In 4.5, we introduced the idea of equations involving two variables being capable of implicitly defining** one or more functions. Tying this concept to the rest of the first half of the unit, we’d like to know how to parametrize these types of functions! 🤔
🏹 Representing Curves in Planes
A **parametrization is a way to express a curve in the form of a set of equations and that depend on a parameter . When these equations are substituted into the corresponding equation that defines the curve implicitly, they should satisfy the equation for every value of in the domain.
For example, the equation describes the unit circle centered at the origin. A parametrization of this circle is , , with domain . When we substitute these equations into the implicit equation we get which is always true.
💡 The key point is that a parametrization gives a specific way to express the curve, in terms of a set of equations, that satisfies the implicit equation for every value of the parameter in the domain.
Moving on, if is a function of and is the graph of , then the curve can be parametrized as , where and . This can be visualized as moving along the x-axis, starting at and ending at and at each point on the x-axis, the corresponding point on the graph of is plotted.
When is invertible, it means that the function has an inverse function, , which is also a function of . In this case, the inverse function can be parametrized as , , for an appropriate interval of . This can be visualized as moving along the y-axis, starting at and ending at and at each point on the y-axis, the corresponding point on the graph of is plotted.
It's important to note that the domain of t for the inverse function will be the range of the original function!
🔔 Representing Conic Sections in Planes
In 4.6, we discussed conic sections, geometric shapes created by intersecting a plane and a cone, such as circles, ellipses, parabolas, and hyperbolas. Recall that each type of conic section can be defined by a specific equation and has its own unique properties.
↪️ Parabolas
A parabola, like any other equation involving two variables, can be represented in a parametric form. One way to do this is to solve for one of the variables and express the other in terms of a parameter.
For example, if the equation of the parabola can be solved for , it can be parametrized as , , , where is expressed in terms of the parameter and is simply .
Similarly, if the equation can be solved for , it can be parametrized as , , where is now expressed in terms of and is .
💡 This way, the parabola can be represented as a set of points that satisfy the original equation for different values of the parameter t in the domain! 📌
🥚 Ellipses
An ellipse can be represented in parametric form by the equations and where , are the center coordinates of the ellipse, and are the semi-major and semi-minor axis respectively.
The domain of the parametric equations is , which represents a complete counterclockwise revolution of the ellipse. These equations use the trigonometric functions cosine and sine, which allow us to express the position of a point on the ellipse at any angle , in terms of the distance from the center and the angle formed by the line segment joining the point on the ellipse to the center and the x-axis.
🔁 Hyperbolas
Recall that a hyperbola is a type of conic section that can be defined by its standard equation, , where and are the lengths of the semi-major and semi-minor axes of the hyperbola, respectively.
For a hyperbola that opens horizontally, with the center located at the point , the parametric equations are given by:
where is the parameter, and and are the lengths of the semi-major and semi-minor axes, respectively. This representation is valid for . ↔️
For a hyperbola that opens vertically, with the center located at the point , the parametric equations are given by:
where is the parameter, and and are the lengths of the semi-major and semi-minor axes, respectively. This representation is valid for . ↕️
It's important to note that the functions and are the reciprocal functions of and respectively.
Frequently Asked Questions
How do I parametrize an implicitly defined function like x² + y² = 25?
Think “circle trig”—use sine and cosine so x(t)² + y(t)² = 25 holds for every t. A standard parametrization is x(t) = 5 cos t, y(t) = 5 sin t, 0 ≤ t < 2π. Substitute to check: (5 cos t)² + (5 sin t)² = 25(cos²t + sin²t) = 25, so it satisfies the implicit equation for all t. This traces the whole circle centered at (0,0) with radius 5 once counterclockwise; changing the sign (x = 5 cos t, y = −5 sin t) reverses orientation. If you want to view it as a function of x or y locally, you can parametrize half-circles as (t, ±√(25 − t²)) with −5 ≤ t ≤ 5, but that won’t cover the full circle in one smooth function. For AP prep and the CED link between implicit curves and trig parametric forms, see the Topic 4.7 study guide (https://library.fiveable.me/ap-pre-calculus/unit-4/parametrization-implicitly-defined-functions/study-guide/iiEy0E9fETps45yV). For more practice, try problems at (https://library.fiveable.me/practice/ap-pre-calculus).
What's the difference between parametrizing a function and finding its implicit derivative?
Parametrizing a curve and finding its implicit derivative are related but different goals. - Parametrization: you produce functions x(t), y(t) so that plugging them into the implicit equation F(x,y)=0 makes it true for every t in some interval (CED 4.7.A.1). That gives a parametric description of the whole curve (location + a parameter t that can encode direction and speed). Example: circle x^2+y^2=1 → (cos t, sin t). - Implicit derivative: you start with F(x,y)=0 and differentiate both sides with respect to x (or implicitly with respect to t and use chain rule) to get dy/dx, the slope of the tangent at a point. For F(x,y)=0 you use ∂F/∂x + ∂F/∂y · (dy/dx)=0 so dy/dx = −(F_x)/(F_y). If you have a parametrization, you can also get the slope as (dy/dt)/(dx/dt). In short: parametrization gives a coordinate description of the curve; implicit differentiation gives the instantaneous slope (rate of change) at points on that curve. For AP prep, make sure you can both parametrize (CED 4.7.A) and compute slopes from implicit equations or parametric forms. For a quick review, see the Topic 4.7 study guide (https://library.fiveable.me/ap-pre-calculus/unit-4/parametrization-implicitly-defined-functions/study-guide/iiEy0E9fETps45yV) and try extra practice problems (https://library.fiveable.me/practice/ap-pre-calculus).
When do I use (t, f(t)) vs (f(t), t) for parametrization?
Use (t, f(t)) when your curve is given as y = f(x) (you solved the equation for y). Let x = t and y = f(t). Use (f(t), t) when the equation is solved for x in terms of y (or when you want to parametrize the inverse): set y = t and x = f(t). This matches the CED essentials: y = f(x) → (t, f(t)); if f is invertible, the inverse can be (f(t), t) (4.7.A.2 and 4.7.B.1). Quick examples: - y = x^2 → parametrize as (t, t^2). - x = y^2 → parametrize as (t^2, t). - If y = e^x and you want the inverse curve x = ln y, an inverse parametrization is (e^t, t) (or swap depending on which variable you set = t). Always verify by substituting x(t), y(t) into the original implicit equation (4.7.A.1). For more practice and AP-aligned guidance, see the Topic 4.7 study guide (https://library.fiveable.me/ap-pre-calculus/unit-4/parametrization-implicitly-defined-functions/study-guide/iiEy0E9fETps45yV) and the Unit 4 overview (https://library.fiveable.me/ap-pre-calculus/unit-4). For lots of practice problems, check (https://library.fiveable.me/practice/ap-pre-calculus).
Can someone explain step by step how to parametrize a circle centered at (3, -2) with radius 4?
Start with the standard trig parametrization of a circle centered at (h, k) with radius a: x(t) = h + a cos t, y(t) = k + a sin t, 0 ≤ t < 2π. For your circle: h = 3, k = −2, a = 4. So x(t) = 3 + 4 cos t, y(t) = −2 + 4 sin t, 0 ≤ t < 2π. Why this works (verification): substitute into the circle equation (x − 3)^2 + (y + 2)^2 = 16. (x(t) − 3)^2 + (y(t) + 2)^2 = (4 cos t)^2 + (4 sin t)^2 = 16(cos^2 t + sin^2 t) = 16, so the equation holds for every t. Notes useful for AP: t in [0, 2π) traces the whole circle once counterclockwise; using x = 3 + 4 cos t, y = −2 − 4 sin t would reverse direction. For more on parametrizing implicitly defined curves and AP-aligned practice, see the Topic 4.7 study guide (https://library.fiveable.me/ap-pre-calculus/unit-4/parametrization-implicitly-defined-functions/study-guide/iiEy0E9fETps45yV) and unit review (https://library.fiveable.me/ap-pre-calculus/unit-4).
I'm confused about parametrizing conic sections - do I always use trig functions?
Short answer: no—not always. Use trig when it gives a neat, full parametrization for closed curves (circles/ellipses) or when trig identities match the conic’s equation (ellipse: x = h + a cos t, y = k + b sin t; hyperbola: sec/tan or tan/sec forms). But for parabolas and any conic you can solve for y (or x) you can parametrize algebraically as (t, f(t)) or (f(t), t)—no trig needed. Why choose one over the other: - Use trig when you want a single-parameter sweep that covers the whole curve smoothly (ellipse/circle) and when trig identities directly satisfy the implicit equation (CED 4.7.B.2–3). - Use nontrig parametrizations when the conic is given explicitly (y = f(x)) or when you only need a piece or simpler algebraic form (CED 4.7.A.2, 4.7.B.1). On the AP exam: show that substitution of x(t), y(t) satisfies the implicit equation (CED 4.7.A.1). Calculator work involving trig should use radian mode per exam instructions. For a focused review and practice, see the Topic 4.7 study guide (https://library.fiveable.me/ap-pre-calculus/unit-4/parametrization-implicitly-defined-functions/study-guide/iiEy0E9fETps45yV) and more unit resources (https://library.fiveable.me/ap-pre-calculus/unit-4). Practice problems are available at (https://library.fiveable.me/practice/ap-pre-calculus).
What's the formula for parametrizing an ellipse that's not centered at the origin?
Use x(t) = h + a cos t, y(t) = k + b sin t for 0 ≤ t ≤ 2π. Here (h, k) is the center, a and b are the semi-major and semi-minor axes (absolute values), and substituting x(t), y(t) into the ellipse equation ((x − h)^2 / a^2) + ((y − k)^2 / b^2) = 1 makes it hold for every t (CED 4.7.B.2). If the ellipse is rotated by an angle φ, use the rotated parametrization: x(t) = h + a cos t cos φ − b sin t sin φ y(t) = k + a cos t sin φ + b sin t cos φ. Verify your parametrization by substitution (CED 4.7.A.1). For a quick refresher and examples, see the Topic 4.7 study guide (https://library.fiveable.me/ap-pre-calculus/unit-4/parametrization-implicitly-defined-functions/study-guide/iiEy0E9fETps45yV) and practice problems (https://library.fiveable.me/practice/ap-pre-calculus).
How do I know if a hyperbola opens left/right or up/down when parametrizing?
Look at the standard form of the hyperbola. Which squared term is positive tells you the transverse (opening) direction: - (x − h)^2/a^2 − (y − k)^2/b^2 = 1 → opens left/right (transverse axis horizontal). Parametrize with x(t)=h + a sec t, y(t)=k + b tan t. - (y − k)^2/b^2 − (x − h)^2/a^2 = 1 → opens up/down (transverse axis vertical). Parametrize with x(t)=h + a tan t, y(t)=k + b sec t. Quick checks you can use: vertices lie a units from the center (h,k) along the transverse axis (so if vertices are left/right, it’s horizontal); asymptotes have slopes ±b/a for horizontal-transverse form and ±a/b for vertical-transverse form. Substitute your parametric x(t),y(t) back into the implicit equation to verify (CED 4.7.B.3). For a quick review, see the Topic 4.7 study guide (https://library.fiveable.me/ap-pre-calculus/unit-4/parametrization-implicitly-defined-functions/study-guide/iiEy0E9fETps45yV). For extra practice, try problems at (https://library.fiveable.me/practice/ap-pre-calculus).
Why do we use sec t and tan t for hyperbolas instead of sin and cos?
Good question. We use sec t and tan t because they satisfy the identity that matches a hyperbola. For the standard hyperbola x^2/a^2 − y^2/b^2 = 1, substituting x = a sec t and y = b tan t gives (a sec t)^2/a^2 − (b tan t)^2/b^2 = sec^2 t − tan^2 t = 1, so the parametrization automatically satisfies the equation for every t in its domain. Sin and cos fit ellipses because sin^2 + cos^2 = 1, which matches x^2/a^2 + y^2/b^2 = 1. Also, sec and tan naturally reflect hyperbola behavior: they blow up (go to ±∞) at certain t-values, which corresponds to the hyperbola’s vertical/horizontal asymptotes—useful when sketching or reasoning about limits and asymptotes on the AP exam (CED Topic 4.7.B). Remember to avoid t where sec or tan are undefined. For more examples and checks you can verify by substitution, see the Topic 4.7 study guide (https://library.fiveable.me/ap-pre-calculus/unit-4/parametrization-implicitly-defined-functions/study-guide/iiEy0E9fETps45yV). For extra practice, try problems at (https://library.fiveable.me/practice/ap-pre-calculus).
When parametrizing y = x³, do I write it as (t, t³) or (t³, t)?
Use (t, t^3). For y = x^3 you want x(t) = t and y(t) = f(t) = t^3, because when you substitute x(t) and y(t) into the equation you get y(t) = (x(t))^3 → t^3 = t^3, which holds for every t (CED 4.7.A.1–4.7.A.2). (t^3, t) does not parametrize y = x^3—it parametrizes the inverse relation x = y^3 (equivalently y = ∛x). That form (f(t), t) is only correct when you want the inverse function as your parametric curve (CED 4.7.A.2). Always check by substitution: plug x(t) and y(t) into the defining equation to verify. For a quick refresher on these rules and examples, see the Topic 4.7 study guide (https://library.fiveable.me/ap-pre-calculus/unit-4/parametrization-implicitly-defined-functions/study-guide/iiEy0E9fETps45yV). For more practice, try problems in the Unit 4 practice set (https://library.fiveable.me/practice/ap-pre-calculus).
I don't understand how to check if my parametrization is correct - what do I substitute where?
Check your parametrization by direct substitution: wherever the curve’s equation has x, plug in x(t); wherever it has y, plug in y(t). If the resulting identity holds for every t in the parameter domain, the parametrization is correct (this is 4.7.A.1 in the CED). Quick checklist: - Write the implicit equation F(x,y)=0. - Replace x with x(t) and y with y(t) to get F(x(t),y(t)). - Simplify. If you get 0 (or an identity true for all t in the allowed domain), you’re done. If not, it’s wrong or the domain is too big/narrow. Examples to keep in mind: - If y = f(x), param as (t, f(t)). Substitute to get f(t) = f(t) (trivial check). - For a circle x^2 + y^2 = r^2 and (x(t),y(t)) = (r cos t, r sin t), substitute and use cos^2+sin^2 = 1 to verify. - For an ellipse use x = h + a cos t, y = k + b sin t and check ((x−h)/a)^2 + ((y−k)/b)^2 = 1. - For a hyperbola use sec/tan or tan/sec forms and verify with sec^2−tan^2 = 1. Remember the AP requirement: it must hold for every t in the domain you specify (and pick domain so you trace the intended part of the curve). For extra practice and examples see the Topic 4.7 study guide (https://library.fiveable.me/ap-pre-calculus/unit-4/parametrization-implicitly-defined-functions/study-guide/iiEy0E9fETps45yV) and Unit 4 resources (https://library.fiveable.me/ap-pre-calculus/unit-4). For lots of practice problems, go to (https://library.fiveable.me/practice/ap-pre-calculus).
What's the domain for t when parametrizing a parabola vs an ellipse?
For a parabola: you parametrize it like any solve-for-y (or x) curve. If y = f(x) use (x(t), y(t)) = (t, f(t)) and the domain for t is whatever x’s domain is (often all real numbers, (−∞, ∞)). If you parametrize one branch after solving for x = g(y), use t over that branch’s domain (for example t ≥ 0 for a parabola opening right). So t’s domain = the function’s domain (CED 4.7.A, 4.7.B.1). For an ellipse: use trig parametrization x(t)=h+a cos t, y(t)=k+b sin t. To trace the ellipse once choose 0 ≤ t ≤ 2π (CED 4.7.B.2). You can extend t to all reals to trace it repeatedly, but AP convention is 0 to 2π for one full traversal. Note: on the AP exam use radian mode for trig parametrizations (exam guidance). Want more examples and practice? See the Topic 4.7 study guide (https://library.fiveable.me/ap-pre-calculus/unit-4/parametrization-implicitly-defined-functions/study-guide/iiEy0E9fETps45yV) and Unit 4 overview (https://library.fiveable.me/ap-pre-calculus/unit-4). For more practice problems try (https://library.fiveable.me/practice/ap-pre-calculus).
How do I parametrize x² - 4y² = 16 if it's a hyperbola?
Write it like a standard hyperbola with center (0,0): x^2/16 − y^2/4 = 1 (so a = 4, b = 2). Two common parametric forms: - Trig (covers both branches): x(t) = ±4 sec t, y(t) = 2 tan t, with t in any interval avoiding cos t = 0 (e.g., t ∈ (−π/2, π/2) gives the right branch; add the negative sign for the left branch). Substitute into x^2 − 4y^2 = 16 to check it satisfies the equation for all allowed t (this is required by the CED 4.7.A.1). - Hyperbolic (single smooth parametrization of each branch): x(u) = ±4 cosh u, y(u) = 2 sinh u, u ∈ (−∞, ∞). This also satisfies x^2 − 4y^2 = 16 because cosh^2 − sinh^2 = 1. Both are AP-appropriate: use trig/sec & tan for left/right hyperbolas per CED 4.7.B.3 or cosh/sinh if you prefer. For extra practice and worked examples, check the Topic 4.7 study guide (https://library.fiveable.me/ap-pre-calculus/unit-4/parametrization-implicitly-defined-functions/study-guide/iiEy0E9fETps45yV) and more practice problems at (https://library.fiveable.me/practice/ap-pre-calculus).
Can I parametrize any curve or are there restrictions on which functions can be parametrized?
Short answer: most plane curves can be parametrized, but there are important restrictions and choices. What you need to know (CED-aligned) - If a curve is given explicitly y = f(x), parametrize as (x(t),y(t)) = (t, f(t)). If f is invertible you can use (f(t), t) for the inverse (4.7.A.2). - For conics use standard trig or sec/tan forms: circle/ellipse with cos/sin, hyperbola with sec/tan, parabola by solving for x or y (4.7.B.1–3). - For an implicitly defined equation F(x,y)=0 you must pick x(t), y(t) so F(x(t),y(t)) = 0 for all t (4.7.A.1). When parametrization is tricky - Some curves can’t be written globally as y = f(x) (vertical tangents, loops, self-intersections) so you need piecewise or nonstandard param (e.g., use x(t)=cos t, y(t)=sin 2t). - Smooth, single-valued parametrizations may fail at cusps or where the implicit function theorem doesn’t hold. - Many algebraic curves have parametrizations; some require rational or trig choices, and some higher-degree curves aren’t parametrizable by elementary functions. For examples and practice, see the Topic 4.7 study guide (https://library.fiveable.me/ap-pre-calculus/unit-4/parametrization-implicitly-defined-functions/study-guide/iiEy0E9fETps45yV) and try practice problems (https://library.fiveable.me/practice/ap-pre-calculus).
What happens when I try to parametrize a function that's not invertible?
If a curve comes from a function that isn’t invertible, you can still parametrize the curve—you just can’t use the simple inverse trick (x(t), y(t)) = (f(t), t) globally. Two common approaches: - Parametrize the curve directly as (t, f(t)) (always works if y = f(x)). That gives the original graph but not the inverse. - If the relation is implicitly defined or not a function (or a function with repeats), choose a parameter that traces the curve (e.g., circle: x = cos t, y = sin t). That works even when no single-valued inverse exists or when the curve crosses itself. If you want an inverse parametrization, restrict the domain to an interval where f is one-to-one; then (f(t), t) gives the inverse on that interval. Otherwise you may need piecewise parametrizations or a parameter that traces the curve multiple times. Always check your parametrization by substituting x(t), y(t) into the defining equation (CED 4.7.A.1). For more examples and practice problems, see the Topic 4.7 study guide (https://library.fiveable.me/ap-pre-calculus/unit-4/parametrization-implicitly-defined-functions/study-guide/iiEy0E9fETps45yV) and more practice at (https://library.fiveable.me/practice/ap-pre-calculus).
Step by step, how do I go from the equation (x-1)²/9 + (y+2)²/4 = 1 to parametric form?
This is an ellipse in standard form with center (h,k) = (1, −2), a = 3 (horizontal semi-axis), b = 2 (vertical semi-axis). Use the trig parametrization from the CED (4.7.B.2): x(t) = 1 + 3 cos t y(t) = −2 + 2 sin t, 0 ≤ t < 2π Check: substitute into the equation: ((x−1)^2)/9 + ((y+2)^2)/4 = ( (3 cos t)^2 / 9 ) + ( (2 sin t)^2 / 4 ) = cos^2 t + sin^2 t = 1, so it satisfies the curve for every t. Notes: t runs 0 to 2π for one full counterclockwise trace; you can switch cos/sin or use negative signs to change start point or direction. This follows the AP parametrization rule for ellipses (Topic 4.7). For more review, see the Topic 4.7 study guide (https://library.fiveable.me/ap-pre-calculus/unit-4/parametrization-implicitly-defined-functions/study-guide/iiEy0E9fETps45yV) and extra practice (https://library.fiveable.me/practice/ap-pre-calculus).