2 Higher derivatives

Having found the first derivative $\frac{dy}{dx}$ using parametric differentiation we now ask how we might determine the second derivative $\frac{d^{2}y}{d{x}^2}$ .

By definition:

$\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)$

But

$\frac{dy}{dx}=\frac{ẏ}{ẋ}\text{and so}\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{ẏ}{ẋ}\right)$

Now $\frac{ẏ}{ẋ}$ is a function of $t$ so we can change the derivative with respect to $x$ into a derivative with respect to $t$ since

$\frac{d}{dx}\left(\frac{dy}{dx}\right)=\left\{\frac{d}{dt}\left(\frac{dy}{dx}\right)\right\}\frac{dt}{dx}$

from the function of a function rule (Key Point 11 in Section 11.5).

But, differentiating the quotient $ẏ∕ẋ$ , we have

$\frac{d}{dt}\left(\frac{ẏ}{ẋ}\right)=\frac{ẋÿ-ẏẍ}{{ẋ}^2}\text{and}\frac{dt}{dx}=\frac{1}{\left(\frac{dx}{dt}\right)}=\frac{1}{ẋ}$

so finally:

$\frac{d^{2}y}{d{x}^2}=\frac{ẋÿ-ẏẍ}{{ẋ}^3}$

Example 14

If the equations of a curve are    $x=2t,y=t^2-3,$  determine $\frac{dy}{dx}$ and $\frac{d^{2}y}{d{x}^2}.$

Solution

Here $ẋ=2,ẏ=2t$ $\therefore \frac{dy}{dx}=\frac{ẏ}{ẋ}=\frac{2t}{2}=t.$

Also $ẍ=0,ÿ=2$ $\therefore \frac{d^{2}y}{d{x}^2}=\frac{2\left(2\right)-2t\left(0\right)}{{\left(2\right)}^3}=\frac{1}{2}$ .

These results can easily be checked since $t=\frac{x}{2}$ and $y=t^2-3$ which imply $y=\frac{x^2}{4}-3$ . Therefore the derivatives can be obtained directly: $\frac{dy}{dx}=\frac{2x}{4}=\frac{x}{2}\text{and}\frac{d^{2}y}{d{x}^2}=\frac{1}{2}.$

Exercises

1. For the following sets of parametric equations find $\frac{dy}{dx}$ and $\frac{d^{2}y}{d{x}^2}$
   1. $x=3t^{2}y=4t^3$
   2. $x=4-t^{2}y=t^2+4t$
   3. $x=t^2{e}^{t}y=t$
2. Find the equation of the tangent line to the curve

   $x=1+3sinty=2-5cost\text{at}t=\frac{\pi }{6}$

Answer