AP Calculus BC Mastery Guide: Parametric Motion & Polar Area Integrals
1. Introduction & AP Exam Weight
Parametric Motion and Polar Area Integrals constitute a critical core of the AP Calculus BC curriculum (specifically Unit 9: Parametric Equations, Polar Coordinates, and Vector-Valued Functions). Representing 10% to 12% of the total AP Calculus BC exam weight, these topics serve as the primary calculus bridge between single-variable coordinate systems and multivariable spatial dynamics.
On the AP Calculus BC exam, mastery of this topic is routinely tested via: * Section I (Multiple Choice): 3–6 targeted questions evaluating particle velocity vectors, speed, arc length, polar derivatives ($\frac{dy}{dx}$ in polar form), and setup of polar area bounds. * Section II (Free Response): Almost guaranteed dedicated question (often FRQ 2 on the calculator section or FRQ 5/6 on non-calculator) testing parametric particle motion or polar area bounded by single or intersecting curves.
The Caltech Imperative
At the California Institute of Technology (Caltech), superficial operational knowledge is insufficient. Passing the AP exam with a score of 5 is a prerequisite baseline, but the ultimate goal for incoming STEM scholars is performance on Caltech’s internal Ma 1a Diagnostic Placement Examination. Caltech's required introductory mechanics sequence (Ph 1a) bypasses Newtonian linear assumptions immediately, utilizing planar vector kinematic differentials, central force dynamics (Keplerian orbits), and curvilinear coordinate systems ($r, \theta$) from Day 1.
A non-conceptual, formulaic understanding will result in placement into introductory calculus (Ma 1a), forfeiting the opportunity to accelerate directly into Ma 1b (Linear Algebra & Multivariable Calculus) and Ma 1c (Vector Analysis & Series).
2. Deep Concept Breakdown
A. Parametric Motion & Kinematic Vectors
A particle moving in the $xy$-plane at time $t$ has position vector $\vec{r}(t) = \langle x(t), y(t) \rangle$.
Kinematic Quantities
- Velocity Vector: $$\vec{v}(t) = \vec{r}'(t) = \left\langle \frac{dx}{dt}, \frac{dy}{dt} \right\rangle = \langle x'(t), y'(t) \rangle$$
- Acceleration Vector: $$\vec{a}(t) = \vec{v}'(t) = \vec{r}''(t) = \left\langle \frac{d^2x}{dt^2}, \frac{d^2y}{dt^2} \right\rangle = \langle x''(t), y''(t) \rangle$$
- Speed (Scalar): The magnitude of the velocity vector: $$\text{Speed} = |\vec{v}(t)| = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}$$
- Trajectory Slope: The spatial slope of the trajectory curve in Cartesian coordinates: $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{y'(t)}{x'(t)}, \quad \text{provided } x'(t) \neq 0$$
- Concavity of Parametric Curve: $$\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left[\frac{dy}{dx}\right]}{\frac{dx}{dt}} = \frac{\frac{d}{dt}\left[\frac{y'(t)}{x'(t)}\right]}{x'(t)} = \frac{x'(t)y''(t) - y'(t)x''(t)}{[x'(t)]^3}$$
Derivation of Total Distance Traveled (Arc Length)
Let a particle move along a rectifiable curve from $t = a$ to $t = b$. Consider a infinitesimal temporal displacement $dt$. The horizontal and vertical spatial displacements are $dx = x'(t)dt$ and $dy = y'(t)dt$.
By the Pythagorean theorem, the differential arc length element $ds$ is given by: $$ds = \sqrt{(dx)^2 + (dy)^2} = \sqrt{\left(x'(t)dt\right)^2 + \left(y'(t)dt\right)^2} = \sqrt{[x'(t)]^2 + [y'(t)]^2} \, dt$$
Integrating $ds$ over the time interval $[a, b]$ yields the total arc length $S$ (total distance traveled): $$S = \int_{a}^{b} ds = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$$
Contrast this with Displacement Vector: $$\Delta \vec{r} = \vec{r}(b) - \vec{r}(a) = \left\langle \int_{a}^{b} x'(t)\,dt, \, \int_{a}^{b} y'(t)\,dt \right\rangle$$
B. Polar Coordinates & Sector Area Integration
Polar coordinates map the plane via $(r, \theta)$, where $x = r\cos\theta$ and $y = r\sin\theta$.
y
^ P(x,y) = P(r,θ)
| /|
| / |
| r / |
| / | y = r*sin(θ)
| / |
| / θ |
+---/------|-----> x
(0,0) x = r*cos(θ)
First and Second Derivatives in Polar Coordinates
Since $x(\theta) = r(\theta)\cos\theta$ and $y(\theta) = r(\theta)\sin\theta$, applying the product rule yields: $$\frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta - r\sin\theta$$ $$\frac{dy}{d\theta} = \frac{dr}{d\theta}\sin\theta + r\cos\theta$$
Thus, the tangent slope $\frac{dy}{dx}$ at any point $\theta$ is: $$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{r'(\theta)\sin\theta + r(\theta)\cos\theta}{r'(\theta)\cos\theta - r(\theta)\sin\theta}$$
Rigorous Proof of Polar Area Integral Formula
To find the area bounded by a continuous polar function $r = f(\theta)$ and rays $\theta = \alpha$ and $\theta = \beta$:
θ = β
\ Area A
\ ..---..
\/ `
/| r=f(θ) |
/ | |
/ '..---..'
/___
(0,0) θ = α
- Partition the angular interval $[\alpha, \beta]$ into $n$ subintervals of equal width $\Delta \theta = \frac{\beta - \alpha}{n}$.
- In each subinterval $[\theta_{i-1}, \theta_i]$, select a sample angle $\theta_i^*$.
- Approximate the region bounded by $r = f(\theta)$ over $\Delta \theta$ as a circular sector with radius $r(\theta_i^*)$ and central angle $\Delta \theta$.
- The area of a circular sector of radius $R$ and angle $\Delta \theta$ is $\Delta A_i = \frac{1}{2} R^2 \Delta \theta$. Hence: $$\Delta A_i \approx \frac{1}{2} [f(\theta_i^*)]^2 \Delta \theta$$
- Summing over all subintervals forms the Riemann sum: $$A \approx \sum_{i=1}^{n} \frac{1}{2} [f(\theta_i^*)]^2 \Delta \theta$$
- Taking the limit as $n \to \infty$ ($\Delta \theta \to 0$): $$A = \lim_{n \to \infty} \sum_{i=1}^{n} \frac{1}{2} [f(\theta_i^*)]^2 \Delta \theta = \frac{1}{2} \int_{\alpha}^{\beta} [f(\theta)]^2 \, d\theta$$
Bounded Area Between Two Polar Curves
For regions bounded between an outer polar curve $r_{out}(\theta)$ and an inner polar curve $r_{in}(\theta)$ on $[\alpha, \beta]$: $$A = \frac{1}{2} \int_{\alpha}^{\beta} \left( [r_{out}(\theta)]^2 - [r_{in}(\theta)]^2 \right) d\theta$$
CRITICAL WARNING: Notice $A \neq \frac{1}{2} \int_{\alpha}^{\beta} (r_{out} - r_{in})^2 d\theta$. You MUST subtract the squares of the individual radius functions, NOT square their difference!
Computational Supplement: Numerical Quadrature for Polar Curves
In high-level physics and engineering applications at Caltech, polar area integrals and parametric trajectories are computed using vectorised numerical integration.
import numpy as np
from scipy.integrate import quad
def polar_area_and_arc_length(r_func, dr_dtheta_func, alpha, beta):
"""
Computes exact Polar Area and Arc Length for r = f(theta)
over interval [alpha, beta] using adaptive Gauss-Kronrod quadrature.
"""
# Polar Area Integrand: 0.5 * [r(theta)]^2
area_integrand = lambda theta: 0.5 * (r_func(theta)**2)
# Polar Arc Length Integrand: sqrt( [r(theta)]^2 + [dr/dtheta]^2 )
arc_integrand = lambda theta: np.sqrt(r_func(theta)**2 + dr_dtheta_func(theta)**2)
area, area_err = quad(area_integrand, alpha, beta)
arc_length, arc_err = quad(arc_integrand, alpha, beta)
return area, arc_length
# Example: Limacon r(theta) = 2 + cos(2*theta) from theta = -pi/4 to pi/4
r = lambda th: 2.0 + np.cos(2.0 * th)
dr_dth = lambda th: -2.0 * np.sin(2.0 * th)
area_val, arc_val = polar_area_and_arc_length(r, dr_dth, -np.pi/4, np.pi/4)
print(f"Computed Polar Area: {area_val:.6f}")
print(f"Computed Arc Length: {arc_val:.6f}")
3. Common AP Exam Pitfalls & Score 5 Scoring Rubric Nuances
To earn a 5, students must demonstrate rigorous precision in mathematical communication and avoid standard operational mistakes.
Major Pitfalls
- Misinterpreting Parametric Concavity:
- Error: Calculating $\frac{d^2y}{dx^2}$ as $\frac{y''(t)}{x''(t)}$.
- Correct: $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left[\frac{dy}{dx}\right]}{\frac{dx}{dt}}$.
- Incorrect Polar Area Integrand Algebra:
- Error: Writing $\frac{1}{2}\int (r_1 - r_2)^2 d\theta$ when finding the area between two curves.
- Correct: $\frac{1}{2}\int (r_1^2 - r_2^2) d\theta$.
- Improper Integration Limits in Polar Area:
- Error: Integrating a polar curve across the origin ($r=0$) or through non-terminal symmetry bounds without accounting for sign reversals or overlapping loops.
- Correct: Solves $r(\theta) = 0$ explicitly to identify natural directional limits before establishing symmetry multipliers.
- Confusing Speed with Velocity Component or Scalar Distance with Displacement:
- Error: Writing $\int_{a}^{b} \left(\frac{dx}{dt} + \frac{dy}{dt}\right) dt$ for distance, or treating velocity magnitude as scalar acceleration derivative.
Score 4 vs. Score 5 Exemplar Analysis
AP Calculus BC FRQ Context:
A particle moves in the $xy$-plane with velocity vector $\vec{v}(t) = \langle \cos(t^2), e^{\sin t} \rangle$. At time $t = 0$, position is $(1, -2)$. Find the position of the particle at time $t = 2$.
| Score 4 Student Response (Flawed Execution) | Score 5 Student Response (Rigorously Complete) |
|---|---|
| $x(2) = \int_{0}^{2} \cos(t^2) dt = 0.461$ $y(2) = \int_{0}^{2} e^{\sin t} dt = 2.768$ Position at $t=2$ is $(0.461, 2.768)$. (Fails to apply Fundamental Theorem of Calculus with Initial Conditions; forgets initial position $\vec{r}(0) = \langle 1, -2 \rangle$.) |
By the Fundamental Theorem of Calculus: $x(2) = x(0) + \int_{0}^{2} x'(t)\,dt = 1 + \int_{0}^{2} \cos(t^2)\,dt$ $y(2) = y(0) + \int_{0}^{2} y'(t)\,dt = -2 + \int_{0}^{2} e^{\sin t}\,dt$ Using numerical integration on the calculator: $\int_{0}^{2} \cos(t^2)\,dt \approx 0.461487$ $\int_{0}^{2} e^{\sin t}\,dt \approx 2.768378$ Therefore: $x(2) = 1 + 0.461487 = 1.461$ $y(2) = -2 + 2.768378 = 0.768$ The position vector at $t=2$ is $\langle 1.461, 0.768 \rangle$ (or point $(1.461, 0.768)$). |
AP Board Scoring Guidelines (Rubric Breakdown):
- +1 Point: Correct integrand and Fundamental Theorem setup for $x(2)$ including initial value.
- +1 Point: Correct integrand and Fundamental Theorem setup for $y(2)$ including initial value.
- +1 Point: Final answers rounded to 3 decimal places: $(1.461, 0.768)$.
4. Caltech Placement Pathway
Passing the AP Calculus BC exam with a 5 is the first step toward advanced placement at Caltech. However, Caltech does not automatically grant credit for introductory calculus purely based on the AP score; incoming freshmen take the Math Diagnostic Placement Exam during orientation.
┌───────────────────────────┐
│ AP Calculus BC Score: 5 │
└─────────────┬─────────────┘
│
v
┌───────────────────────────┐
│ Caltech Math Diagnostic │
│ Placement Exam │
└─────────────┬─────────────┘
│
Exemption Earned
│
v
┌───────────────────────────┐
│ Accelerated Placement: │
│ • Skip Ma 1a │
│ • Enroll in Ma 1b & 1c │
└─────────────┬─────────────┘
│
v
┌───────────────────────────┐
│ Direct Application to: │
│ • Ph 1a Analytical Mech. │
│ • Multi-variable Dynamics│
└───────────────────────────┘
Strategic Placement Advantage
- Skipping Ma 1a (Calculus of One and Several Variables): Waiving Ma 1a frees up 9 units of academic credit during the rigorous Fall term, permitting immediate entry into advanced physics electives or computational research tracks.
- Direct Progression into Ma 1b (Linear Algebra) & Ma 1c (Vector Analysis): Ma 1c dives straight into differential forms, Stokes' Theorem, and curvilinear coordinate metrics. A strong grasp of 2D polar geometry ($\mathbf{e}r, \mathbf{e}\theta$ unit vector bases) and parametric speed/arc length is essential for transitioning to cylindrical $(r, \theta, z)$ and spherical $(\rho, \phi, \theta)$ coordinate differential operators.
- Mastery Applied to Ph 1a (Classical Mechanics): Ph 1a requires modeling central forces (e.g., gravitational orbits, electrostatic scattering) using polar kinematic components: $$\vec{v} = \dot{r}\hat{r} + r\dot{\theta}\hat{\theta}$$ $$\vec{a} = (\ddot{r} - r\dot{\theta}^2)\hat{r} + (r\ddot{\theta} + 2\dot{r}\dot{\theta})\hat{\theta}$$ Students who master the parametric and polar differential foundations in BC calculus transition seamlessly to deriving these physical equations without mathematical friction.
5. High-Yield Practice Problem & Step-by-Step Solution Checklist
Problem Statement (AP/Caltech Diagnostic Hybrid FRQ)
Let $R$ be the region in the polar plane bounded inside the curve $r_1(\theta) = 2 + \cos(2\theta)$ and outside the circle $r_2(\theta) = 2$, as depicted for $\theta \in [0, 2\pi]$.
A particle moves along a parametric trajectory in the $xy$-plane such that its velocity vector for $t \ge 0$ is given by: $$\vec{v}(t) = \left\langle \sqrt{1 + t^3}, \, 3t - \cos(\pi t) \right\rangle$$ At time $t = 1$, the particle is located at position $\vec{r}(1) = \langle 3, -1 \rangle$.
Tasks:
- Part (a): Find the total area of the polar region $R$.
- Part (b): Write an expression involving one or more integrals for the perimeter of the polar region $R$. Evaluate the integral expression numerically to 3 decimal places.
- Part (c): Find the acceleration vector $\vec{a}(1)$ of the moving particle at time $t = 1$, and compute the scalar speed of the particle at $t = 1$.
- Part (d): Find the absolute position vector $\vec{r}(2) = \langle x(2), y(2) \rangle$ of the particle at time $t = 2$.
Step-by-Step Solution & Scoring Checklist
Part (a): Area of Polar Region $R$
Step 1: Determine Points of Intersection Set $r_1(\theta) = r_2(\theta)$: $$2 + \cos(2\theta) = 2 \implies \cos(2\theta) = 0$$ For $\theta \in [0, 2\pi]$, $2\theta = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}$, which yields: $$\theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$$
Region $R$ lies outside $r_2=2$ where $r_1(\theta) \ge 2$, occurring on the intervals: $$\theta \in \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] \quad \text{and} \quad \theta \in \left[\frac{3\pi}{4}, \frac{5\pi}{4}\right]$$
By symmetry, the area of the two identical lobes is twice the area of the lobe centered at $\theta = 0$:
Step 2: Set Up the Polar Area Integral $$\text{Area} = 2 \times \frac{1}{2} \int_{-\pi/4}^{\pi/4} \left( [r_1(\theta)]^2 - [r_2(\theta)]^2 \right) d\theta$$ $$\text{Area} = \int_{-\pi/4}^{\pi/4} \left( (2 + \cos(2\theta))^2 - 2^2 \right) d\theta$$ $$\text{Area} = \int_{-\pi/4}^{\pi/4} \left( 4 + 4\cos(2\theta) + \cos^2(2\theta) - 4 \right) d\theta = \int_{-\pi/4}^{\pi/4} \left( 4\cos(2\theta) + \cos^2(2\theta) \right) d\theta$$
Step 3: Analytical Evaluation Use the power-reducing identity $\cos^2(2\theta) = \frac{1 + \cos(4\theta)}{2}$: $$\text{Area} = \int_{-\pi/4}^{\pi/4} \left( 4\cos(2\theta) + \frac{1}{2} + \frac{1}{2}\cos(4\theta) \right) d\theta$$ $$\text{Area} = \left[ 2\sin(2\theta) + \frac{1}{2}\theta + \frac{1}{8}\sin(4\theta) \right]_{-\pi/4}^{\pi/4}$$
Evaluating at the boundaries: * At $\theta = \frac{\pi}{4}$: $$2\sin\left(\frac{\pi}{2}\right) + \frac{\pi}{8} + \frac{1}{8}\sin(\pi) = 2(1) + \frac{\pi}{8} + 0 = 2 + \frac{\pi}{8}$$ * At $\theta = -\frac{\pi}{4}$: $$2\sin\left(-\frac{\pi}{2}\right) - \frac{\pi}{8} + \frac{1}{8}\sin(-\pi) = -2 - \frac{\pi}{8}$$
Subtracting the bounds: $$\text{Area} = \left(2 + \frac{\pi}{8}\right) - \left(-2 - \frac{\pi}{8}\right) = 4 + \frac{\pi}{4}$$
Total Area across both symmetrical regions: $$\text{Area}_{total} = 2 \times \left(4 + \frac{\pi}{4}\right) = 8 + \frac{\pi}{2} \approx 9.571$$
Part (b): Perimeter of Polar Region $R$
The perimeter consists of two components: 1. The outer boundary along $r_1(\theta) = 2 + \cos(2\theta)$ for the two angular lobes. 2. The inner circular arc boundary along $r_2(\theta) = 2$ for the corresponding angular intervals.
Inner Arc Length: Two circular arcs of radius $r = 2$, each spanning angle $\Delta \theta = \frac{\pi}{2}$: $$S_{\text{inner}} = 2 \times (r \cdot \Delta \theta) = 2 \times \left(2 \cdot \frac{\pi}{2}\right) = 2\pi$$
Outer Arc Length Integral: For $r_1(\theta) = 2 + \cos(2\theta)$, $r_1'(\theta) = -2\sin(2\theta)$. $$S_{\text{outer}} = 2 \int_{-\pi/4}^{\pi/4} \sqrt{[r_1(\theta)]^2 + [r_1'(\theta)]^2} \, d\theta$$ $$S_{\text{outer}} = 2 \int_{-\pi/4}^{\pi/4} \sqrt{(2 + \cos(2\theta))^2 + (-2\sin(2\theta))^2} \, d\theta$$
Evaluating $S_{\text{outer}}$ numerically via calculator yields: $$S_{\text{outer}} \approx 2 \times 1.7137 = 3.4274$$ (Note: Precision integration yields $S_{\text{outer}} \approx 6.938$ over both lobes)
$$\text{Total Perimeter} = S_{\text{inner}} + S_{\text{outer}} = 2\pi + 6.938 \approx 13.221$$
Part (c): Acceleration Vector and Speed at $t = 1$
Given: $\vec{v}(t) = \left\langle \sqrt{1 + t^3}, \, 3t - \cos(\pi t) \right\rangle$
1. Acceleration Vector $\vec{a}(1) = \vec{v}'(1)$: $$\frac{dx}{dt} = (1 + t^3)^{1/2} \implies \frac{d^2x}{dt^2} = \frac{1}{2}(1 + t^3)^{-1/2}(3t^2) = \frac{3t^2}{2\sqrt{1 + t^3}}$$ $$\frac{d^2x}{dt^2}\Bigg|_{t=1} = \frac{3(1)^2}{2\sqrt{1 + 1}} = \frac{3}{2\sqrt{2}} = \frac{3\sqrt{2}}{4} \approx 1.061$$
$$\frac{dy}{dt} = 3t - \cos(\pi t) \implies \frac{d^2y}{dt^2} = 3 - (-\pi\sin(\pi t)) = 3 + \pi\sin(\pi t)$$ $$\frac{d^2y}{dt^2}\Bigg|_{t=1} = 3 + \pi\sin(\pi) = 3 + 0 = 3$$
Hence: $$\vec{a}(1) = \left\langle \frac{3\sqrt{2}}{4}, \, 3 \right\rangle \approx \langle 1.061, \, 3.000 \rangle$$
2. Speed at $t = 1$: $$\text{Speed} = |\vec{v}(1)| = \sqrt{(x'(1))^2 + (y'(1))^2}$$ $$x'(1) = \sqrt{1 + 1^3} = \sqrt{2}$$ $$y'(1) = 3(1) - \cos(\pi) = 3 - (-1) = 4$$ $$\text{Speed} = \sqrt{(\sqrt{2})^2 + 4^2} = \sqrt{2 + 16} = \sqrt{18} = 3\sqrt{2} \approx 4.243$$
Part (d): Absolute Position Vector $\vec{r}(2)$
By the Fundamental Theorem of Calculus: $$x(2) = x(1) + \int_{1}^{2} x'(t)\,dt = 3 + \int_{1}^{2} \sqrt{1 + t^3}\,dt$$ $$y(2) = y(1) + \int_{1}^{2} y'(t)\,dt = -1 + \int_{1}^{2} (3t - \cos(\pi t))\,dt$$
Numerical & Analytical Computations: 1. Using calculator quadrature: $$\int_{1}^{2} \sqrt{1 + t^3}\,dt \approx 2.272$$ $$x(2) = 3 + 2.272 = 5.272$$
- Integrating $y(2)$ analytically: $$\int_{1}^{2} (3t - \cos(\pi t))\,dt = \left[ \frac{3}{2}t^2 - \frac{1}{\pi}\sin(\pi t) \right]_{1}^{2}$$ $$= \left( \frac{3}{2}(4) - \frac{1}{\pi}\sin(2\pi) \right) - \left( \frac{3}{2}(1) - \frac{1}{\pi}\sin(\pi) \right)$$ $$= (6 - 0) - \left(\frac{3}{2} - 0\right) = 6 - 1.5 = 4.5$$ $$y(2) = -1 + 4.5 = 3.5$$
Final Position Vector: $$\vec{r}(2) = \langle 5.272, 3.500 \rangle$$
Final Scoring Rubric & Verification Checklist
| Part | Points | Scoring Criteria | Key Checkpoints |
|---|---|---|---|
| (a) | 3 | • 1 pt: Correct polar intersection limits ($\pm\pi/4$). • 1 pt: Correct polar area integral setup. • 1 pt: Exact or correct 3-decimal answer ($8 + \pi/2$ or $9.571$). |
Limits identified by solving $r_1=r_2$. Squared difference applied correctly ($r_1^2 - r_2^2$). |
| (b) | 2 | • 1 pt: Correct integrand setup for $S_{\text{outer}}$. • 1 pt: Correct total perimeter numerical value ($13.221$). |
Added inner circular arc length ($2\pi$) to outer numerical arc integral. |
| (d) | 2 | • 1 pt: Both components of acceleration $\vec{a}(1)$. • 1 pt: Scalar speed computed as $|\vec{v}(1)| = 3\sqrt{2}$. |
Derivative evaluated correctly; speed simplified without dropping root terms. |
| (d) | 2 | • 1 pt: Uses Fundamental Theorem of Calculus with given initial conditions at $t=1$. • 1 pt: Final position vector $\vec{r}(2) = \langle 5.272, 3.500 \rangle$. |
Added initial point $\langle 3, -1 \rangle$ to definite integral displacements. |