Need help w/horizontal&vertical tangent pts for parametrics

[commanderbuttons]

I solved the problem correctly (I believe), but I am having trouble because I have a coordinate point that shows up as the location of a vertical and horizontal tangent line.

The problem:
Find the points to the curve where the tangent line to the curve is vertical or horizontal.
C: x=t^3-3t^2 , y=t^3-3t
dy/dt=3t^2-3
dx/dt=3t^2-6t

Horizontal:
0=3t^2-3
0=(t^2-1)
t=+/-1
x(1)=-2
y(1)=-2
(-2,-2)
x(-1)=-4
y(-1)=2
(-4,2)

Vertical:
0=3t^2-6t
0=t^2-2t
0=t(t-2)...t=0,2
x(0)=0
y(0)=0
(0,0)
x(2)=8-12=-4
y(2)=8-6=2
(-4,2)

I am confused because the point (-4,2) appears to be both a vertical and horizontal tangent which doesn't look or sound right to me. What am I missing here?

 

[galactus]

Re: Need help w/horizontal&vertical tangent pts for parametr

Since \(\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\), the vertical tangents occur where \(\displaystyle \frac{dx}{dt}=0\), but \(\displaystyle \frac{dy}{dt}\neq 0\).

\(\displaystyle x'(t)=3t^{2}-6t\)

\(\displaystyle 3t^{2}-6t=0\)

\(\displaystyle t=0, \;\ t=2\)

 

[commanderbuttons]

Re: Need help w/horizontal&vertical tangent pts for parametr

yes I know that, but what if the coordinate pair appears as both a vertical and horizontal tangent?

 

[galactus]

Re: Need help w/horizontal&vertical tangent pts for parametr

Here is the graph. The horizontal and vertical occur at the same location, but the t value is different.

For t=-1 and t=1, the line is horizontal. For t=2, the line is vertical.

\(\displaystyle \frac{dy}{dt}=3t^{2}-3=0\)

\(\displaystyle t=-1, \;\ t=1\)

The horizontal and vertical lines occur at the same location on the graph. It's just that the t values differ.

 

[commanderbuttons]

Re: Need help w/horizontal&vertical tangent pts for parametr

Okay everything is much more clear now. Thank you for the graph and the explanation!

 

[Deleted member 4993]

Re: Need help w/horizontal&vertical tangent pts for parametr

For these types of problems - it is always advisable to use your graphing calculator and graph the given expression. Note that the given expression is not a function - it fails vertical line test.