Question

Find all points (if any) of horizontal and vertical tangency to the curve. Use a graphing utility to confirm your results. $$ x=1-t, \quad y=t^{2} $$

Short answer

The curve \(x=1-t, y=t^{2}\) only has one point of horizontal tangency which is at (1,0) and no points of vertical tangency.

Step-by-step solution

Finding Dy/Dx

First, find the derivative of the function. As we have two parametric equations \(x = 1 - t\) and \(y = t^{2}\), we need to find \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\). Here, \(\frac{dx}{dt} = -1\) and \(\frac{dy}{dt} = 2t\). Now we can find \(\frac{dy}{dx}\) by dividing \(\frac{dy}{dt}\) by \(\frac{dx}{dt} = -2t\).

Find Horizontal Tangent

To find horizontal tangency, set \(\frac{dy}{dx} = 0\). This gives the equation \(-2t=0\), which results in \(t=0\). Substitute \(t=0\) into the original parametric equations to find the coordinating point \(x=1, y=0\).

Find Vertical Tangent

The vertical tangent occurs when \(\frac{dx}{dt} = 0\), but in this case \(\frac{dx}{dt}\) is a constant \(-1\) that doesn’t depend on \(t\), so there’re no points of vertical tangency in this case.