical Tangent

No idea on this one. Nothing seems to work.