# Second derivative with parametric equations

1. Sep 8, 2015

http://tutorial.math.lamar.edu/Classes/CalcII/ParaTangent.aspx

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

provided $\frac{dx}{dt} \neq 0$.

In example 4, $\frac{dx}{dt} = -2t$, which is zero when $t$ is zero. In simplifying $\frac{dy}{dx}$ the author even divides the numerator and denominator by $t$ which is only possible if $t \neq 0$.
This is all consistent with the requirement $\frac{dx}{dt} \neq 0$.
The author then obtains an expression for the second derivative in terms of $t$, plugs in zero, and finds out that the second derivative is zero at $t = 0$.
How is this consistent with the assumption that $\frac{dx}{dt} \neq 0$? What's going on here?

2. Sep 8, 2015

3. Sep 8, 2015

### Staff: Mentor

He doesn't actually "plug in" zero. Toward the bottom of the example he shows this work:
$$\frac{d^2y}{dx^2} = \frac{-\frac 1 2 (35t^4 + 15t^2) }{-2t} = \frac 1 4 (3t^3 + 15t)$$
The expression in the middle is undefined if t = 0, but as t approaches zero, the value of this expression approaches zero as well. The 2nd and 3rd expressions are exactly the same except at a single point -- the origin. He could have elaborated on this fact a bit more.

The graph he shows doesn't make it clear that there is a "hole" at (0, 0). At this point both dy/dx and d2y/dx2 are undefined.