Q.Find $\frac{dy}{dx}$, if $x = at^2$, $y = 2at$.

Concept: Parametric Differentiation — when both $x$ and $y$ are given in terms of a parameter $t$, we use the chain rule:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$, provided $dx/dt \neq 0$.

Step 1: Differentiate $x = at^2$ with respect to $t$:

$\frac{dx}{dt} = 2at$.

Step 2: Differentiate $y = 2at$ with respect to $t$:

$\frac{dy}{dt} = 2a$.

Step 3: Apply the formula:

$\frac{dy}{dx} = \frac{2a}{2at} = \frac{1}{t}$.

When a curve is given in parametric form — with both $x$ and $y$ expressed in terms of a third variable (here $t$) — you cannot directly differentiate $y$ with respect to $x$. Instead, you use the chain rule in reverse:

Why does this work? Because $dy/dx$ is the rate of change of $y$ with respect to $x$, and both change with $t$. Dividing their individual rates with respect to $t$ cancels the $t$ dependence, leaving the slope of the curve in the $xy$-plane.

Let’s apply it step by step.

1. Differentiate $x$ with respect to $t$

   $x = at^2$

   $\frac{dx}{dt} = a \cdot 2t = 2at$

2. Differentiate $y$ with respect to $t$

   $y = 2at$

   $\frac{dy}{dt} = 2a$

3. Form the ratio

4. Simplify Cancel $2a$ (provided $a \neq 0$, which is true for a non-degenerate parabola):

Method: Differentiating Parametric Equations for dy/dx

This method applies whenever $x$ and $y$ are each given as separate functions of a parameter (commonly $t$ or $\theta$), rather than $y$ being written directly in terms of $x$.

Steps

Step 1: Write down the parametric-derivative formula

This comes from the chain rule applied in reverse: since $\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}$, solving for $\frac{dy}{dx}$ gives the ratio of the two parameter-derivatives.

Step 2: Differentiate $x(t)$ and $y(t)$ independently

Apply the ordinary power/product/chain rules to each expression with respect to $t$, keeping the results as two separate expressions — never try to eliminate $t$ before differentiating.

Step 3: Divide and cancel common factors

Form $\dfrac{dy/dt}{dx/dt}$ and cancel any common numerical or variable factor (a constant like $a$, or a shared power of $t$) that appears in both derivatives — this is what produces the clean final answer.

Common Mistakes

Mistake 1: Trying to differentiate $y$ directly with respect to $x$

Why it's wrong: $x=at^2$ means $x$ itself depends on $t$ — treating $x$ as a plain independent variable and differentiating $y=2at$ term-by-term as though $t$ were simply proportional to $x$ skips the actual relationship and gives a wrong slope. Correct approach: always go through the parameter — compute $dy/dt$ and $dx/dt$ separately, then divide.

Mistake 2: Cancelling incorrectly when simplifying the ratio

Why it's wrong: $\dfrac{2a}{2at}$ is sometimes mis-simplified by cancelling $t$ instead of the constant $2a$, or by cancelling only part of a factor. Correct approach: cancel the full common factor ($2a$) first, leaving $\dfrac{1}{t}$, and check the result is a pure function of $t$ with no leftover constant.