4.7 - Parametrization of Implicitly Defined Functions

Overview

In Topic 4.4 you parametrized circles and line segments. This topic extends the idea to curves described by equations: graphs of functions, inverse functions, and every conic section from Topic 4.6. A parametrization turns an equation into motion along the curve, one point for each value of

t

. It's how you graph a sideways parabola or a hyperbola on a calculator in parametric mode, and it's how calculus and physics describe paths.

What a parametrization has to do

A parametric function

(x(t), y(t))

is a parametrization of a curve defined by an equation in

x

and

y

when this is true:

Substituting

x(t)

for

x

and

y(t)

for

y

makes the equation true for every value of

t

in the domain.

"Every value" is the key phrase. Checking one or two points isn't enough. You need the equation to simplify to something that's always true, usually by algebra or by a Pythagorean identity like

\cos^2 t + \sin^2 t = 1

. You also usually want the parametrization to cover the whole curve (or the part you care about) as

t

runs through its domain.

Example 1: Checking a parametrization

(a) Show that

(x(t), y(t)) = (2t, \, t^2 + 1)

is a parametrization of

y = \frac{x^2}{4} + 1

. (b) Is

(x(t), y(t)) = (\cos t, \, 2\sin t)

a parametrization of the circle

x^2 + y^2 = 4

(a) Substitute

x = 2t

into the right side:

\frac{(2t)^2}{4} + 1 = \frac{4t^2}{4} + 1 = t^2 + 1 = y(t)

Answer: The equation holds for every

t

, so this is a parametrization. As

t

runs through all real numbers,

x = 2t

does too, so it covers the whole parabola.

(b) Substitute into the left side:

(\cos t)^2 + (2\sin t)^2 = \cos^2 t + 4\sin^2 t = 1 + 3\sin^2 t

Answer: No. This equals 4 only when

\sin^2 t = 1

, not for every

t

. At

t = 0

the point is

(1, 0)

, which is 1 unit from the origin, not 2. (This function actually traces the ellipse

x^2 + \frac{y^2}{4} = 1

Graphs of functions and their inverses

Any function

f

is a function of

x

, the graph of

y = f(x)

can be parametrized by letting

x

itself be the parameter:

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

Check:

x = t

, so

y = f(t) = f(x)

for every

t

. As

t

increases, the point moves left to right along the graph.

Inverses: Swap the coordinates

The graph of an inverse function is the graph of

f

reflected across the line

y = x

: every point

(a, b)

becomes

(b, a)

. So if

f

is invertible, swap the two coordinates:

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

This works even when you can't solve for

f^{-1}(x)

with algebra, so it's often the easiest way to graph an inverse.

Sometimes this only works on an appropriate interval of

t

. If

f

is only one-to-one on part of its domain, restrict

t

to that part. For example,

f(x) = x^2

isn't invertible on all real numbers, but it is for

x \ge 0

. Its inverse there,

f^{-1}(x) = \sqrt{x}

, is parametrized by

(t^2, \, t)

for

t \ge 0

Example 2: An inverse you can't solve for

Let

f(x) = x^3 + x

. (a) Parametrize the graph of

f

. (b) Explain why

f

is invertible, and parametrize the graph of

f^{-1}

. (c) Use the parametrization to find

f^{-1}(2)

and

f^{-1}(10)

(a) Use

(t, f(t))

(x(t), \, y(t)) = (t, \, t^3 + t)

(b) Both

x^3

and

x

are increasing for all

x

, so their sum is increasing too. An increasing function never repeats an output, so

f

is one-to-one and invertible on all real numbers. Swap the coordinates:

(x(t), \, y(t)) = (t^3 + t, \, t)

To find a formula for

f^{-1}

, you'd have to solve

x = y^3 + y

for

y

, a cubic equation with no easy algebra. The parametrization skips that step entirely.

4.7 - Parametrization of Implicitly Defined Functions

(c)

f^{-1}(2)

is the

y

-value on the inverse's graph where

x = 2

. Set

x(t) = 2

, so

t^3 + t = 2

. Trying

t = 1

gives

1 + 1 = 2

, and since

t^3 + t

is increasing, that's the only solution. The point is

(2, 1)

. In the same way,

t^3 + t = 10

t = 2

(since

8 + 2 = 10

), giving the point

(10, 2)

Answer:

f^{-1}(2) = 1

and

f^{-1}(10) = 2

. Check:

f(1) = 2

and

f(2) = 10

Parametrizing parabolas

A parabola, like any equation you can solve for one variable, is parametrized the same way as a function graph:

If you can solve for $y$ : write $y = f(x)$ , then use $(x(t), y(t)) = (t, \, f(t))$ . This replaces $x$ with $t$ .

If you can solve for $x$ : write $x = f(y)$ , then use $(x(t), y(t)) = (f(t), \, t)$ . This replaces $y$ with $t$ .

Example 3: Two parabolas

Parametrize (a)

y - 1 = -2(x + 3)^2

and (b)

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

, the parabola from Example 1 in Topic 4.6.

(a) Solve for

y

y = -2(x + 3)^2 + 1

. Replace

x

with

t

(x(t), \, y(t)) = (t, \, -2(t + 3)^2 + 1)

(b) Solve for

x

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

. Replace

y

with

t

(x(t), \, y(t)) = (2(t - 2)^2 - 3, \, t)

Check (b):

t = 2

gives

(-3, 2)

, the vertex, and

t = 0

gives

(2(4) - 3, 0) = (5, 0)

, which matches the graph in Topic 4.6. Since

y = t

, the point moves up the parabola as

t

increases.

Answer: (a)

(t, \, -2(t + 3)^2 + 1)

and (b)

(2(t - 2)^2 - 3, \, t)

, each for all real

t

Parametrizing ellipses

The ellipse with center

(h, k)

, horizontal radius

a

, and vertical radius

b

is parametrized by

x(t) = h + a\cos t, \qquad y(t) = k + b\sin t, \qquad 0 \le t \le 2\pi

Why it works: $\frac{x - h}{a} = \cos t$ and $\frac{y - k}{b} = \sin t$ , so $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = \cos^2 t + \sin^2 t = 1$ for every $t$ .

Start and direction: $t = 0$ gives $(h + a, k)$ , the right end. As $t$ goes from $0$ to $2\pi$ , the point goes once around counterclockwise.

Connection to circles: this is the circle parametrization from Topic 4.4, stretched by $a$ horizontally and by $b$ vertically. When $a = b = r$ , it's exactly the circle $(h + r\cos t, \, k + r\sin t)$ .

Heads up: unless the ellipse is a circle, $t$ is not the angle from the center to the point. It's just the parameter.

Example 4: Tracing an ellipse

Parametrize the ellipse

\frac{(x - 2)^2}{9} + \frac{(y + 3)^2}{4} = 1

(from Example 2 in Topic 4.6), make a table of points, and describe the motion.

Step 1: Read the center and radii. The center is

(2, -3)

a = 3

, and

b = 2

x(t) = 2 + 3\cos t, \qquad y(t) = -3 + 2\sin t, \qquad 0 \le t \le 2\pi

Step 2: Check.

\frac{(3\cos t)^2}{9} + \frac{(2\sin t)^2}{4} = \cos^2 t + \sin^2 t = 1

Step 3: Table.

$t$	$0$	$\frac{\pi}{2}$	$\pi$	$\frac{3\pi}{2}$	$2\pi$
$(x(t), y(t))$	$(5, -3)$	$(2, -1)$	$(-1, -3)$	$(2, -5)$	$(5, -3)$

Answer: The point starts at the right end,

(5, -3)

, moves counterclockwise through the top

(2, -1)

, the left end

(-1, -3)

, and the bottom

(2, -5)

, and returns to

(5, -3)

t = 2\pi

. To trace it clockwise instead, use

y(t) = -3 - 2\sin t

Parametrizing hyperbolas

Hyperbolas use secant and tangent, because of the Pythagorean identity

\sec^2 t - \tan^2 t = 1

which comes from dividing

\cos^2 t + \sin^2 t = 1

\cos^2 t

to get

1 + \tan^2 t = \sec^2 t

. A minus sign between two squares is exactly what a hyperbola's equation has.

Hyperbola	Parametrization, $0 \le t \le 2\pi$
opens left and right: $\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$	$x(t) = h + a\sec t$ , $y(t) = k + b\tan t$
opens up and down: $\frac{(y - k)^2}{b^2} - \frac{(x - h)^2}{a^2} = 1$	$x(t) = h + a\tan t$ , $y(t) = k + b\sec t$

Secant goes with the positive term. The variable whose squared term is positive gets $\sec t$ . Since $|\sec t| \ge 1$ , that coordinate never gets closer to the center than a vertex, which is exactly how a hyperbola behaves.

Check (left and right): $\frac{(a\sec t)^2}{a^2} - \frac{(b\tan t)^2}{b^2} = \sec^2 t - \tan^2 t = 1$ .

Two branches: when $\cos t > 0$ , $\sec t \ge 1$ and the point is on one branch (right or upper). When $\cos t < 0$ , $\sec t \le -1$ and the point is on the other branch (left or lower).

Gaps at $t = \frac{\pi}{2}$ and $t = \frac{3\pi}{2}$ : $\sec t$ and $\tan t$ are undefined there, so those two values are left out. As $t$ gets close to them, the point runs off along an asymptote, then comes back on the other branch.

Example 5: Parametrizing a hyperbola

Parametrize the hyperbola

\frac{(y - 2)^2}{9} - \frac{(x + 1)^2}{4} = 1

from Example 4 in Topic 4.6. Then find the points at

t = 0

t = \pi

, and

t = \frac{\pi}{3}

, and say which branch each is on.

Step 1: Set up. It opens up and down, with center

(-1, 2)

a = 2

, and

b = 3

. Secant goes with

y

x(t) = -1 + 2\tan t, \qquad y(t) = 2 + 3\sec t, \qquad 0 \le t \le 2\pi, \ t \neq \frac{\pi}{2}, \frac{3\pi}{2}

Step 2: Check.

\frac{(3\sec t)^2}{9} - \frac{(2\tan t)^2}{4} = \sec^2 t - \tan^2 t = 1

Step 3: Find the points.

$t = 0$ : $\tan 0 = 0$ and $\sec 0 = 1$ , so the point is $(-1, 5)$ , the upper vertex.

$t = \pi$ : $\tan \pi = 0$ and $\sec \pi = -1$ , so the point is $(-1, -1)$ , the lower vertex.

$t = \frac{\pi}{3}$ : $\tan\frac{\pi}{3} = \sqrt{3}$ and $\sec\frac{\pi}{3} = 2$ , so the point is $(-1 + 2\sqrt{3}, \, 8) \approx (2.46, 8)$ . Check: $\frac{(8 - 2)^2}{9} - \frac{(2\sqrt{3})^2}{4} = 4 - 3 = 1$ .

Answer:

(x(t), y(t)) = (-1 + 2\tan t, \, 2 + 3\sec t)

. The points

(-1, 5)

and

(-1 + 2\sqrt{3}, 8)

are on the upper branch (where

\sec t > 0

), and

(-1, -1)

is on the lower branch (where

\sec t < 0

). On each branch the point moves left to right as

t

increases, because

\tan t

increases on every interval where it's defined.

Example 6: Going backward

Identify each curve and give its equation in standard form: (a)

x(t) = -2 + 5\cos t

y(t) = 4 + 5\sin t

; (b)

x(t) = 1 + 2\cos t

y(t) = 3\sin t

; (c)

x(t) = 3 + 4\sec t

y(t) = -1 + 5\tan t

Strategy: isolate the trig functions, then use a Pythagorean identity to eliminate

t

(a)

\cos t = \frac{x + 2}{5}

and

\sin t = \frac{y - 4}{5}

. Since

\cos^2 t + \sin^2 t = 1

\frac{(x + 2)^2}{25} + \frac{(y - 4)^2}{25} = 1 \quad \Rightarrow \quad (x + 2)^2 + (y - 4)^2 = 25

Answer: a circle with center

(-2, 4)

and radius

5

(b)

\cos t = \frac{x - 1}{2}

and

\sin t = \frac{y}{3}

\frac{(x - 1)^2}{4} + \frac{y^2}{9} = 1

Answer: an ellipse with center

(1, 0)

, horizontal radius 2, and vertical radius 3, so it's taller than it is wide.

(c)

\sec t = \frac{x - 3}{4}

and

\tan t = \frac{y + 1}{5}

. Since

\sec^2 t - \tan^2 t = 1

\frac{(x - 3)^2}{16} - \frac{(y + 1)^2}{25} = 1

Answer: a hyperbola with center

(3, -1)

that opens left and right, with vertices

(7, -1)

and

(-1, -1)

and asymptotes

y + 1 = \pm\frac{5}{4}(x - 3)

Summary

Curve	Parametrization
graph of $y = f(x)$	$(t, \, f(t))$
graph of $f^{-1}$	$(f(t), \, t)$ , on an interval where $f$ is one-to-one
parabola solved for $x$ , $x = f(y)$	$(f(t), \, t)$
ellipse	$(h + a\cos t, \, k + b\sin t)$ , $0 \le t \le 2\pi$
hyperbola, left and right	$(h + a\sec t, \, k + b\tan t)$ , $0 \le t \le 2\pi$
hyperbola, up and down	$(h + a\tan t, \, k + b\sec t)$ , $0 \le t \le 2\pi$

Common mistakes

Checking only a few points. A parametrization has to satisfy the equation for every $t$ , not just at $t = 0$ .

Swapping secant and tangent. Secant goes with the variable whose squared term is positive.

Forgetting the center. $a\cos t$ alone is centered at $0$ . Add $h$ to $x$ and $k$ to $y$ .

Using $a^2$ instead of $a$ . For $\frac{(x - 2)^2}{9}$ , use $3\cos t$ , not $9\cos t$ .

Ignoring the domain of an inverse. $(f(t), t)$ gives an inverse only where $f$ is one-to-one. Restrict $t$ if you need to.

Plugging in $t = \frac{\pi}{2}$ for a hyperbola. Secant and tangent are undefined there.

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