Short on time? This page is the recap. For the chain-rule derivation, the graphs and the worked examples, use the full study guide. BC-only material. Prerequisites (basic derivatives 2.5–2.7, chain rule 3.1, tangent lines) are on the Calculus BC roadmap.
Recap
- A parametric curve gives x = f(t) and y = g(t). Each t gives one point; as t increases the point moves, so the curve has a direction.
- The curve can turn back or cross itself, so one point (x, y) may come from two values of t.
- By the chain rule, dy/dt = (dy/dx)(dx/dt). Divide by dx/dt to get the slope.
- dy/dx is usually a function of t. Substitute a value of t, not x.
Key relationships
| Idea | Result | Note |
|---|---|---|
| Slope of tangent | dy/dx = (dy/dt) ÷ (dx/dt) | Only where dx/dt ≠ 0 |
| Tangent line at t = a | y − y(a) = m(x − x(a)), with m = dy/dx at t = a | Find the point and the slope from the same t |
| Horizontal tangent | dy/dt = 0 and dx/dt ≠ 0 | Check both conditions |
| Vertical tangent | dx/dt = 0 and dy/dt ≠ 0 | No sideways motion at that instant |
| Both zero | 0 ÷ 0: no conclusion | Simplify first or look at the graph |
Meaning in context
- dy/dx is the slope of the path, in units of y per unit of x.
- It is not a speed and does not tell you the direction of travel. A positive dy/dx with dx/dt < 0 and dy/dt < 0 means moving down and left along a line that rises to the right.
Mistakes to avoid
- Upside-down formula. dy/dx is dy/dt on top, dx/dt underneath.
- Putting an x-value in for t. Solve x(t) = x₀ and y(t) = y₀ for t first.
- Missing a second t. Check every t that gives the point; each can have its own slope.
- Skipping the check that dx/dt ≠ 0 at a horizontal-tangent candidate.
- Swapping horizontal and vertical: dx/dt = 0 gives a vertical tangent.
- Chain-rule slips inside dx/dt or dy/dt, for example d/dt [sin 3t] = 3 cos 3t.
Quick self-check
- For x = 2t − 1, y = t³, find dy/dx at t = 2. (3t²/2 = 6)
- For x = cos t, y = sin t, find the slope at t = π/6. (cos t ÷ (−sin t) = −√3)
- For x = t² − 6t, y = 2t + 1, where is the tangent vertical? (dx/dt = 2t − 6 = 0 at t = 3, dy/dt = 2 ≠ 0; point (−9, 7))
Next: practice questions.