Parametric differentiation

Curves defined parametrically, finding dy/dx as dy/dt divided by dx/dt, gradients and tangents at a point given by a parameter, and checking with the Cartesian equation.

Parametric curves

A curve can be described by giving both $x$ and $y$ in terms of a third variable $t$, the parameter: $x = f(t)$ and $y = g(t)$. Each value of $t$ gives one point. The path of a projectile is an example, with $t$ as time.

$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}, \quad \text{provided } \frac{dx}{dt} \ne 0$$

This is the chain rule, $\dfrac{dy}{dt} = \dfrac{dy}{dx} \times \dfrac{dx}{dt}$, rearranged.

Example 1●●

A parabola has parametric equations $x = 2t$ and $y = t^2$. Find the gradient at the point where $t = 3$, and the equation of the tangent there.

$\dfrac{dy}{dx} = \dfrac{2t}{2} = t$, so the gradient is $3$.

The point is $(6, 9)$, so the tangent is $y - 9 = 3(x - 6)$, that is, $y = 3x - 9$.

Check: eliminating $t$ gives $y = \tfrac14 x^2$, so $\dfrac{dy}{dx} = \tfrac12 x = 3$ at $x = 6$.

Example 2: a circle●●

Find $\dfrac{dy}{dx}$ for $x = 3\cos t$, $y = 3\sin t$, and its value at $t = \frac{\pi}{4}$.

$\dfrac{dy}{dx} = \dfrac{3\cos t}{-3\sin t} = -\cot t$.

At $t = \frac{\pi}{4}$, the gradient is $-1$.

Flashcards

$dfrac{dy}{dx}$ for a parametric curve

$\dfrac{dy/dt}{dx/dt}$

Vertical tangent

$\dfrac{dx}{dt} = 0$ and $\dfrac{dy}{dt} \ne 0$.

Parameter

A third variable, such as $t$, that gives both $x$ and $y$.

Practice

For $x = 2t$, $y = t^2$, find $\dfrac{dy}{dx}$ at $t = 3$.
The tangent at $t = 3$ is $y = 3x + c$. Find $c$.
For $x = 3\cos t$, $y = 3\sin t$, find $\dfrac{dy}{dx}$ at $t = \frac{\pi}{4}$.
For $x = t^2$, $y = t^3$, find $\dfrac{dy}{dx}$ at $t = 1$.
For $x = t + 1$, $y = t^2$, find $\dfrac{dy}{dx}$ at $t = 1$.
For $x = 3\cos t$, $y = 3\sin t$, find $\dfrac{dy}{dx}$ at $t = \frac{\pi}{2}$.
Eliminating $t$ from $x = 2t$, $y = t^2$ gives:
A parametric curve has a vertical tangent where:
The formula $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ comes from: