`x=t+1 , y=t^2+3t` Find all points (if any) of horizontal and vertical tangency to the curve.

Parametric curve (x(t),y(t)) has a horizontal tangent if its slope `dy/dx` is zero i.e. when `dy/dt=0` and `dx/dt!=0`

It has a vertical tangent, if its slope approaches infinity i.e. `dx/dt=0` and `dy/dt!=0`

Given parametric equations are:

`x=t+1`

`y=t^2+3t`

`dx/dt=1`

`dy/dt=2t+3`

For Horizontal tangents,

`dy/dt=0`

`2t+3=0`

`=>2t=-3`

`=>t=-3/2`

Corresponding point on the curve can be found by plugging in the value of t in the equations,

`x=-3/2+1=-1/2`

`y=(-3/2)^2+3(-3/2)`

`y=9/4-9/2`

`y=(9-18)/4=-9/4`

Horizontal tangent is at the point `(-1/2,-9/4)`

For vertical tangents,

`dx/dt=0`

However `dx/dt=1!=0`

So there are no vertical tangents.

Comments

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