Parametric differentiation
Objective. Calculate dy/dx from derivatives with respect to t.
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- 2. Worked example
- 3. Practice
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Learn: the key idea
Parametric differentiation finds dy/dx for curves given as x(t) and y(t) using the chain rule result dy/dx=(dy/dt)÷(dx/dt), without needing to eliminate the parameter first. Example: for x=t²+1, y=t³, dx/dt=2t and dy/dt=3t², so dy/dx=3t²/2t=3t/2. This gradient can then be evaluated at a specific t value to find the tangent's slope at the corresponding point on the curve.
Worked example
Find dy/dx for the parametric curve x=2t³, y=3t² using parametric differentiation.
- 1Differentiate x with respect to t: dx/dt=6t²
- 2Differentiate y with respect to t: dy/dt=6t
- 3Apply dy/dx=(dy/dt)÷(dx/dt)=6t/6t²
- 4Simplify the result: dy/dx=1/t
Answer: dy/dx=1/t
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Start practisingCommon mistake. Students compute dy/dx as (dy/dt)×(dx/dt) instead of dividing, confusing it with a product rule; always divide dy/dt by dx/dt for parametric gradients.
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