Math Core

Many curves can't be written as $y = f(x)$ at all: a path that loops back on itself or crosses itself fails the vertical line test. Parametric equations handle these curves by giving $x$ and $y$ separately as functions of a third variable, $t$ . This lesson shows how to find the slope of such a curve without ever solving for $y$ in terms of $x$ .

Curves described by a parameter

A parametric curve is a pair of equations

x = x(t), \qquad y = y(t),

where each value of the parameter $t$ gives one point $(x(t), y(t))$ . Often $t$ is time, and you can picture a particle tracing the curve as $t$ increases.

For example, $x = t^2$ and $y = t^3 - 3t$ trace the curve below for $-2 \le t \le 2$ . At $t = -2$ the point is $(4, -2)$ ; at $t = 0$ it is $(0, 0)$ ; at $t = 2$ it is $(4, 2)$ . The curve passes through $(3, 0)$ twice, once at $t = -\sqrt{3}$ and once at $t = \sqrt{3}$ , so it is not the graph of any function of $x$ .

x = t², y = t³ − 3t for −2 ≤ t ≤ 2. The marked points are where the tangent line is horizontal (1, ±2) or vertical (0, 0).Open in grapher →

The slope formula

You want $\dfrac{dy}{dx}$ , the slope of the tangent line. Both $x$ and $y$ depend on $t$ , so the chain rule says

\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}.

Solving for $\dfrac{dy}{dx}$ gives the key formula.

Slope of a parametric curve

If $x = x(t)$ and $y = y(t)$ are differentiable and $\dfrac{dx}{dt} \ne 0$ , then

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}.

The result is usually an expression in $t$ . To find the slope at a particular point, plug in the value of $t$ that produces that point.

Think of it as a ratio of rates: $\dfrac{dy}{dt}$ is how fast the particle moves vertically, $\dfrac{dx}{dt}$ is how fast it moves horizontally, and their ratio is the steepness of the path.

Worked example: Slope in terms of t, then at a point

For $x = t^2$ and $y = t^3 - 3t$ , find $\dfrac{dy}{dx}$ and the slope of the curve at $t = 2$ .

Differentiate each coordinate: $\dfrac{dx}{dt} = 2t$ and $\dfrac{dy}{dt} = 3t^2 - 3$ . So

\frac{dy}{dx} = \frac{3t^2 - 3}{2t}.

At $t = 2$ : $\dfrac{dy}{dx} = \dfrac{3(4) - 3}{2(2)} = \dfrac{9}{4}$ . The curve passes through $(4, 2)$ there with slope $\dfrac{9}{4}$ .

Tangent lines

A tangent line needs a point and a slope. Both come from the same value of $t$ : plug $t$ into $x(t)$ and $y(t)$ for the point, and into $\dfrac{dy}{dx}$ for the slope. Then use point-slope form.

Worked example: Equation of a tangent line

Find the tangent line to $x = t^2 + 1$ , $y = t^3$ at $t = 1$ .

Point: $x(1) = 2$ , $y(1) = 1$ , so the point is $(2, 1)$ .

Slope: $\dfrac{dx}{dt} = 2t$ and $\dfrac{dy}{dt} = 3t^2$ , so $\dfrac{dy}{dx} = \dfrac{3t^2}{2t}$ . At $t = 1$ this is $\dfrac{3}{2}$ .

Line: $y - 1 = \dfrac{3}{2}(x - 2)$ .

The same method works with trig functions. For the ellipse $x = 3\cos t$ , $y = 2\sin t$ , you get $\dfrac{dy}{dx} = \dfrac{2\cos t}{-3\sin t}$ , and at $t = \dfrac{\pi}{4}$ the sines and cosines cancel to give a slope of $-\dfrac{2}{3}$ .

Horizontal and vertical tangents

The fraction $\dfrac{dy/dt}{dx/dt}$ tells you where the tangent line is flat or straight up and down.

Horizontal tangent: $\dfrac{dy}{dt} = 0$ and $\dfrac{dx}{dt} \ne 0$ . The particle is momentarily moving only sideways.
Vertical tangent: $\dfrac{dx}{dt} = 0$ and $\dfrac{dy}{dt} \ne 0$ . The particle is momentarily moving only up or down.

If both derivatives are $0$ at the same $t$ , the formula gives $\dfrac{0}{0}$ and tells you nothing on its own. The curve might have a cusp, a corner, or an ordinary tangent there, and you would need more analysis.

Worked example: Finding horizontal and vertical tangents

For $x = t^2$ , $y = t^3 - 3t$ , find every point where the tangent line is horizontal or vertical.

Horizontal: $\dfrac{dy}{dt} = 3t^2 - 3 = 0$ gives $t = 1$ and $t = -1$ . At both, $\dfrac{dx}{dt} = 2t \ne 0$ . The points are $(1, -2)$ at $t = 1$ and $(1, 2)$ at $t = -1$ .

Vertical: $\dfrac{dx}{dt} = 2t = 0$ gives $t = 0$ . There $\dfrac{dy}{dt} = -3 \ne 0$ , so the tangent is vertical at $(0, 0)$ .

These match the three marked points on the graph above.

Common mistake

Don't write $\dfrac{dx/dt}{dy/dt}$ . The quantity you are differentiating on top is the one you want on top: $\dfrac{dy}{dx}$ has $y$ in the numerator, so $\dfrac{dy}{dt}$ goes in the numerator. A quick check is that setting the numerator to zero should give horizontal tangents.

When you only know the rates

AP free-response questions often skip the formulas entirely and give you values like $x'(3) = 4$ and $y'(3) = -6$ . The slope at $t = 3$ is still just the ratio: $\dfrac{-6}{4} = -\dfrac{3}{2}$ . You don't need $x(t)$ or $y(t)$ themselves.

Tip

To check a slope, you can sometimes eliminate the parameter. For $x = t^2 + 1$ , $y = t^3$ with $t > 0$ , you have $t = \sqrt{x - 1}$ and $y = (x - 1)^{3/2}$ , so $\dfrac{dy}{dx} = \dfrac{3}{2}(x - 1)^{1/2}$ . At $x = 2$ that's $\dfrac{3}{2}$ , matching the example. Eliminating the parameter is usually harder, though, so treat it as a check, not the main method.

Practice

Practice 1

A curve is given by $x = t^3$ and $y = t^2 + 1$ . Find $\dfrac{dy}{dx}$ at $t = 2$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

For $x = t^2 + 3t$ and $y = t^3 - 2t$ , find $\dfrac{dy}{dx}$ in terms of $t$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 3

A particle moves along the curve $x = \cos t$ , $y = \sin(2t)$ . What is the slope of the curve at $t = \dfrac{\pi}{6}$ ?

Practice 4

The curve $x = t^2 + 1$ , $y = t^3 - 3t$ has horizontal tangent lines at which values of $t$ ? List all of them.

Separate answers with commas, e.g. 2, -5

Practice 5

Find an equation of the line tangent to $x = 2t + 1$ , $y = t^2 - t$ at $t = 2$ . Give $y$ in terms of $x$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 6

The curve $x = t^3 - 3t$ , $y = t^2$ has a vertical tangent line at one point with $t > 0$ . Find that point $(x, y)$ .

Enter a point like (2, -3)

Practice 7

A particle moves in the plane so that $x'(3) = 4$ and $y'(3) = -6$ . The particle is at $(1, 5)$ when $t = 3$ . Which is an equation of the line tangent to its path at that point?