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Parametric Differentiation

Parametric Differentiation

Differentiation of a parametric curve is simply an application of the chain rule. Given a curve $C$, defined parametrically by $x = x(t),\; y = y(t),\; t \in S$, the derivative $\frac{dy}{dx}$ can be expressed as $$ \frac{dy}{dx} = \frac{dy}{dt}\times\frac{dt}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} $$

Parametric Differentiation 1

Consider the curve defined by $x(t) = 2\sin(t)$, $y(t) = 4 - \cos(t)$, $t \in [0,2\pi]$. Find an expression for $\frac{dy}{dx}$ in terms of $t$
solution - press button to display
$$ \frac{dy}{dt} = \sin(t),\; \frac{dx}{dt} = 2\cos(t) $$ $$ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\sin(t)}{2\cos(t)} = \frac{1}{2}\tan(t) $$

Parametric Differentiation 2

The curve $C$ is an oblique ellipse, centred on the origin, and can be defined parametrically by $$\begin{align*}x(t) &= \sqrt{2}\sin t - \frac{1}{\sqrt{2}}\cos t \\ y(t) &= \sqrt{2}\sin t +\frac{1}{\sqrt{2}}\cos t\end{align*}$$oblique ellipse

$$0\lt t\lt 2\pi$$

By using parametric differentiation, determine the coordinates of the highest point on the curve.

 

solution - press button to display
The derivative, $\frac{dy}{dx}$ is given by $$ \frac{dy}{dx} = \frac{dy}{dt}\times \frac{dt}{dx} $$

In this instance, $\frac{dy}{dt} = \sqrt{2}\cos t -\frac{1}{\sqrt{2}}\sin t$.

Similarly $\frac{dx}{dt} = \sqrt{2}\cos t +\frac{1}{\sqrt{2}}\sin t $

As a consequence then, we get $$ \frac{dy}{dx} = \frac{\sqrt{2}\cos t -\frac{1}{\sqrt{2}}\sin t}{\sqrt{2}\cos t +\frac{1}{\sqrt{2}}\sin t} $$

At the highest point of the curve, $\frac{dy}{dx} = \frac{\sqrt{2}\cos t -\frac{1}{\sqrt{2}}\sin t}{\sqrt{2}\cos t +\frac{1}{\sqrt{2}}\sin t}=0$. It follows that $\sqrt{2}\cos t -\frac{1}{\sqrt{2}}\sin t =0$. Rearranging this equation gives us $$ \tan t = 2 $$

Parametric Differentiation 3

algebraic curveThe curve $C$ is defined parametrically by $$\begin{align*}x(t) & =t^2-1 \\ y(t) & = t^4 +2t\end{align*}$$

$$-\frac{3}{2}\lt t\lt 1 $$

By using parametric differentiation, determine the coordinates of the highest point on the curve.

 

solution - press button to display

The curve is generated by $x(t) = t^2 - 1$, $y(t) = t^4 + 2t$. Differentiating these with respect to $t$ yields $$ \frac{dx}{dt} = 2t,\; \frac{dy}{dt} = 4t^3 + 2 $$ The derivative of $y$ with regards to $x$ is then given by $$ \frac{dy}{dx} = \frac{4t^3 + 2}{2t} $$ We seek the point at which $\frac{dy}{dx} = 1$, so $$ \begin{align*} \frac{4t^3 + 2}{2t} &= 1 \\ 2t^3 + 1 &= t \\ 2t^3 - t+ 1 &=0 \end{align*} $$ Note that $t = -1$ is a solution of this equation. $$ 2t^3 - t + 1 = (t+1)(2t^2 -2t +1) $$ Now observe that $2t^2 - 2t + 1 =0$ has no real solutions since $b^2 - 4ac = (-2)^2 - 4(2)(1) \lt 0$. Therefore, there is only one point where $\frac{dy}{dx} = 1$. That point is $(0,-1)$