medium difficulty7 min

Parametric and Second-Order Derivatives

For parametric equations x = f(t), y = g(t), the derivative dy/dx is found as (dy/dt)/(dx/dt), provided dx/dt is not zero. The second-order derivative measures the rate of change of dy/dx with respect to x.

In one line: First derivative is y-rate over x-rate; second derivative needs one more division by x-rate.

Practice This Concept

Understand it

Parametric differentiation is used when x and y are both expressed in terms of a third variable t. For the second derivative, do not simply differentiate dy/dx with respect to t; divide that derivative by dx/dt to convert it into differentiation with respect to x.

Example

If x = t² and y = t³, then dy/dx = (3t²)/(2t) = 3t/2 for t ≠ 0.

Formula and rule

For x = f(t), y = g(t): dy/dx = (dy/dt)/(dx/dt), where dx/dt ≠ 0.

Second derivative: d2y/dx2 = [d/dt(dy/dx)]/(dx/dt).

Interpretation: dy/dx is slope, while d2y/dx2 indicates how the slope changes with x.

See it worked out

Follow the example line by line, then use the same method on new questions.

Worked example

Let x = a cos t and y = a sin t.

1Then dx/dt = -a sin t and dy/dt = a cos t.
2Hence dy/dx = (a cos t)/(-a sin t) = -cot t.
3Now d/dt(dy/dx) = cosec² t.
4Therefore d2y/dx2 = cosec² t/(-a sin t) = -1/(a sin³ t).

Method you can reuse

1Differentiate x and y separately with respect to t.
2Divide dy/dt by dx/dt to get dy/dx.
3For second derivative, differentiate dy/dx with respect to t.
4Divide the result by dx/dt.
5State any condition such as dx/dt ≠ 0.

Visual tip

A curve sketch may help interpret slope or concavity, but standard exam questions usually require derivative calculation from parametric equations.

Check yourself

Answer the question in your head before opening the answer.

Try it yourself

If x = t and y = t², find dy/dx and d2y/dx2.

Think first, then show answer

dy/dx = 2t/1 = 2t. Then d/dt(dy/dx) = 2 and dx/dt = 1, so d2y/dx2 = 2.

Common wrong answer

Cancelling the parameter incorrectly or forgetting that dx/dt must be non-zero before dividing.

Don't mix this up

Students calculate d/dt(dy/dx) and call it d2y/dx2 without dividing by dx/dt.

Exam tip

For second derivative in parametric form, use d2y/dx2 = [d/dt(dy/dx)]/(dx/dt).

Write it in the exam

Each point is worth about one mark. Write point 1 for a 1-mark question and keep adding points as the marks go up.

How CBSE asks it

Asked as direct parametric differentiation, tangent slope calculation, or finding the second derivative from equations involving trigonometric parameters.

1
Enough for 1 mark
For parametric equations x = f(t), y = g(t), the derivative dy/dx is found as (dy/dt)/(dx/dt), provided dx/dt is not zero. The second-order derivative measures the rate of change of dy/dx with respect to x.
2
Add for 2 marks
Example: If x = t² and y = t³, then dy/dx = (3t²)/(2t) = 3t/2 for t ≠ 0.
3
Add for 3 marks
Parametric differentiation is used when x and y are both expressed in terms of a third variable t.

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