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Parametric and Second-Order Derivatives

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Student-friendly explanation

How to write this in exams

1
Start with the exact idea
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
Then show how to use it
Differentiate x and y separately with respect to t. Divide dy/dt by dx/dt to get dy/dx. For second derivative, differentiate dy/dx with respect to t. Divide the result by dx/dt. State any condition such as dx/dt != 0.
3
Add one concrete example
If x = t^2 and y = t^3, then dy/dx = (3t^2)/(2t) = 3t/2 for t != 0.
4
Avoid this incomplete answer
Cancelling the parameter incorrectly or forgetting that dx/dt must be non-zero before dividing.

Definition

Example

If x = t^2 and y = t^3, then dy/dx = (3t^2)/(2t) = 3t/2 for t != 0.

Rule to remember

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.

Memory hook

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

Examples and method

Worked example

Let x = a cos t and y = a sin t. Then dx/dt = -a sin t and dy/dt = a cos t. Hence dy/dx = (a cos t)/(-a sin t) = -cot t. Now d/dt(dy/dx) = cosec^2 t. Therefore d2y/dx2 = cosec^2 t/(-a sin t) = -1/(a sin^3 t).

Method to apply

Differentiate x and y separately with respect to t. Divide dy/dt by dx/dt to get dy/dx. For second derivative, differentiate dy/dx with respect to t. Divide the result by dx/dt. State any condition such as dx/dt != 0.

Diagram support

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

How CBSE asks it

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

Avoid common mistakes

Common confusion

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

Common wrong answer

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

Exam tip

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

Quick check

If x = t and y = t^2, find dy/dx and d2y/dx2.

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

Answer writing and exam use

1-mark answer

2-mark answer

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. 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. If x = t^2 and y = t^3, then dy/dx = (3t^2)/(2t) = 3t/2 for t != 0.

3-mark answer

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. 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. Let x = a cos t and y = a sin t. Then dx/dt = -a sin t and dy/dt = a cos t. Hence dy/dx = (a cos t)/(-a sin t) = -cot t. Now d/dt(dy/dx) = cosec^2 t. Therefore d2y/dx2 = cosec^2 t/(-a sin t) = -1/(a sin^3 t). Asked as direct parametric differentiation, tangent slope calculation, or finding the second derivative from equations involving trigonometric parameters. Cancelling the parameter incorrectly or forgetting that dx/dt must be non-zero before dividing.

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