Tangent to a Parametric Curve

Let a parametric curve be given by

x = x(t), \quad y = y(t).

At $t=t_0$ , the curve passes through the point

P = \big(x(t_0), y(t_0)\big).

If $x'(t_0) \neq 0$ , then the tangent slope is

\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{y'(t_0)}{x'(t_0)}.

Method 1: Chain rule via a local graph

Suppose locally the curve can be written as a function $y = F(x)$ . Then by the chain rule,

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

So when $x'(t_0) \neq 0$ ,

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

This is the usual tangent formula for a parametric curve.

The formula breaks down when $x'(t_0)=0$ . In that case we cannot divide by $dx/dt$ , and the curve may fail to be a local graph $y=F(x)$ .

For example, consider the heart-shaped curve

x = 16\sin^3 t, \quad y = 13\cos t - 5\cos(2t) - 2\cos(3t) - \cos(4t).

At $t=\pi/2$ ,

x = 16, \quad y = 4, \quad x'(\pi/2)=0,

so the tangent line is vertical.

Heart curve with a vertical tangent

Method 2: Right-triangle approximation

Now take two nearby parameter values $t_0$ and $t_0 + \Delta t$ .

The change in position is approximately

\Delta x \approx x'(t_0)\,\Delta t, \quad \Delta y \approx y'(t_0)\,\Delta t.

Those two changes form the legs of a small right triangle, so the secant slope is approximately

\frac{\Delta y}{\Delta x} \approx \frac{y'(t_0)\Delta t}{x'(t_0)\Delta t} = \frac{y'(t_0)}{x'(t_0)}.

As $\Delta t \to 0$ , the secant line becomes the tangent line.

For the same heart curve, choose $t_0 = \pi/4$ . Then

P \approx (5.6569, 11.6066), \quad \frac{dy}{dx} \approx 0.2976.

So the tangent line is approximately

y - 11.6066 = 0.2976(x - 5.6569),

or equivalently

y \approx 0.2976x + 9.9232.

Tangent line from a local secant approximation

Second Derivative and Concavity

The first derivative of a parametric curve is

\frac{dy}{dx} = \frac{y'(t)}{x'(t)}, \quad x'(t) \neq 0.

To differentiate again with respect to $x$ , use

\frac{d}{dx} = \frac{1}{x'(t)}\frac{d}{dt}.

Therefore,

\frac{d^2y}{dx^2} = \frac{1}{x'(t)}\frac{d}{dt}\left(\frac{y'(t)}{x'(t)}\right) = \frac{x'(t)y''(t) - y'(t)x''(t)}{(x'(t))^3}.

This is the standard second-derivative formula for a parametric curve.

Concavity

If $\dfrac{d^2y}{dx^2} > 0$ , the curve is concave upward.
If $\dfrac{d^2y}{dx^2} < 0$ , the curve is concave downward.

For example, take

x = t, \quad y = t^3 - 3t.

Here $x'(t)=1$ , so the formula reduces to the usual calculus rule:

\frac{d^2y}{dx^2} = 6t.

So the curve is concave downward for $t<0$ , concave upward for $t>0$ , and changes concavity at $t=0$ .

Second derivative and concavity

Arc Length

For a tiny piece of curve, the change in position can be approximated by a right triangle.

If the small changes are $\Delta x$ and $\Delta y$ , then by the Pythagorean theorem,

(\Delta s)^2 \approx (\Delta x)^2 + (\Delta y)^2.

So the small arc length is approximately

\Delta s \approx \sqrt{(\Delta x)^2 + (\Delta y)^2}.

If the curve is written as $y=f(x)$ , then

\Delta y \approx \frac{dy}{dx}\Delta x,

\Delta s \approx \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,\Delta x.

Passing to the limit gives the arc length formula

L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx.

For a parametric curve $x=x(t)$ , $y=y(t)$ ,

dx = x'(t)\,dt, \quad dy = y'(t)\,dt.

Then the same Pythagorean idea gives

ds = \sqrt{(x'(t))^2 + (y'(t))^2}\,dt,

and therefore

L = \int_a^b \sqrt{(x'(t))^2 + (y'(t))^2}\,dt.

A curve used to visualize tiny arc segments

Exercises

For the heart curve

x=16\sin^3 t, \quad y=13\cos t - 5\cos(2t) - 2\cos(3t) - \cos(4t),

find the tangent line at $t=\pi/4$ .

Hint: compute $x'(t)$ and $y'(t)$ , then use $\dfrac{dy}{dx}=\dfrac{y'}{x'}$ .

Answer: $\dfrac{dy}{dx} \approx 0.2976$ , so the tangent line is

y \approx 0.2976x + 9.9232.

On the same heart curve, what happens at $t=\pi/2$ ?

Hint: compute $x'(\pi/2)$ .

Answer: $x'(\pi/2)=0$ , so the quotient $\dfrac{y'}{x'}$ is not usable there. The tangent is vertical, and the curve is not locally a graph $y=F(x)$ .

Let $x=t$ and $y=t^3-3t$ . Compute $\dfrac{d^2y}{dx^2}$ and determine where the curve is concave upward and concave downward.

Answer: Since $x'(t)=1$ and $y''(t)=6t$ ,

\frac{d^2y}{dx^2}=6t.

The curve is concave downward for $t<0$ and concave upward for $t>0$ .

Let $x=3t$ and $y=4t$ for $0\le t\le 1$ . Find the arc length.

Hint: use $L=\int_0^1\sqrt{(x')^2+(y')^2}\,dt$ .

Answer: $x'=3$ and $y'=4$ , so

L=\int_0^1\sqrt{3^2+4^2}\,dt=\int_0^1 5\,dt=5.

Summary

For a parametric curve, the tangent slope is $\dfrac{dy}{dx}=\dfrac{y'}{x'}$ when $x'\neq 0$ .
The chain rule explains the formula, but it fails when $x'=0$ .
The second derivative is $\dfrac{d^2y}{dx^2}=\dfrac{x'y''-y'x''}{(x')^3}.$
Concavity is determined by the sign of $\dfrac{d^2y}{dx^2}$ .
Arc length comes from a tiny right triangle and the Pythagorean theorem.

Calculus of the Curves