Calculus with Vector Functions — Question 7

Question 7 – Solution

Strategy A zero derivative makes the given parametrization nonregular, but the geometric curve may still have a tangent.

Regularity $\vec r'(t)=\left\langle 2t,3t^2\right\rangle$, so $\vec r'(0)=\vec 0$ and this polynomial parametrization is smooth but not regular at 0. Eliminating $t$ gives $y^2=x^3$, a semicubical cusp.

Tangent direction For $t\ne 0$, the secant direction from the origin is proportional to $\left\langle t^2,t^3\right\rangle=t^2\left\langle 1,t\right\rangle$, which approaches $\left\langle 1,0\right\rangle$ from both sides. Thus the geometric tangent line exists and is $\boxed{y=0}$.

Distinction The curve has a tangent at the cusp, but no nonzero velocity there.