Unit 9: Parametric Equations, Polar Coordinates, and Vector-Valued Functions

Understanding Parametric Equations and Converting Between Parametric and Cartesian Forms

Definition of Parametric Equations

• Parametric equations describe a curve by expressing both \(x\) and \(y\) as functions of a third variable, usually \(t\) (called the parameter). This allows more flexibility in describing motion or paths that cannot be represented as a single function \(y=f(x)\). For example, a circle can be written as \(x = r\cos t\), \(y = r\sin t\), which avoids the problem of failing the vertical line test in Cartesian form.
• The parameter \(t\) often represents time, but it can represent any quantity that drives both \(x\) and \(y\) simultaneously. This means we can track the position of an object in the plane at any moment without requiring one coordinate to be the independent variable. Because of this, parametric equations are especially useful in physics, engineering, and motion problems.
• By having both coordinates dependent on \(t\), you can describe curves that loop, intersect themselves, or have vertical tangents without breaking the function into multiple pieces. For example, projectile motion under gravity naturally produces parametric equations for horizontal and vertical positions over time. This capability is one reason AP Calculus BC requires fluency in both interpreting and converting these equations.
• Parametric equations also allow you to represent direction along a curve, which is essential in velocity and acceleration problems. The curve itself is the set of points \((x(t), y(t))\), but the parameter \(t\) tells you the sequence in which those points are traced. Without parametric form, direction information is often lost when switching to Cartesian form.
• In summary, parametric form is a more general representation than Cartesian form, making it a fundamental skill to both interpret and manipulate these equations. Mastery of this topic allows you to tackle complex motion and curve problems efficiently. You must understand the relationship between \(t\), \(x(t)\), and \(y(t)\) to apply calculus correctly.

Converting from Parametric to Cartesian Form

• To convert from parametric form to Cartesian form, your goal is to eliminate the parameter \(t\) to obtain a direct relationship between \(x\) and \(y\). The first step is to solve one of the parametric equations for \(t\) in terms of \(x\) or \(y\). For example, if \(x = 3t + 2\), then \(t = (x - 2)/3\).
• Once you have \(t\) expressed in terms of \(x\) or \(y\), substitute it into the other parametric equation. For instance, with \(y = t^2 - 1\) and \(t = (x - 2)/3\), we substitute to get \(y = ((x - 2)/3)^2 - 1\). This gives you a single equation in \(x\) and \(y\) that represents the same curve without the parameter.
• If the parameter is inside a trigonometric function, you may need to use Pythagorean identities to eliminate \(t\). For example, if \(x = r\cos t\) and \(y = r\sin t\), use \(\sin^2 t + \cos^2 t = 1\) to get \(x^2 + y^2 = r^2\). Recognizing when to apply these identities is a critical skill for AP questions.
• When converting, be aware that the Cartesian equation may not preserve direction or restricted intervals of \(t\). Always check the domain and range of \(t\) to understand which portion of the curve is actually represented. Ignoring parameter restrictions can lead to incorrect graph interpretations.
• After conversion, verify the resulting Cartesian equation by plugging in a few \(t\)-values from the original parametric equations to ensure the same \((x,y)\) pairs are produced. This step confirms your conversion is correct and helps avoid sign or algebra errors.

Converting from Cartesian to Parametric Form

• To convert from Cartesian form to parametric form, you choose a parameter \(t\) and express both \(x\) and \(y\) in terms of \(t\) so that eliminating \(t\) recovers the original Cartesian equation. There is often more than one valid way to do this, which makes this process flexible. Choosing a convenient \(x(t)\) can simplify \(y(t)\) significantly.
• A common approach is to let \(x = t\) and then express \(y\) in terms of \(t\) using the Cartesian equation. For example, if \(y = x^2 + 1\), setting \(x = t\) gives \(y = t^2 + 1\). This is the simplest method but may not be best if the problem context suggests a different parameter choice.
• In cases involving circular or elliptical curves, using trigonometric parameterizations makes the process more natural. For instance, \(x^2 + y^2 = r^2\) can be parameterized as \(x = r\cos t\), \(y = r\sin t\), where \(t\) runs from 0 to \(2\pi\). This choice directly reflects the geometry of the curve and is common in physics and engineering.
• For motion problems, parametric form often directly represents horizontal and vertical components of velocity. For example, if horizontal velocity is constant \(v_x\) and vertical velocity changes with time, you can write \(x(t) = v_x t + x_0\) and \(y(t) = v_y t - \frac12 g t^2 + y_0\). This form is more physically meaningful than Cartesian form alone.
• Once you choose a parameterization, verify it by eliminating \(t\) to see if you return to the original Cartesian equation. This double-check ensures your parametric equations are valid and equivalent to the given Cartesian relationship.

Finding Derivatives of Parametric Functions and Vector-Valued Functions

Derivative of a Parametric Function

• For a curve defined by \( x = f(t) \) and \( y = g(t) \), the slope of the tangent line is given by \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \), provided \( \frac{dx}{dt} \neq 0 \). This formula comes from the chain rule, where \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \). Always compute both derivatives with respect to \( t \) before dividing to avoid algebra mistakes.
• When \( \frac{dx}{dt} = 0 \), the tangent line is vertical, and the slope is undefined. In this case, \( t \) values where \( \frac{dx}{dt} = 0 \) should be solved separately to find vertical tangents. Always check if \( \frac{dy}{dt} = 0 \) at the same \( t \) to determine if a cusp or stationary point exists.
• For second derivatives, use \( \frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}} \). This requires differentiating \( \frac{dy}{dx} \) with respect to \( t \) and then dividing by \( \frac{dx}{dt} \). This is essential for concavity analysis in parametric motion problems.
• Units remain consistent: \( t \) often represents time, \( x(t) \) is horizontal position, and \( y(t) \) is vertical position. Differentiating with respect to \( t \) produces velocities in each direction, and dividing them gives the slope of motion at a given moment.
• Always check for domain restrictions on \( t \) before differentiating, as undefined expressions for \( dx/dt \) or \( dy/dt \) can create points of discontinuity. Graphing the parametric curve can help confirm where slopes match the derivative calculations.

Derivatives of Vector-Valued Functions

• A vector-valued function is written as \( \vec{r}(t) = \langle x(t), y(t), z(t) \rangle \), where each component is a function of \( t \). The derivative is \( \vec{r}'(t) = \langle x'(t), y'(t), z'(t) \rangle \), representing the velocity vector of the particle at time \( t \).
• The magnitude of \( \vec{r}'(t) \) gives the speed: \( \text{Speed} = \sqrt{[x'(t)]^2 + [y'(t)]^2 + [z'(t)]^2} \). This formula comes from the Pythagorean theorem applied to the velocity components.
• To find the slope of the path in the \( xy \)-plane, use \( \frac{dy}{dx} = \frac{y'(t)}{x'(t)} \) exactly as with regular parametric functions. The \( z \)-component is ignored unless slope in a 3D plane is specifically required.
• Acceleration is the derivative of velocity: \( \vec{a}(t) = \vec{r}''(t) = \langle x''(t), y''(t), z''(t) \rangle \). This vector indicates both the rate of change of speed and the change in direction of motion.
• Unit tangent vectors can be found by \( \vec{T}(t) = \frac{\vec{r}'(t)}{\|\vec{r}'(t)\|} \). These are useful in curvature problems and motion along a path, ensuring direction is preserved while magnitude is normalized to 1.

Calculating the Accumulation of Change in Length Over an Interval Using a Definite Integral

Arc Length for Parametric Functions

• The arc length of a curve given by \( x = f(t) \) and \( y = g(t) \) from \( t = a \) to \( t = b \) is calculated using the formula: \[ L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \] This formula comes from summing up infinitely small straight-line segments along the curve, where each segment length is given by the Pythagorean Theorem. It is crucial that you compute \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) first before plugging them into the square root.
• To apply this formula, first differentiate \( x(t) \) and \( y(t) \) with respect to \( t \) separately. Then square each derivative, add them together, and simplify before taking the square root. Simplifying before integrating reduces mistakes and makes integration easier. You must ensure your derivatives are correct, because even a sign error can make the arc length incorrect.
• After simplifying under the square root, set up the definite integral from the given \( t \)-interval. If the expression inside the square root can be simplified using factoring or trigonometric identities, do so before integrating. Many AP problems intentionally include expressions that simplify to avoid extremely difficult integrals. If no simplification is possible, be prepared to use numerical methods or a calculator in the "calculator-allowed" section of the exam.
• When evaluating the integral, use exact values when possible to avoid rounding errors. If using a calculator, input the entire square root expression in one go rather than rounding intermediate values. Always check that the interval limits match the given range of \( t \), not \( x \) or \( y \), unless the problem specifically asks for arc length over an \( x \)-interval, in which case a different formula is needed.
• If the function is given in Cartesian form \( y = f(x) \), the arc length formula changes to: \[ L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \] In this case, you differentiate \( y \) with respect to \( x \), square it, add 1, then integrate over the \( x \)-interval. The same simplification rules apply here, and trig substitutions often help for integrals involving square roots.

Arc Length for Polar Functions

• For curves given in polar form \( r = f(\theta) \), the arc length from \( \theta = \alpha \) to \( \theta = \beta \) is: \[ L = \int_\alpha^\beta \sqrt{\left( \frac{dr}{d\theta} \right)^2 + r^2} \, d\theta \] This comes from converting polar coordinates into parametric form and applying the parametric arc length formula. You must compute \(\frac{dr}{d\theta}\) accurately to avoid errors.
• Differentiate \( r(\theta) \) carefully, then square it. Add \( r^2 \) directly to \(\left( \frac{dr}{d\theta} \right)^2\). Ensure parentheses are correct when squaring, especially if \( r(\theta) \) has more than one term. Simplifying this step before integrating is often the difference between a solvable and unsolvable problem.
• Set up the definite integral over the given \( \theta \)-interval. Watch out for cases where the curve is symmetric—sometimes you can integrate over half or a quarter of the curve and multiply by 2 or 4, which saves time and avoids unnecessary work. However, only use symmetry if you are absolutely certain of the curve’s symmetry properties.
• As with other forms, simplify the integrand when possible using trig identities like \( \sin^2\theta + \cos^2\theta = 1 \) or double-angle formulas. If no simplification is possible, numerical integration with a calculator is often necessary, especially for non-polynomial or oscillating functions.
• Always double-check whether the problem specifies "total distance traveled" for a moving particle or "length of a curve." These are calculated the same way in arc length problems, but total distance implies using absolute values if the speed function could be negative, whereas curve length is inherently non-negative due to the square root.

Determining the Position of a Particle Moving in a Plane

• When a particle moves in a plane, its position is given by a **position vector** \( \vec{r}(t) = \langle x(t), y(t) \rangle \), where \( x(t) \) and \( y(t) \) describe horizontal and vertical positions at time \( t \). To determine the position at a given time, you plug that value of \( t \) into both \( x(t) \) and \( y(t) \) separately. If initial position information is given as coordinates, ensure your \( x(t) \) and \( y(t) \) functions account for that initial displacement before solving.
• If you are given **velocity functions** instead of position functions, you must integrate each component separately to get position: \[ x(t) = x(t_0) + \int_{t_0}^t v_x(u) \, du, \quad y(t) = y(t_0) + \int_{t_0}^t v_y(u) \, du \] where \( x(t_0) \) and \( y(t_0) \) are initial coordinates. Always include the constant from integration by using the initial position conditions provided.
• For problems involving displacement between two times \( t_1 \) and \( t_2 \), compute the change in position vector: \[ \Delta \vec{r} = \langle x(t_2) - x(t_1), \ y(t_2) - y(t_1) \rangle \] This gives the exact horizontal and vertical changes. If they ask for total distance traveled, you will need to compute the arc length using velocity magnitude (explained later), not just the straight-line displacement.

Calculating Velocity, Speed, and Acceleration of a Particle Moving Along a Curve

• The **velocity vector** is the derivative of the position vector: \[ \vec{v}(t) = \frac{d\vec{r}}{dt} = \langle x'(t), y'(t) \rangle \] You differentiate each component separately. The velocity vector points in the direction of motion and its components represent instantaneous rates of change in \( x \) and \( y \) positions. Always pay attention to units — velocity is measured in units of length per unit of time.
• The **speed** of the particle is the magnitude of its velocity vector: \[ \text{Speed} = |\vec{v}(t)| = \sqrt{(x'(t))^2 + (y'(t))^2} \] This gives the instantaneous scalar rate of motion. It is always non-negative and ignores direction. For motion problems, you often integrate speed over a time interval to get total distance traveled.
• The **acceleration vector** is the derivative of velocity: \[ \vec{a}(t) = \frac{d\vec{v}}{dt} = \langle x''(t), y''(t) \rangle \] It shows how velocity is changing in each direction. A positive acceleration component increases speed in that direction, while a negative component decreases it. You can also find **tangential** and **normal** components of acceleration in advanced problems by projecting acceleration onto the velocity and normal direction vectors.

Vector-Valued Functions: Notation, Operations, and Applications

• A **vector-valued function** maps time \( t \) to a vector in the plane (or 3D space). In AP Calculus BC, it is usually written as: \[ \vec{r}(t) = \langle f(t), g(t) \rangle \quad \text{or} \quad \vec{r}(t) = f(t)\mathbf{i} + g(t)\mathbf{j} \] where \( f(t) \) and \( g(t) \) are continuous differentiable functions. The brackets \( \langle \ \rangle \) indicate vector components in Cartesian form.
• Common vector operations include:
  • **Addition**: \( \langle a_1, b_1 \rangle + \langle a_2, b_2 \rangle = \langle a_1+a_2, b_1+b_2 \rangle \)
  • **Scalar multiplication**: \( k\langle a, b \rangle = \langle ka, kb \rangle \)
  • **Magnitude**: \( |\langle a, b \rangle| = \sqrt{a^2 + b^2} \)
  You should be comfortable computing these because they are needed in velocity, speed, and acceleration problems.
• Applications include **modeling motion** (position, velocity, acceleration), **describing forces** in physics, and **calculating displacement/distance traveled**. In many problems, you start with parametric definitions \( x(t), y(t) \) and use calculus on each component separately — differentiation, integration, and magnitude calculations are all done component-wise.

Finding Derivatives of Functions Written in Polar Coordinates

• In polar coordinates, a point is given as \( (r, \theta) \), where \( r \) is the radius and \( \theta \) is the angle in radians. To find the derivative \( \frac{dy}{dx} \), first convert the polar equations into parametric form: \( x = r(\theta) \cos\theta \) and \( y = r(\theta) \sin\theta \). This step ensures you can differentiate \( x \) and \( y \) with respect to \( \theta \) before forming \( \frac{dy/d\theta}{dx/d\theta} \).
• Compute \( \frac{dx}{d\theta} \) by differentiating \( x = r\cos\theta \) using the product rule: \( \frac{dx}{d\theta} = \frac{dr}{d\theta} \cos\theta - r\sin\theta \). Similarly, compute \( \frac{dy}{d\theta} = \frac{dr}{d\theta} \sin\theta + r\cos\theta \). Be meticulous with product rule signs because a mistake here will make the slope calculation wrong.
• Once you have \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \), divide them to get \( \frac{dy}{dx} \). The slope formula becomes: \[ \frac{dy}{dx} = \frac{\frac{dr}{d\theta} \sin\theta + r\cos\theta}{\frac{dr}{d\theta} \cos\theta - r\sin\theta} \] This gives the instantaneous slope of the tangent to the polar curve at a specific \( \theta \).
• If you need horizontal or vertical tangent lines, set the numerator or denominator of \( \frac{dy}{dx} \) to zero, respectively. For a horizontal tangent, solve \( \frac{dy}{d\theta} = 0 \) and ensure \( \frac{dx}{d\theta} \neq 0 \). For a vertical tangent, solve \( \frac{dx}{d\theta} = 0 \) with \( \frac{dy}{d\theta} \neq 0 \).
• Always interpret your slope in the context of the polar graph. This means once you find \( \theta \) values of special slopes, substitute back into \( r(\theta) \) to find the coordinates in polar form, then optionally convert to Cartesian if needed. This ensures your answers are both numerically correct and geometrically meaningful.

Finding the Area of Regions Bounded by Polar Curves

• The formula for the area enclosed by a polar curve \( r = f(\theta) \) from \( \theta = a \) to \( \theta = b \) is: \[ A = \frac{1}{2} \int_a^b \left[ r(\theta) \right]^2 d\theta \] This formula comes from summing up infinitely thin sectors of circles with radius \( r(\theta) \) over the given interval.
• When finding the area between two polar curves \( r_1(\theta) \) and \( r_2(\theta) \), integrate the difference of their squared radii: \[ A = \frac{1}{2} \int_a^b \left[ r_{\text{outer}}^2 - r_{\text{inner}}^2 \right] d\theta \] Always identify which curve is outer and which is inner over each interval.
• Set the bounds \( a \) and \( b \) carefully based on intersection points of the curves. To find intersections, set \( r_1(\theta) = r_2(\theta) \) and solve for \( \theta \). This step is crucial because incorrect limits will give extra or missing area.
• If the curve is symmetric (about the polar axis, the vertical line through the pole, or the origin), you can integrate over one symmetric section and multiply by the number of repetitions. This saves time and reduces calculation complexity. Always verify symmetry by checking how \( r(\theta) \) behaves under transformations like \( \theta \to -\theta \) or \( \theta \to \pi - \theta \).
• When evaluating the integral, ensure that \( r(\theta) \) is squared before integrating. Forgetting to square will completely change the area because the formula derives from the sector area formula \( \frac{1}{2} r^2 \Delta\theta \). Double-check units — the result should be in square units.

Applications of Polar Equations in Modeling Curves

• Polar equations model natural spirals, petals, and radial growth patterns more effectively than Cartesian equations. For example, \( r = a\theta \) models an Archimedean spiral, while \( r = a\cos(n\theta) \) can model flower petals. Recognizing the type of polar equation lets you predict the shape before graphing.
• Rose curves \( r = a\cos(n\theta) \) or \( r = a\sin(n\theta) \) produce \( n \) petals if \( n \) is odd and \( 2n \) petals if \( n \) is even. This makes them useful in physics and engineering to describe periodic radial patterns. Always note whether the curve starts along the polar axis or at a rotated angle.
• Limaçons, given by \( r = a \pm b\cos\theta \) or \( r = a \pm b\sin\theta \), can model closed loops, cardioid shapes, or dimpled curves. The relationship between \( a \) and \( b \) determines whether there is an inner loop. These are important in acoustics, optics, and antenna theory for directional patterns.
• Spiral curves like \( r = ae^{b\theta} \) can model growth patterns in nature, such as seashells or galaxies. In these cases, the exponential change in \( r \) with \( \theta \) captures scaling properties. Engineers also use spirals for mechanical cams and gear teeth profiles.
• In applications, polar equations often simplify integration or differentiation compared to Cartesian forms, especially for rotationally symmetric problems. For instance, computing the area swept by a rotating arm is more direct in polar form. Always assess whether polar coordinates reduce the complexity of a problem before defaulting to Cartesian.

Practice Problem 1: Derivatives in Polar Coordinates

Problem: A curve is given by the polar equation \( r = 2\theta \) for \( 0 \leq \theta \leq \pi \). Find \( \frac{dy}{dx} \) at \( \theta = \frac{\pi}{4} \).

Solution:

• Step 1: Recall that in polar coordinates, \( x = r\cos\theta \) and \( y = r\sin\theta \). For \( r = 2\theta \), we have \( x = 2\theta \cos\theta \) and \( y = 2\theta \sin\theta \).
• Step 2: Differentiate \( x(\theta) \) and \( y(\theta) \) with respect to \( \theta \): \[ \frac{dx}{d\theta} = 2\cos\theta - 2\theta\sin\theta, \quad \frac{dy}{d\theta} = 2\sin\theta + 2\theta\cos\theta \]
• Step 3: Use the chain rule for parametric derivatives: \[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \]
• Step 4: At \( \theta = \frac{\pi}{4} \): \[ \frac{dx}{d\theta} = 2\cos\frac{\pi}{4} - 2\left(\frac{\pi}{4}\right)\sin\frac{\pi}{4} = \sqrt{2} - \frac{\pi\sqrt{2}}{4} \] \[ \frac{dy}{d\theta} = 2\sin\frac{\pi}{4} + 2\left(\frac{\pi}{4}\right)\cos\frac{\pi}{4} = \sqrt{2} + \frac{\pi\sqrt{2}}{4} \]
• Step 5: Therefore: \[ \frac{dy}{dx} = \frac{\sqrt{2} + \frac{\pi\sqrt{2}}{4}}{\sqrt{2} - \frac{\pi\sqrt{2}}{4}} = \frac{1 + \frac{\pi}{4}}{1 - \frac{\pi}{4}} \]

Practice Problem 2: Area Enclosed by a Polar Curve

Problem: Find the area enclosed by one loop of the polar curve \( r = \sin(2\theta) \).

Solution:

• Step 1: The formula for area in polar coordinates is: \[ A = \frac{1}{2} \int_{\theta_1}^{\theta_2} r^2 \, d\theta \] For one loop of \( r = \sin(2\theta) \), the bounds are \( \theta = 0 \) to \( \theta = \frac{\pi}{2} \).
• Step 2: Substitute \( r = \sin(2\theta) \) into the formula: \[ A = \frac{1}{2} \int_{0}^{\pi/2} \sin^2(2\theta) \, d\theta \]
• Step 3: Use the identity \( \sin^2x = \frac{1 - \cos(2x)}{2} \): \[ A = \frac{1}{2} \int_{0}^{\pi/2} \frac{1 - \cos(4\theta)}{2} \, d\theta \] \[ A = \frac{1}{4} \int_{0}^{\pi/2} \left[ 1 - \cos(4\theta) \right] \, d\theta \]
• Step 4: Integrate: \[ A = \frac{1}{4} \left[ \theta - \frac{\sin(4\theta)}{4} \right]_{0}^{\pi/2} \] Since \( \sin(2\pi) = \sin(0) = 0 \), this simplifies to: \[ A = \frac{1}{4} \left[ \frac{\pi}{2} - 0 \right] = \frac{\pi}{8} \]
• Step 5: The area enclosed by one loop is: \[ \boxed{\frac{\pi}{8}} \]