Differentiation – Parametric differentiation

A curve can be given by two equations:

\[ x=f(t), \qquad y=g(t) \]

where \(t\) is the parameter.

To find \(\dfrac{dy}{dx}\), first differentiate both \(x\) and \(y\) with respect to \(t\).

This works provided that \(\dfrac{dx}{dt}\ne 0\).

1. Find \(\dfrac{dx}{dt}\).
2. Find \(\dfrac{dy}{dt}\).
3. Use \[ \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}. \]
4. Simplify the result.

Given

\[ x=t^2+1, \qquad y=t^3-2 \]

Find \(\dfrac{dy}{dx}\).

\[ \frac{dx}{dt}=2t \]

\[ \frac{dy}{dt}=3t^2 \]

So

Given

\[ x=\cos t, \qquad y=\sin t \]

Find \(\dfrac{dy}{dx}\).

\[ \frac{dx}{dt}=-\sin t \]

\[ \frac{dy}{dt}=\cos t \]

Therefore

After finding \(\dfrac{dy}{dx}\), substitute the value of the parameter \(t\).

Example: For

\[ x=t^2+1,\qquad y=t^3-2 \]

find the gradient when \(t=2\).

From earlier, \[ \frac{dy}{dx}=\frac{3t}{2} \]

To find the second derivative, use:

Differentiate \(\dfrac{dy}{dx}\) with respect to \(t\), then divide by \(\dfrac{dx}{dt}\).

Given

\[ x=t^2,\qquad y=t^4 \]

Find \(\dfrac{d^2y}{dx^2}\).

First derivative:

\[ \frac{dx}{dt}=2t,\qquad \frac{dy}{dt}=4t^3 \]

\[ \frac{dy}{dx}=\frac{4t^3}{2t}=2t^2 \]

Differentiate with respect to \(t\):

\[ \frac{d}{dt}\left(\frac{dy}{dx}\right)=\frac{d}{dt}(2t^2)=4t \]

Now divide by \(\dfrac{dx}{dt}\):

Once the gradient is known, the equation of the tangent can be found using the point on the curve.

First find the coordinates by substituting the value of \(t\), then use

\[ y-y_1=m(x-x_1) \]

• Using \(\dfrac{dx}{dt}\div\dfrac{dy}{dt}\) instead of \(\dfrac{dy}{dt}\div\dfrac{dx}{dt}\).
• Substituting the parameter too early.
• Forgetting to divide by \(\dfrac{dx}{dt}\) again when finding \(\dfrac{d^2y}{dx^2}\).
• Not finding the point on the curve before writing the tangent equation.