Parametric Derivatives: Why dy/dx Equals (dy/dt)/(dx/dt)

If x = x(t) and y = y(t) are differentiable and dx/dt ≠ 0, then dy/dx = (dy/dt)/(dx/dt). Evaluate both derivatives at the same parameter value.

The parameter controls two coordinates

A parametric curve specifies the x- and y-coordinates separately in terms of a parameter t. The quantity dy/dt tells how y changes with that parameter, while dx/dt tells how x changes with it. Their ratio gives change in y per change in x.

This is a chain-rule relationship, not an instruction to divide the coordinate values y/x. A point’s ratio y/x is the slope from the origin to that point, which need not equal the curve’s tangent slope there.

Worked example

Let x = t² + 1 and y = t³ − t. Find the slope when t = 2. The curve point is (5, 6). Differentiate separately: dx/dt = 2t and dy/dt = 3t² − 1. At t = 2, these are 4 and 11.

Therefore dy/dx = 11/4. The tangent equation is y − 6 = (11/4)(x − 5). The coordinate ratio would have been 6/5, demonstrating why y/x is not the desired derivative.

Check by eliminating the parameter locally

Near t = 2, t is positive, so x = t² + 1 gives t = √(x − 1). Substituting into y gives y = (x − 1)^(3/2) − (x − 1)^(1/2). Differentiating and evaluating at x = 5 yields 3 − 1/4 = 11/4, confirming the parametric calculation.

Elimination is a check, not a compulsory first step. It can introduce branch choices or complicated expressions. The derivative ratio often avoids that work while retaining the parameter’s direction information.

When the denominator is zero

If dx/dt = 0 but dy/dt ≠ 0, a regular curve may have a vertical tangent. If both derivatives vanish, the ratio is 0/0 and further analysis is required; a cusp, regular limiting tangent or another behaviour may occur. Do not label every such point automatically.

For second derivatives, first differentiate dy/dx with respect to t and then divide by dx/dt again, where allowed. Simply taking y″(t)/x″(t) is not the general second-derivative rule.

Common mistake

Use the same t in both coordinate derivatives, and check whether the question asks for the parameter value or the point. A curve can pass through the same coordinate more than once with different parameter values, potentially giving different local directions.

dy/dt ko dx/dt se divide karo; y ko x se nahi.

Three practice questions with worked answers

1. For x = 3t + 1 and y = 6t − 2, find dy/dx.

   The derivatives are dx/dt = 3 and dy/dt = 6. Their ratio is 6/3 = 2 for every real t.

2. For x = t² and y = t³, find the slope at t = 1.

   At t = 1, dx/dt = 2 and dy/dt = 3, so the slope is 3/2. The point is (1, 1).

3. For x = 3 cos t and y = 3 sin t, find the slope at t = π/6.

   The ratio is (3 cos t)/(−3 sin t) = −cot t where sin t ≠ 0. At π/6 it equals −√3; angles are in radians.

Related study guides

• Differentiation Formulas Class 12: Complete List with Examples
• Slope of a Line Through Two Points

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References and authorship

Original examples and practice questions, prepared with AI assistance and editorial arithmetic review. Related definitions and methods: OpenStax: Calculus of Parametric Curves. No textbook or exam-board affiliation is claimed.