Derivatives of Parametric Equations

To find the derivative of parametric equations x = x(t) and y = y(t), divide the two rates: dy/dx = (dy/dt)/(dx/dt), as long as dx/dt is not zero. For the second derivative, differentiate dy/dx with respect to t, then divide by dx/dt again.

Before this lesson: Parametric Equations, Chain Rule and Equation of a Tangent Line

The idea: slope is rise rate over run rate

In a parametric curve, a point moves as t (often time) changes. At each moment it has a vertical speed, dy/dt, and a horizontal speed, dx/dt.

The slope of the path is how much it rises for each unit it runs. If the point rises 6 units per second while it moves right 2 units per second, the path rises 3 units for every 1 unit across. So the slope is 6/2 = 3.

The chain rule says the same thing with symbols: dy/dt = (dy/dx)(dx/dt). Divide both sides by dx/dt to get the formula below.

Parametric derivative formulas

\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} \qquad \dfrac{d^2y}{dx^2} = \dfrac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{dx/dt}

• Horizontal tangent: dy/dt = 0 and dx/dt ≠ 0.
• Vertical tangent: dx/dt = 0 and dy/dt ≠ 0.
• If both are 0 at the same t, the formula doesn’t decide. The curve may have a cusp or a sharp turn there.
• The second derivative tells concavity: positive means concave up, negative means concave down.

The second derivative is not d²y/dt² divided by d²x/dt². It is the derivative of the slope with respect to t, divided by dx/dt.

Derivatives of parametric equations examples

Example 1: Slope and tangent line

For x = t² + 1 and y = t³ − 3t, find dy/dx and the tangent line at t = 2.

\dfrac{dx}{dt} = 2t, \quad \dfrac{dy}{dt} = 3t^2 - 3

\dfrac{dy}{dx} = \dfrac{3t^2 - 3}{2t}

\dfrac{3(2)^2 - 3}{2(2)} = \dfrac{9}{4}

x(2) = 2^2 + 1 = 5

y(2) = 2^3 - 3(2) = 2

y - 2 = \tfrac{9}{4}(x - 5)

Answer: dy/dx = (3t² − 3)/(2t), and the tangent line at t = 2 is y − 2 = (9/4)(x − 5).

Example 2: Second derivative

For the same curve, x = t² + 1 and y = t³ − 3t, find d²y/dx² and the concavity at t = 2.

\dfrac{dy}{dx} = \dfrac{3t^2 - 3}{2t} = \dfrac{3}{2}t - \dfrac{3}{2}t^{-1}

\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right) = \dfrac{3}{2} + \dfrac{3}{2}t^{-2}

\dfrac{d^2y}{dx^2} = \dfrac{\tfrac{3}{2} + \tfrac{3}{2}t^{-2}}{2t}

\dfrac{d^2y}{dx^2} = \dfrac{3t^2 + 3}{4t^3}

\dfrac{3(2)^2 + 3}{4(2)^3} = \dfrac{15}{32}

Answer: d²y/dx² = (3t² + 3)/(4t³), which is 15/32 at t = 2. It is positive, so the curve is concave up there.

Example 3: A trig curve

For the ellipse x = 3 cos t, y = 2 sin t, find dy/dx and the tangent line at t = π/4.

\dfrac{dx}{dt} = -3\sin t, \quad \dfrac{dy}{dt} = 2\cos t

\dfrac{dy}{dx} = \dfrac{2\cos t}{-3\sin t} = -\dfrac{2}{3}\cot t

-\dfrac{2}{3}\cot\dfrac{\pi}{4} = -\dfrac{2}{3}

\left(3 \cdot \dfrac{\sqrt{2}}{2},\; 2 \cdot \dfrac{\sqrt{2}}{2}\right) = \left(\dfrac{3\sqrt{2}}{2},\; \sqrt{2}\right)

y - \sqrt{2} = -\tfrac{2}{3}\left(x - \tfrac{3\sqrt{2}}{2}\right)

Answer: dy/dx = −(2/3)cot t, and the tangent line at t = π/4 is y − √2 = −(2/3)(x − 3√2/2).

Example 4: Horizontal and vertical tangents

Find all points where x = t² − 4t, y = t³ − 3t has a horizontal or a vertical tangent.

\dfrac{dx}{dt} = 2t - 4, \quad \dfrac{dy}{dt} = 3t^2 - 3

3t^2 - 3 = 0

t = 1 \text{ or } t = -1

\dfrac{dx}{dt} = -2 \text{ and } -6

(-3, -2) \text{ and } (5, 2)

2t - 4 = 0

t = 2

\dfrac{dy}{dt} = 3(2)^2 - 3 = 9

(2^2 - 4(2),\; 2^3 - 3(2)) = (-4, 2)

Answer: horizontal tangents at (−3, −2) and (5, 2); a vertical tangent at (−4, 2).

Example 5: The path of a ball (word problem)

A ball’s position in meters after t seconds is x = 30t, y = 40t − 4.9t². Find the slope of its path at t = 2, the highest point, and d²y/dx².

\dfrac{dy}{dx} = \dfrac{40 - 9.8t}{30}

\dfrac{40 - 9.8(2)}{30} = \dfrac{20.4}{30} = 0.68

40 - 9.8t = 0

t = \dfrac{40}{9.8} \approx 4.082

y \approx 40(4.082) - 4.9(4.082)^2 \approx 81.63

x \approx 30(4.082) \approx 122.45

\dfrac{d^2y}{dx^2} = \dfrac{-9.8/30}{30} \approx -0.01089

Answer: the slope is 0.68 at t = 2; the highest point is about (122.4, 81.6) meters, reached after about 4.08 seconds; and d²y/dx² ≈ −0.0109, so the path is always concave down.

Common mistakes

Related methods

Implicit differentiation

The ellipse in Example 3 can also be written as x²/9 + y²/4 = 1. Implicit differentiation gives the same slope, −4x/(9y), at the same point.

Eliminating the parameter

Sometimes you can solve for t and write y as a function of x, as in parametric equations. For x = eᵗ, y = e²ᵗ, that gives y = x². The parametric formula is still faster when elimination is messy.

Motion along a path

Parametric curves also give speed and distance traveled. Those are the next lesson, Arc Length of Parametric Curves, and Vector-Valued Functions, later in this unit. In one dimension, the same ideas appear in position, velocity and acceleration.

Practice problems

1. Find dy/dx for x = 2t + 1, y = t² − 4.
2. Find dy/dx for x = t³, y = t² + t, and its value at t = 1.
3. Find the tangent line to the curve in problem 2 at t = 1.
4. Find dy/dx for x = eᵗ, y = e²ᵗ. Write it in terms of x.
5. Find the slope of x = 4 cos t, y = 4 sin t at t = π/6.
6. Find d²y/dx² for x = 2t + 1, y = t² − 4.
7. Find d²y/dx² for x = t², y = t³, with t positive.
8. Find the horizontal and vertical tangents of x = t² − 2t, y = t² + 2t.
9. For the cycloid x = t − sin t, y = 1 − cos t, find dy/dx at t = π/2. What happens at t = 0?
10. A drone flies along x = 5t, y = 20 + 6t − t² (meters, seconds). When is it flying level, and how high is it then?

\dfrac{dx}{dt} = 2, \quad \dfrac{dy}{dt} = 2t

\dfrac{dy}{dx} = \dfrac{2t}{2} = t

\dfrac{dx}{dt} = 3t^2, \quad \dfrac{dy}{dt} = 2t + 1

\dfrac{dy}{dx} = \dfrac{2t + 1}{3t^2}

\dfrac{2(1) + 1}{3(1)^2} = 1

(1^3,\; 1^2 + 1) = (1, 2)

y - 2 = 1(x - 1)

y = x + 1

\dfrac{dy}{dx} = \dfrac{2e^{2t}}{e^{t}} = 2e^{t}

2e^{t} = 2x

\dfrac{dy}{dx} = \dfrac{4\cos t}{-4\sin t} = -\cot t

-\cot\dfrac{\pi}{6} = -\sqrt{3}

\dfrac{d}{dt}(t) = 1

\dfrac{d^2y}{dx^2} = \dfrac{1}{2}

\dfrac{dy}{dx} = \dfrac{3t^2}{2t} = \dfrac{3}{2}t

\dfrac{d}{dt}\left(\dfrac{3}{2}t\right) = \dfrac{3}{2}

\dfrac{d^2y}{dx^2} = \dfrac{3/2}{2t} = \dfrac{3}{4t}

\dfrac{dx}{dt} = 2t - 2, \quad \dfrac{dy}{dt} = 2t + 2

2t + 2 = 0 \Rightarrow t = -1

(1 + 2,\; 1 - 2) = (3, -1)

2t - 2 = 0 \Rightarrow t = 1

(1 - 2,\; 1 + 2) = (-1, 3)

\dfrac{dy}{dx} = \dfrac{\sin t}{1 - \cos t}

\dfrac{\sin(\pi/2)}{1 - \cos(\pi/2)} = \dfrac{1}{1} = 1

\dfrac{\sin 0}{1 - \cos 0} = \dfrac{0}{0}

\dfrac{dy}{dt} = 6 - 2t = 0

t = 3

y(3) = 20 + 18 - 9 = 29

Parametric derivatives FAQ

How do you find dy/dx for parametric equations?

Differentiate x and y with respect to t, then divide: dy/dx = (dy/dt)/(dx/dt). Substitute a value of t to get the slope at a point.

How do you find the second derivative of parametric equations?

Take the derivative of dy/dx with respect to t, then divide by dx/dt. In symbols, d²y/dx² = [d/dt (dy/dx)] / (dx/dt).

How do you find horizontal and vertical tangents of a parametric curve?

Horizontal tangents are where dy/dt = 0 and dx/dt is not 0. Vertical tangents are where dx/dt = 0 and dy/dt is not 0.

What if dy/dt and dx/dt are both zero?

Then the formula gives 0/0 and doesn’t tell you the slope. Look at the curve or take a limit; the point is often a cusp or a sharp turn.

Summary

• dy/dx = (dy/dt)/(dx/dt), as long as dx/dt is not 0.
• d²y/dx² = [d/dt (dy/dx)] / (dx/dt). Don’t forget the second division.
• Horizontal tangent: dy/dt = 0. Vertical tangent: dx/dt = 0.
• Both rates 0: no conclusion; look for a cusp.
• For a tangent line, find the point (x(t), y(t)) and the slope at the same t.

Previous lesson: Logistic Differential Equations

Next lesson: Arc Length of Parametric Curves

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