Question

Find all points (if any) of horizontal and vertical tangency to the curve. Use a graphing utility to confirm your results. (If an answer does not exist, enter DNE.)
x=1-t , y=t^2

Horizontal tangent
(x,y)=________
Vertical tangent
(x,y)=________

Answer 1

Answer:

Horizontal tangent

(x, y) = (1, 0)

Vertical tangent

(x, y) = DNE

Step-by-step explanation:

The equation for the slope (m) of the tangent line at any point of a parametric curve is:

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

Where $\frac{dx}{dt}$ and $\frac{dy}{dt}$ are the first derivatives of the horizontal and vertical components of the parametric curves. Now, the first derivatives are now obtained:

$\frac{dx}{dt} = -1$ and $\frac{dy}{dt} = 2\cdot t$

The equation of the slope is:

$m = -2\cdot t$

As resulting expression is a linear function, there are no discontinuities and for that reason there are no vertical tangents. However, there is one horizontal tangent, which is:

$-2\cdot t = 0$

$t = 0$

The point associated with the horizontal tangent is:

$x = 1 - 0$

$x = 1$

$y = 0^{2}$

$y = 0$

The answer is:

Horizontal tangent

(x, y) = (1, 0)

Vertical tangent

(x, y) = DNE