Find parametric gradients from two derivatives with respect to the parameter
Calculate a curve point and its coordinate rates, then use their ratio for a tangent or normal when the horizontal rate is non-zero.
Exam scope and assessment notes
Pure Mathematics 2 (Paper 2, AS only) and Pure Mathematics 3 (Paper 3, A Level) · Mathematics 9709, 2026–2027. Paper 1 knowledge is assumed. The AS route using Papers 1 + 2 cannot be carried forward to A Level. Shared sections 2.4 and 3.4 cover exponential, natural logarithmic and trigonometric derivatives, composites, products and quotients, and first derivatives of parametric and implicit curves, including tangents and normals. Paper 3 additionally requires the derivative of inverse tangent; use its separate lesson for that method. Derivatives of inverse sine and inverse cosine are not required in section 3.4. Trigonometric differentiation uses radians. Keep original domains and check zero derivative denominators separately. Written practice is self-checked.
Before you begin
- Differentiate natural and radian trigonometric composites.
- Build point-gradient lines and distinguish a horizontal coordinate from a parameter.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which ratio gives dy/dx when dx/dt is non-zero?
What you will learn
- Find dy/dx = (dy/dt)/(dx/dt) for a parametrically defined curve where dx/dt ≠ 0.
- Use the specified parameter value to calculate the point, tangent and normal and check any original parameter domain.
Differentiate both coordinates before taking their ratio
A parametric curve supplies x = f(t) and y = g(t). The parameter t identifies a point (f(t), g(t)); it need not equal x or represent time. Where dx/dt ≠ 0, the chain rule gives dy/dx = (dy/dt)/(dx/dt). The numerator is the vertical coordinate rate and the denominator the horizontal one. Their ratio is the curve's Cartesian gradient, not either coordinate rate alone or the inverted ratio dx/dy.
For x = cos t and y = sin t with radian parameter t, dx/dt = −sin t and dy/dt = cos t. Hence dy/dx = −cot t where sin t ≠ 0. At t = π/4 the point is (√2/2, √2/2), the rates are −√2/2 and √2/2, and the gradient is −1. Increasing t moves the point left and up locally, consistent with that negative gradient. The parametrisation traces the unit circle and may revisit the same point after a full turn.
Use the given parameter to find both the point and line
At t = π/4 on the unit circle, the tangent is y − √2/2 = −(x − √2/2), or x + y = √2. The normal has gradient 1 through the same point, so it is y = x. Check the parameter domain and dx/dt before using a finite point-gradient rule. If dx/dt = 0, inspect dy/dt separately: a non-zero vertical rate gives a vertical tangent, while two zero rates leave the ratio indeterminate and need a separate argument.
For x = t − e^(2t), y = t + e^(2t), differentiate to get dx/dt = 1 − 2e^(2t) and dy/dt = 1 + 2e^(2t). At t = 0 the point is (−1, 1), the rates are −1 and 3, and the gradient is −3. Thus the tangent is y = −3x − 2 and the normal is x − 3y + 4 = 0. Evaluate the two original coordinate rules at t = 0 for the point, rather than mistake their derivatives for coordinates.
Pause and explain
For x = cos t, y = sin t at t = π/4, why is the tangent gradient negative?
Worked example
For x = cos t, y = sin t, find the tangent and normal at t = π/4, using radians.
Show the worked solution
- Evaluate both coordinate rules to find P(√2/2, √2/2).
- Differentiate with respect to t: dx/dt = −sin t, dy/dt = cos t; at π/4 their ratio is −1.
- Use that gradient for tangent x + y = √2; the normal's negative reciprocal gradient is 1, giving y = x.
Answer dy/dx = −1 at P(√2/2, √2/2); tangent x + y = √2; normal y = x.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
For x = t² + 1 and y = t³ + t, find tangent and normal equations at t = 1.
Give me a hint
Find the point from the coordinates and the gradient from their parameter derivatives.
Compare my reasoning
- At t = 1 the original point is (2, 2).
- The coordinate rates are dx/dt = 2t and dy/dt = 3t² + 1; their ratio at 1 is 4/2 = 2.
- The tangent is y − 2 = 2(x − 2), or y = 2x − 2; the normal is y − 2 = −(x − 2)/2, or x + 2y = 6.
Point (2, 2); gradient 2; tangent y = 2x − 2; normal x + 2y = 6.
Look for these in your work
- Evaluate coordinates and coordinate rates separately.
- Build both lines through the same point with perpendicular finite gradients.
For x = t + 1 and y = ln t with t > 0, find the gradient, tangent and normal at t = 1. Check the gradient by eliminating t.
Give me a hint
The horizontal coordinate rate is 1; retain the original positive log parameter.
Compare my reasoning
- At t = 1 the point is (2, 0), and (dy/dt)/(dx/dt) = (1/t)/1 gives gradient 1.
- The tangent is y = x − 2 and the normal is y = −x + 2.
- Eliminating t gives y = ln(x − 1) for x > 1, whose derivative 1/(x − 1) also gives 1 at x = 2.
Gradient 1 at (2, 0); tangent y = x − 2; normal y = −x + 2; Cartesian check y = ln(x − 1), x > 1.
Look for these in your work
- Respect the log parameter domain and convert it when eliminating t.
- Check the parameter-gradient ratio against a valid Cartesian derivative.
For x = t − e^(2t) and y = t + e^(2t), find tangent and normal equations at t = 0 and describe the signs of the coordinate rates there.
Give me a hint
Differentiate each exponential composite separately before forming the vertical-over-horizontal ratio.
Compare my reasoning
- At zero the original point is (−1, 1).
- The rates are dx/dt = 1 − 2e^(2t) = −1 and dy/dt = 1 + 2e^(2t) = 3, so dy/dx = −3.
- The tangent is y = −3x − 2; the normal is x − 3y + 4 = 0. The negative horizontal and positive vertical rates describe local left-and-up motion as the parameter increases.
Point (−1, 1); rates −1 and 3; gradient −3; tangent y = −3x − 2; normal x − 3y + 4 = 0.
Look for these in your work
- Use the exponential inner factor in both coordinate derivatives.
- Check both line equations and the rate signs at the original point.
Common mistakes
- Differentiate both coordinates with respect to the same parameter before taking the ratio.
- Do not invert the gradient ratio or use dy/dt as dy/dx.
- Check dx/dt before dividing and evaluate the original coordinate rules for the point.
What can you explain now?
At a parameter value, dy/dt = 6 and dx/dt = 2. What is dy/dx?
Compare with the explanation
3.
The curve gradient is vertical rate divided by horizontal rate, 6/2.
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Make it stick
Explain how two coordinate rates produce a Cartesian gradient. Find a parameter point, tangent and normal, and state what must be checked before dividing by the horizontal rate.
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