Differentiation of Parametric Functions

Neetesh Kumar | February 08, 2025 $\space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space$ Share this Page on:

• 1. Definition of Derivatives of Parametric Functions
• 2. Formula Used for Parametric Derivatives
• 3. Solved Examples
• 4. Conclusion

Parametric Equations:
In parametric form, the equations of a curve are given by two functions:

• $x = f(t)$
• $y = g(t)$

where $t$ is a parameter. The derivative of $y$ with respect to $x$, denoted $\dfrac{dy}{dx}$, is found using the following formula:

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

This formula allows us to differentiate parametric equations by first differentiating $y$ and $x$ with respect to $t$, and then dividing $\dfrac{dy}{dt}$ by $\dfrac{dx}{dt}$.

Example 1: Finding the Slope of a Circle

Find the derivative of the parametric equations for a circle: $x = 3 \cos t$, $y = 3 \sin t$

Solution:

Here, $x = f(t) = 3 \cos t$ and $y = g(t) = 3 \sin t$.

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

$\dfrac{dx}{dt} = -3 \sin t$

$\dfrac{dy}{dt} = 3 \cos t$

Now, apply the formula:

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

Thus, the slope of the tangent at any point on the circle is: $\boxed{-\cot t}$.

Example 2: Differentiating a Cycloid

Find the derivative of the parametric equations of a cycloid: $x = r(t - \sin t)$, $y = r(1 - \cos t)$

Solution:

Here, $x = f(t) = r(t - \sin t)$ and $y = g(t) = r(1 - \cos t)$.

Differentiating both functions with respect to $t$:

$\dfrac{dx}{dt} = r(1 - \cos t)$

$\dfrac{dy}{dt} = r \sin t$

Now, apply the chain rule:

$\dfrac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \dfrac{r \sin t}{r(1 - \cos t)}$

Simplifying:

$\dfrac{dy}{dx} = \dfrac{\sin t}{1 - \cos t}$

Thus, the derivative of the cycloid is: $\boxed{\frac{\sin t}{1 - \cos t}}$

Example 3: Velocity of a Parametric Function

The position of a particle is given by the parametric equations: $x = 4t^3 + 2t$, $y = 3t^2 - t$

Find the velocity of the particle at time $t = 1$.

Solution:

Here, $x = f(t) = 4t^3 + 2t$ and $y = g(t) = 3t^2 - t$.

To find the velocity, we first find the derivatives:

$\dfrac{dx}{dt} = 12t^2 + 2$

$\dfrac{dy}{dt} = 6t - 1$

The velocity vector $\mathbf{v}(t)$ is given by:

$\mathbf{v}(t) = \left( \dfrac{dx}{dt}, \dfrac{dy}{dt} \right)$

At $t = 1$, substitute into the derivatives:

$\dfrac{dx}{dt} = 12(1)^2 + 2 = 14$

$\dfrac{dy}{dt} = 6(1) - 1 = 5$

Thus, the velocity vector at $t = 1$ is:

$\mathbf{v}(1) = (14, 5)$

The magnitude of the velocity is:

$|\mathbf{v}(1)| = \sqrt{14^2 + 5^2} = \sqrt{196 + 25} = \sqrt{221}$

Thus, the magnitude of the velocity at $t = 1$ is: $\boxed{\sqrt{221}}$

Example 4: Parametric Equation of a Spirograph

Find the derivative of the parametric equations for a spirograph:
$x = (3 + 2 \cos t) \cos 2t$, $y = (3 + 2 \cos t) \sin 2t$

Solution:

Here, $x = f(t) = (3 + 2 \cos t) \cos 2t$ and $y = g(t) = (3 + 2 \cos t) \sin 2t$.

We differentiate both functions:

$\dfrac{dx}{dt} = -2 \sin t \cos 2t - (3 + 2 \cos t)(2 \sin 2t)$

$\dfrac{dy}{dt} = -2 \sin t \sin 2t + (3 + 2 \cos t)(2 \cos 2t)$

Now, applying the chain rule:

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

This is a complex spirograph function, and solving it involves simplifying these expressions further for specific values of $t$.

Thus, the derivative can be found as:

$\boxed{\frac{dy}{dx} = \frac{-2 \sin t \sin 2t + (3 + 2 \cos t)(2 \cos 2t)}{-2 \sin t \cos 2t - (3 + 2 \cos t)(2 \sin 2t)}}$