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Chapter 9: Parametric, Polar, and Vector-Valued Functions BC Only

Updated February 2026 · 20 min read · AP Calculus Past Papers

This chapter covers three powerful frameworks for describing curves and motion that go beyond the standard $y = f(x)$ approach. Parametric equations let you describe curves using an independent parameter. Polar coordinates offer a natural way to express curves with rotational symmetry. Vector-valued functions combine both components into a single object ideal for modeling motion. These topics appear exclusively on the AP Calculus BC exam and frequently account for 10–15% of the test.

Table of Contents

  1. Parametric Equations
  2. Arc Length, Speed, and Area for Parametric Curves
  3. Polar Coordinates
  4. Area and Arc Length in Polar Coordinates
  5. Vector-Valued Functions
  6. Cycloid Curves
  7. Practice Problems

9.1 Parametric Equations

In many real-world scenarios, both $x$ and $y$ depend on a third variable, typically time. Instead of writing $y$ as a function of $x$, we describe the curve with a pair of functions that share a common parameter.

Definition: Parametric Equations

A parametric curve in the plane is defined by

$$x = f(t), \qquad y = g(t), \qquad a \le t \le b$$

where $t$ is the parameter (often representing time). As $t$ increases from $a$ to $b$, the point $(x, y)$ traces a path called the parametric curve. The direction of travel is called the orientation of the curve.

Eliminating the Parameter

Sometimes we can eliminate $t$ to obtain a standard Cartesian equation. This is done by solving one parametric equation for $t$ and substituting into the other, or by using algebraic identities (such as $\cos^2 t + \sin^2 t = 1$).

For example, given $x = 2\cos t$ and $y = 2\sin t$, we have $\cos t = x/2$ and $\sin t = y/2$. Using the Pythagorean identity: $(x/2)^2 + (y/2)^2 = 1$, which gives $x^2 + y^2 = 4$, a circle of radius 2.

Derivatives of Parametric Curves

Theorem: First and Second Derivatives for Parametric Curves

If $x = f(t)$ and $y = g(t)$ are differentiable and $dx/dt \neq 0$, then:

First derivative:

$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)}$$

Second derivative:

$$\frac{d^2 y}{dx^2} = \frac{\frac{d}{dt}\!\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$$

That is, you first compute $dy/dx$ as a function of $t$, then differentiate that result with respect to $t$, and finally divide by $dx/dt$.

A common mistake is computing the second derivative as $\frac{d^2y/dt^2}{d^2x/dt^2}$. This formula is incorrect. You must differentiate $dy/dx$ (which is a function of $t$) with respect to $t$, then divide by $dx/dt$.

Example 9.1.1

Problem: A curve is defined by $x = t^2 - 1$ and $y = t^3 - 3t$. Find $dy/dx$ and the equation of the tangent line at $t = 2$.

Solution:

Compute the derivatives with respect to $t$:

$$\frac{dx}{dt} = 2t, \qquad \frac{dy}{dt} = 3t^2 - 3$$

Therefore:

$$\frac{dy}{dx} = \frac{3t^2 - 3}{2t}$$

At $t = 2$: $\dfrac{dy}{dx} = \dfrac{3(4) - 3}{2(2)} = \dfrac{9}{4}$

The point at $t = 2$ is $(x, y) = (2^2 - 1,\; 2^3 - 6) = (3, 2)$.

The tangent line is $y - 2 = \dfrac{9}{4}(x - 3)$, or equivalently $y = \dfrac{9}{4}x - \dfrac{19}{4}$.

Example 9.1.2

Problem: For $x = e^t$ and $y = e^{-2t}$, eliminate the parameter and find $d^2y/dx^2$.

Solution:

Eliminating the parameter: Since $x = e^t$, we have $t = \ln x$. Then $y = e^{-2t} = e^{-2\ln x} = x^{-2} = 1/x^2$.

Second derivative:

$$\frac{dx}{dt} = e^t, \qquad \frac{dy}{dt} = -2e^{-2t}$$ $$\frac{dy}{dx} = \frac{-2e^{-2t}}{e^t} = -2e^{-3t}$$

Now differentiate $dy/dx$ with respect to $t$:

$$\frac{d}{dt}\!\left(\frac{dy}{dx}\right) = \frac{d}{dt}(-2e^{-3t}) = 6e^{-3t}$$

Divide by $dx/dt$:

$$\frac{d^2y}{dx^2} = \frac{6e^{-3t}}{e^t} = 6e^{-4t}$$

Since $e^t = x$, we can write $\dfrac{d^2y}{dx^2} = \dfrac{6}{x^4}$, which is consistent with differentiating $y = x^{-2}$ twice directly.

Example 9.1.3

Problem: Find all points where the curve $x = t^3 - 3t$, $y = t^2 - 4$ has a horizontal or vertical tangent.

Solution:

$$\frac{dx}{dt} = 3t^2 - 3, \qquad \frac{dy}{dt} = 2t$$

Horizontal tangent occurs when $dy/dt = 0$ and $dx/dt \neq 0$: $2t = 0 \Rightarrow t = 0$. At $t = 0$: $dx/dt = -3 \neq 0$, so the point $(0, -4)$ has a horizontal tangent.

Vertical tangent occurs when $dx/dt = 0$ and $dy/dt \neq 0$: $3t^2 - 3 = 0 \Rightarrow t = \pm 1$.

Graphing Parametric Curves: Table of Values

One of the most reliable ways to understand a parametric curve is the table of values method. The procedure is:

  1. Choose representative values of $t$ spread across the domain (multiples of $\pi/4$ or $\pi/6$ work well for trig-based curves).
  2. For each $t$, compute both $x(t)$ and $y(t)$ to obtain a point $(x, y)$.
  3. Plot all the points in order of increasing $t$ and connect them with a smooth curve.
  4. Draw arrows to show the orientation (direction of travel as $t$ increases).

Good choices for $t$ include values where the curve reaches its extremes, crosses the axes, or self-intersects. You can always use technology (Desmos, a graphing calculator) to verify, but building the table by hand gives you a deeper understanding of how $x$ and $y$ change together.

Example 9.1.4 — Graphing a Figure-Eight by Table of Values

Problem: Sketch the parametric curve $x = \cos t$, $y = \sin 2t$ for $0 \le t \le 2\pi$ by building a table of values. Identify the shape and describe the orientation.

Step 1: Build the table.

We evaluate $x(t) = \cos t$ and $y(t) = \sin 2t$ at nine equally-spaced values of $t$:

$t$ $0$ $\dfrac{\pi}{4}$ $\dfrac{\pi}{2}$ $\dfrac{3\pi}{4}$ $\pi$ $\dfrac{5\pi}{4}$ $\dfrac{3\pi}{2}$ $\dfrac{7\pi}{4}$ $2\pi$
$x = \cos t$ $1$ $\tfrac{\sqrt{2}}{2}$ $0$ $-\tfrac{\sqrt{2}}{2}$ $-1$ $-\tfrac{\sqrt{2}}{2}$ $0$ $\tfrac{\sqrt{2}}{2}$ $1$
$y = \sin 2t$ $0$ $1$ $0$ $-1$ $0$ $1$ $0$ $-1$ $0$
$(x,\,y)$ $(1,\,0)$ $(0.71,\,1)$ $(0,\,0)$ $(-0.71,\,-1)$ $(-1,\,0)$ $(-0.71,\,1)$ $(0,\,0)$ $(0.71,\,-1)$ $(1,\,0)$

Step 2: Plot the points and trace the curve.

Plotting these nine points and connecting them with a smooth curve in order of increasing $t$ reveals a figure-eight (self-intersecting curve):

Step 3: Identify the shape. The curve is a figure-eight with two crossing loops. It self-intersects at the origin, which is visited at both $t = \pi/2$ and $t = 3\pi/2$.

Step 4: Verify by eliminating the parameter. Since $x = \cos t$, we have $x^2 = \cos^2 t$, so $\sin^2 t = 1 - x^2$. Also, $y = \sin 2t = 2\sin t\cos t$, giving $y^2 = 4\sin^2 t\cos^2 t = 4(1 - x^2)x^2$. The Cartesian equation is $y^2 = 4x^2(1 - x^2)$, which describes the figure-eight we sketched above.

Figure 9.1 — The figure-eight curve $x = \cos t$, $y = \sin 2t$ from Example 9.1.4. Move the $t_0$ slider to follow the red dot and observe the orientation. The table of values above corresponds to the labeled points.

Parametric curves: (x(t), y(t)) = (a·cos(t), b·sin(t)). Adjust a and b to see how the ellipse shape changes. The red dot traces the curve as t increases.

9.2 Arc Length, Speed, and Area for Parametric Curves

One of the most natural applications of parametric equations is computing the length of a curve and the speed at which a point moves along it.

Theorem: Arc Length of a Parametric Curve

If a curve is defined by $x = f(t)$ and $y = g(t)$ for $a \le t \le b$, and both $f'$ and $g'$ are continuous, then the arc length is:

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

Definition: Speed

The speed of a particle moving along a parametric curve at time $t$ is the magnitude of the velocity vector:

$$\text{speed} = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}$$

Notice that arc length equals the integral of speed over the time interval. This makes intuitive sense: distance = integral of speed.

Example 9.2.1

Problem: Find the arc length of the curve $x = 3\cos t$, $y = 3\sin t$ for $0 \le t \le 2\pi$.

Solution:

$$\frac{dx}{dt} = -3\sin t, \qquad \frac{dy}{dt} = 3\cos t$$ $$\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} = \sqrt{9\sin^2 t + 9\cos^2 t} = \sqrt{9} = 3$$ $$L = \int_0^{2\pi} 3\,dt = 6\pi$$

This confirms the well-known circumference formula $C = 2\pi r$ with $r = 3$.

Example 9.2.2

Problem: A particle moves along the path $x = t^2$, $y = \tfrac{2}{3}t^3$ for $0 \le t \le 3$. Find the total distance traveled and the speed at $t = 2$.

Solution:

$$\frac{dx}{dt} = 2t, \qquad \frac{dy}{dt} = 2t^2$$

Speed at time $t$:

$$\sqrt{(2t)^2 + (2t^2)^2} = \sqrt{4t^2 + 4t^4} = 2t\sqrt{1 + t^2}$$

(We use $2t\sqrt{1 + t^2}$ since $t \ge 0$ on our interval.)

Speed at $t = 2$: $2(2)\sqrt{1 + 4} = 4\sqrt{5}$.

Total distance:

$$L = \int_0^3 2t\sqrt{1 + t^2}\,dt$$

Substitute $u = 1 + t^2$, $du = 2t\,dt$. When $t = 0$, $u = 1$; when $t = 3$, $u = 10$.

$$L = \int_1^{10} \sqrt{u}\,du = \left[\frac{2}{3}u^{3/2}\right]_1^{10} = \frac{2}{3}\left(10\sqrt{10} - 1\right)$$

Area of a Region Bounded by a Parametric Curve

Arc length uses $ds = \sqrt{(dx/dt)^2 + (dy/dt)^2}\,dt$. For area, we convert the standard formula $A = \int y\,dx$ into a $t$-integral using the substitution $dx = x'(t)\,dt$. This is one of the most-tested topics in the BC-only section of the AP exam.

Theorem: Area Under (or Enclosed by) a Parametric Curve

If the curve $x = f(t)$, $y = g(t)$ is traced once as $t$ goes from $\alpha$ to $\beta$, then the area between the curve and the $x$-axis is:

$$A = \int_\alpha^\beta g(t)\,f'(t)\,dt = \int_\alpha^\beta y(t)\,\frac{dx}{dt}\,dt$$

Sign conventions to remember:

Why it works: The area under a Cartesian curve is $A = \int y\,dx$. Substituting $x = f(t)$ changes $dx$ to $f'(t)\,dt$ and changes the limits from $x$-values to $t$-values, giving the formula above.

Example 9.2.3 — Area of an Ellipse

Problem: The standard ellipse is parameterized by $x = a\cos t$, $y = b\sin t$ for $0 \le t \le 2\pi$ (with $a, b > 0$). Use the parametric area formula to find the area enclosed by the ellipse.

Solution:

Compute $dx/dt$:

$$\frac{dx}{dt} = -a\sin t$$

As $t$ increases from $0$ to $2\pi$, the ellipse is traced counterclockwise (the upper half moves from right to left). For a closed counterclockwise curve we use $A = -\oint y\,\frac{dx}{dt}\,dt$:

$$A = -\int_0^{2\pi}(b\sin t)(-a\sin t)\,dt = ab\int_0^{2\pi}\sin^2 t\,dt$$

Apply the half-angle identity $\sin^2 t = \dfrac{1 - \cos 2t}{2}$:

$$A = \frac{ab}{2}\int_0^{2\pi}(1 - \cos 2t)\,dt = \frac{ab}{2}\left[t - \frac{\sin 2t}{2}\right]_0^{2\pi} = \frac{ab}{2}(2\pi - 0) = \pi ab$$

The area of the ellipse is $\boxed{\pi ab}$. When $a = b = r$, this gives $\pi r^2$ — confirming the standard circle area formula. This parametric derivation works for any ellipse without converting to Cartesian form.

Example 9.2.4 — Area of a Self-Intersecting Loop

Problem: The curve $x = t^2 - 1$, $y = t^3 - t$ forms a closed loop. Find the area enclosed by the loop.

Solution:

Step 1: Find the loop interval. The loop closes when the curve passes through the same point twice. Set $f(t_1) = f(t_2)$ and $g(t_1) = g(t_2)$ for $t_1 \neq t_2$: $$t_1^2 - 1 = t_2^2 - 1 \;\Rightarrow\; t_1 = \pm t_2$$ If $t_2 = -t_1$, then $g(t_1) = t_1^3 - t_1$ and $g(-t_1) = -t_1^3 + t_1 = -(t_1^3 - t_1) = -g(t_1)$. These are equal only when $g(t_1) = 0$, i.e., $t_1(t_1^2 - 1) = 0$, giving $t_1 = 0, \pm 1$.

The self-intersection occurs at $t = 1$ and $t = -1$, both giving the point $(0, 0)$. The loop corresponds to $-1 \le t \le 1$.

Step 2: Apply the area formula. Compute $dx/dt = 2t$. As $t$ goes from $-1$ to $1$:

$$A = \left|\int_{-1}^{1} y(t)\,\frac{dx}{dt}\,dt\right| = \left|\int_{-1}^{1}(t^3 - t)(2t)\,dt\right| = \left|2\int_{-1}^{1}(t^4 - t^2)\,dt\right|$$

Since $(t^4 - t^2)$ is an even function:

$$= \left|4\int_0^1 (t^4 - t^2)\,dt\right| = \left|4\left[\frac{t^5}{5} - \frac{t^3}{3}\right]_0^1\right| = \left|4\left(\frac{1}{5} - \frac{1}{3}\right)\right| = \left|4 \cdot \frac{-2}{15}\right| = \frac{8}{15}$$

The area enclosed by the loop is $\boxed{\dfrac{8}{15}}$.

Note: There is no simple Cartesian equation for this curve, so the parametric approach is essential here. The closed loop in a self-intersecting parametric curve is a classic AP BC topic.

Drag the $t_0$ slider to trace the curve. The blue region shows the enclosed loop area $\tfrac{8}{15}$.

Figure 9.2 — Area enclosed by the loop of $x = t^2 - 1,\; y = t^3 - t$. The shaded region corresponds to the integral $\left|\int_{-1}^{1}(t^3-t)(2t)\,dt\right| = \tfrac{8}{15}$.

9.3 Polar Coordinates

Polar coordinates describe a point in the plane by its distance from the origin and the angle it makes with the positive $x$-axis. Many curves that are complicated in Cartesian form become simple and elegant in polar coordinates.

Definition: Polar Coordinates

A point $P$ in the plane is represented as $(r, \theta)$, where $r$ is the directed distance from the origin (pole) and $\theta$ is the angle measured counterclockwise from the positive $x$-axis (polar axis).

Conversion formulas:

$$x = r\cos\theta, \qquad y = r\sin\theta$$ $$r^2 = x^2 + y^2, \qquad \tan\theta = \frac{y}{x}$$

Common Polar Curves

The AP BC exam expects you to recognize several standard families of polar curves:

Slope of Polar Curves

Theorem: Derivative $dy/dx$ in Polar Coordinates

A polar curve $r = f(\theta)$ can be treated as a parametric curve with $\theta$ as the parameter:

$$x = r\cos\theta = f(\theta)\cos\theta, \qquad y = r\sin\theta = f(\theta)\sin\theta$$

Then by the parametric derivative formula:

$$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{f'(\theta)\sin\theta + f(\theta)\cos\theta}{f'(\theta)\cos\theta - f(\theta)\sin\theta}$$

Equivalently, using $r$ and $dr/d\theta$:

$$\frac{dy}{dx} = \frac{\dfrac{dr}{d\theta}\sin\theta + r\cos\theta}{\dfrac{dr}{d\theta}\cos\theta - r\sin\theta}$$

Example 9.3.1

Problem: Convert the polar equation $r = 4\sin\theta$ to Cartesian form, and identify the curve.

Solution:

Multiply both sides by $r$:

$$r^2 = 4r\sin\theta$$

Using $r^2 = x^2 + y^2$ and $r\sin\theta = y$:

$$x^2 + y^2 = 4y$$ $$x^2 + y^2 - 4y = 0$$ $$x^2 + (y - 2)^2 = 4$$

This is a circle centered at $(0, 2)$ with radius 2.

Example 9.3.2

Problem: Find the slope of the cardioid $r = 1 + \cos\theta$ at $\theta = \pi/3$.

Solution:

We have $r = 1 + \cos\theta$ and $dr/d\theta = -\sin\theta$.

$$\frac{dy}{dx} = \frac{(-\sin\theta)\sin\theta + (1 + \cos\theta)\cos\theta}{(-\sin\theta)\cos\theta - (1 + \cos\theta)\sin\theta}$$

At $\theta = \pi/3$: $\sin(\pi/3) = \sqrt{3}/2$, $\cos(\pi/3) = 1/2$, $r = 3/2$.

Numerator:

$$-\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{3}}{2} + \frac{3}{2}\cdot\frac{1}{2} = -\frac{3}{4} + \frac{3}{4} = 0$$

Denominator:

$$-\frac{\sqrt{3}}{2}\cdot\frac{1}{2} - \frac{3}{2}\cdot\frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{4} - \frac{3\sqrt{3}}{4} = -\sqrt{3}$$

Therefore $\dfrac{dy}{dx} = \dfrac{0}{-\sqrt{3}} = 0$. The tangent line is horizontal at $\theta = \pi/3$.

Polar curves: explore rose curves r = a·cos(nθ) and limaçons r = c + d·cos(θ) by adjusting the parameters.

9.4 Area and Arc Length in Polar Coordinates

Computing area in polar coordinates requires a different approach from Cartesian integration. Instead of rectangles, we use thin circular sectors.

Theorem: Area in Polar Coordinates

The area enclosed by the polar curve $r = f(\theta)$ and the rays $\theta = \alpha$ and $\theta = \beta$ (where $\alpha < \beta$) is:

$$A = \frac{1}{2}\int_\alpha^\beta \left[f(\theta)\right]^2\,d\theta = \frac{1}{2}\int_\alpha^\beta r^2\,d\theta$$

Area between two polar curves: If $f(\theta) \ge g(\theta) \ge 0$ on $[\alpha, \beta]$, the area between $r = f(\theta)$ (outer) and $r = g(\theta)$ (inner) is:

$$A = \frac{1}{2}\int_\alpha^\beta \left([f(\theta)]^2 - [g(\theta)]^2\right)\,d\theta$$

Theorem: Arc Length in Polar Coordinates

The arc length of the polar curve $r = f(\theta)$ from $\theta = \alpha$ to $\theta = \beta$ is:

$$L = \int_\alpha^\beta \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta$$

Example 9.4.1 — Area Enclosed by a Cardioid

Problem: Find the area enclosed by the cardioid $r = 2(1 + \cos\theta)$.

Solution:

The cardioid traces once for $0 \le \theta \le 2\pi$.

$$A = \frac{1}{2}\int_0^{2\pi} [2(1 + \cos\theta)]^2\,d\theta = \frac{1}{2}\int_0^{2\pi} 4(1 + \cos\theta)^2\,d\theta$$ $$= 2\int_0^{2\pi} (1 + 2\cos\theta + \cos^2\theta)\,d\theta$$

Use $\cos^2\theta = \dfrac{1 + \cos 2\theta}{2}$:

$$= 2\int_0^{2\pi} \left(1 + 2\cos\theta + \frac{1}{2} + \frac{\cos 2\theta}{2}\right)d\theta = 2\int_0^{2\pi} \left(\frac{3}{2} + 2\cos\theta + \frac{\cos 2\theta}{2}\right)d\theta$$ $$= 2\left[\frac{3}{2}\theta + 2\sin\theta + \frac{\sin 2\theta}{4}\right]_0^{2\pi} = 2\left(3\pi + 0 + 0\right) = 6\pi$$

Example 9.4.2 — Area Between Two Polar Curves

Problem: Find the area of the region inside $r = 3\cos\theta$ and outside $r = 1 + \cos\theta$.

Solution:

First, find where the curves intersect by setting $3\cos\theta = 1 + \cos\theta$:

$$2\cos\theta = 1 \implies \cos\theta = \frac{1}{2} \implies \theta = \pm\frac{\pi}{3}$$

On $[-\pi/3, \pi/3]$, we have $3\cos\theta \ge 1 + \cos\theta$ (outer curve is the circle). By symmetry about the polar axis:

$$A = 2 \cdot \frac{1}{2}\int_0^{\pi/3} \left[(3\cos\theta)^2 - (1 + \cos\theta)^2\right]d\theta$$ $$= \int_0^{\pi/3} \left[9\cos^2\theta - 1 - 2\cos\theta - \cos^2\theta\right]d\theta$$ $$= \int_0^{\pi/3} \left[8\cos^2\theta - 2\cos\theta - 1\right]d\theta$$

Using $\cos^2\theta = \frac{1 + \cos 2\theta}{2}$:

$$= \int_0^{\pi/3} \left[4 + 4\cos 2\theta - 2\cos\theta - 1\right]d\theta = \int_0^{\pi/3} \left[3 + 4\cos 2\theta - 2\cos\theta\right]d\theta$$ $$= \left[3\theta + 2\sin 2\theta - 2\sin\theta\right]_0^{\pi/3} = \pi + 2\cdot\frac{\sqrt{3}}{2} - 2\cdot\frac{\sqrt{3}}{2} = \pi$$

Example 9.4.3 — Area of One Petal of a Rose

Problem: Find the area of one petal of the rose $r = 4\cos(3\theta)$.

Solution:

Since $n = 3$ is odd, the rose has 3 petals. One petal is traced when $r \ge 0$. For the petal along the positive $x$-axis, $\cos(3\theta) \ge 0$ when $-\pi/6 \le \theta \le \pi/6$.

$$A = \frac{1}{2}\int_{-\pi/6}^{\pi/6} [4\cos(3\theta)]^2\,d\theta = \frac{1}{2}\int_{-\pi/6}^{\pi/6} 16\cos^2(3\theta)\,d\theta$$

By symmetry:

$$= 16\int_0^{\pi/6} \frac{1 + \cos(6\theta)}{2}\,d\theta = 8\left[\theta + \frac{\sin(6\theta)}{6}\right]_0^{\pi/6}$$ $$= 8\left(\frac{\pi}{6} + \frac{\sin\pi}{6}\right) = 8 \cdot \frac{\pi}{6} = \frac{4\pi}{3}$$

9.5 Vector-Valued Functions

Vector-valued functions unify parametric equations and provide a natural language for describing motion. Instead of tracking $x$ and $y$ separately, we combine them into a single vector.

Definition: Vector-Valued Function

A vector-valued function in the plane is

$$\vec{r}(t) = \langle x(t),\, y(t) \rangle = x(t)\,\mathbf{i} + y(t)\,\mathbf{j}$$

where $x(t)$ and $y(t)$ are called the component functions.

Position, Velocity, and Acceleration

When $\vec{r}(t)$ represents the position of a particle at time $t$, we define:

Speed and Distance Traveled

The speed of the particle at time $t$ is the magnitude of the velocity vector:

$$|\vec{v}(t)| = \sqrt{[x'(t)]^2 + [y'(t)]^2}$$

The total distance traveled from $t = a$ to $t = b$ is:

$$\text{distance} = \int_a^b |\vec{v}(t)|\,dt = \int_a^b \sqrt{[x'(t)]^2 + [y'(t)]^2}\,dt$$

Note the distinction: displacement is $\vec{r}(b) - \vec{r}(a)$ (a vector), while distance is the scalar integral above. These are equal only when the particle moves in one direction without turning back.

On the AP BC exam, vector-valued function problems frequently involve finding speed at a given time, computing total distance by integrating speed, or determining when a particle is at rest (when $\vec{v}(t) = \langle 0, 0 \rangle$, meaning both components of velocity are simultaneously zero).

Example 9.5.1

Problem: A particle moves in the $xy$-plane with position $\vec{r}(t) = \langle t^2 - 4t,\; 3t - t^2 \rangle$ for $t \ge 0$. Find the velocity and speed at $t = 1$, and determine when the particle is at rest.

Solution:

Velocity:

$$\vec{v}(t) = \langle 2t - 4,\; 3 - 2t \rangle$$

At $t = 1$: $\vec{v}(1) = \langle -2, 1 \rangle$.

Speed at $t = 1$:

$$|\vec{v}(1)| = \sqrt{(-2)^2 + 1^2} = \sqrt{5}$$

At rest: Both components must be zero simultaneously.

$$2t - 4 = 0 \implies t = 2, \qquad 3 - 2t = 0 \implies t = 3/2$$

Since $t = 2 \neq 3/2$, the particle is never at rest. There is no time when both velocity components vanish simultaneously.

Example 9.5.2

Problem: A particle has velocity $\vec{v}(t) = \langle \cos t,\; e^{-t} \rangle$ and initial position $\vec{r}(0) = \langle 0, 2 \rangle$. Find the position function and the total distance traveled from $t = 0$ to $t = \pi$.

Solution:

Position function: Integrate each component and apply initial conditions.

$$x(t) = \int \cos t\,dt = \sin t + C_1$$

$x(0) = 0 \Rightarrow C_1 = 0$, so $x(t) = \sin t$.

$$y(t) = \int e^{-t}\,dt = -e^{-t} + C_2$$

$y(0) = 2 \Rightarrow -1 + C_2 = 2 \Rightarrow C_2 = 3$, so $y(t) = 3 - e^{-t}$.

Thus $\vec{r}(t) = \langle \sin t,\; 3 - e^{-t} \rangle$.

Distance traveled:

$$\text{distance} = \int_0^\pi \sqrt{\cos^2 t + e^{-2t}}\,dt$$

This integral has no elementary closed form, so on the AP exam you would evaluate it numerically using a calculator. Using numerical integration:

$$\text{distance} \approx 2.504$$

9.6 Cycloid Curves BC

The cycloid is one of the most famous curves in the history of mathematics, studied by Galileo, Pascal, Bernoulli, and many others. It provides a beautiful application of parametric equations, combining geometry, calculus, and physics.

Definition: Cycloid

A cycloid is the curve traced by a point on the rim of a circle of radius $r$ as the circle rolls without slipping along a straight line. Its parametric equations are:

$$x = r(t - \sin t), \qquad y = r(1 - \cos t)$$

where $t$ is the parameter representing the angle (in radians) through which the circle has rotated. One complete arch corresponds to $0 \le t \le 2\pi$.

Derivation of the Parametric Equations

Consider a circle of radius $r$ rolling along the $x$-axis. Let $t$ be the angle through which the circle has rotated from its starting position.

Step 1: Locate the center of the circle. As the circle rolls without slipping, the distance traveled along the $x$-axis equals the arc length on the circle. After rotating through angle $t$, the contact point has moved $rt$ units along the $x$-axis. The center of the circle is always directly above the contact point at height $r$. Therefore, the center is at the point $(rt,\, r)$.

Step 2: Locate the point on the rim. The point $P$ on the rim started at the bottom of the circle (at the origin). After the circle has rotated through angle $t$, the point $P$ is displaced from the center by the vector from the center to $P$. Measuring the angle $t$ clockwise from the downward vertical (since the circle rolls to the right), the displacement from the center is:

$$\Delta x = -r\sin t, \qquad \Delta y = -r\cos t$$

Step 3: Combine. Adding the center coordinates and the displacement:

$$x = rt + (-r\sin t) = r(t - \sin t)$$ $$y = r + (-r\cos t) = r(1 - \cos t)$$

This completes the derivation. When $t = 0$, the point is at the origin $(0, 0)$. When $t = \pi$, the point is at the top of the arch: $(r\pi,\, 2r)$. When $t = 2\pi$, the point returns to the line at $(2\pi r,\, 0)$.

Arc Length of One Arch

Theorem: Arc Length of a Cycloid Arch

The arc length of one complete arch of the cycloid $x = r(t - \sin t)$, $y = r(1 - \cos t)$ for $0 \le t \le 2\pi$ is:

$$L = 8r$$

Proof. We use the parametric arc length formula $L = \int_0^{2\pi}\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt$.

Compute the derivatives:

$$\frac{dx}{dt} = r(1 - \cos t), \qquad \frac{dy}{dt} = r\sin t$$

The integrand becomes:

$$\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = r^2(1 - \cos t)^2 + r^2\sin^2 t$$ $$= r^2\left(1 - 2\cos t + \cos^2 t + \sin^2 t\right) = r^2(2 - 2\cos t) = 2r^2(1 - \cos t)$$

Now we apply the half-angle identity $1 - \cos t = 2\sin^2(t/2)$:

$$\sqrt{2r^2(1 - \cos t)} = r\sqrt{2 \cdot 2\sin^2(t/2)} = 2r\,|\sin(t/2)|$$

For $0 \le t \le 2\pi$, we have $0 \le t/2 \le \pi$, so $\sin(t/2) \ge 0$ and $|\sin(t/2)| = \sin(t/2)$. Therefore:

$$L = \int_0^{2\pi} 2r\sin(t/2)\,dt$$

Substitute $u = t/2$, $du = dt/2$:

$$L = 2r\int_0^{\pi} \sin u \cdot 2\,du = 4r\left[-\cos u\right]_0^{\pi} = 4r\left[-\cos\pi + \cos 0\right] = 4r(1 + 1) = 8r$$

The arc length of one arch is exactly $8r$, which is remarkably simple: it is exactly 4 times the diameter of the generating circle. $\blacksquare$

Area Under One Arch

Theorem: Area Under a Cycloid Arch

The area under one complete arch of the cycloid is:

$$A = 3\pi r^2$$

This is exactly 3 times the area of the generating circle.

Proof. Using the parametric area formula $A = \int y\,dx = \int_0^{2\pi} y(t)\,\frac{dx}{dt}\,dt$:

$$A = \int_0^{2\pi} r(1 - \cos t) \cdot r(1 - \cos t)\,dt = r^2\int_0^{2\pi}(1 - \cos t)^2\,dt$$

Expand the integrand:

$$(1 - \cos t)^2 = 1 - 2\cos t + \cos^2 t = 1 - 2\cos t + \frac{1 + \cos 2t}{2} = \frac{3}{2} - 2\cos t + \frac{\cos 2t}{2}$$

Integrate term by term:

$$\int_0^{2\pi}\left(\frac{3}{2} - 2\cos t + \frac{\cos 2t}{2}\right)dt = \left[\frac{3}{2}t - 2\sin t + \frac{\sin 2t}{4}\right]_0^{2\pi}$$ $$= \frac{3}{2}(2\pi) - 0 + 0 = 3\pi$$

Therefore $A = r^2 \cdot 3\pi = 3\pi r^2$. $\blacksquare$

Properties of the Cycloid

Notable Properties of the Cycloid

Graphing the Cycloid: Table of Values

Applying the table method from Section 9.1 to the cycloid is especially instructive because the $x$-values grow with $t$ (the arch moves to the right) while the $y$-values rise and fall like a hill. Let us build the table for $r = 1$.

Example 9.6.0 — Table of Values for One Arch of the Cycloid

Problem: For the cycloid $x = t - \sin t$, $y = 1 - \cos t$ (with $r = 1$), build a table of eight representative values for $0 \le t \le 2\pi$, plot the points, and describe the shape.

Step 1: Evaluate at $t = 0,\, \pi/4,\, \pi/2,\, 3\pi/4,\, \pi,\, 5\pi/4,\, 3\pi/2,\, 7\pi/4,\, 2\pi$.

$t$ $0$ $\dfrac{\pi}{4}$ $\dfrac{\pi}{2}$ $\dfrac{3\pi}{4}$ $\pi$ $\dfrac{5\pi}{4}$ $\dfrac{3\pi}{2}$ $\dfrac{7\pi}{4}$ $2\pi$
$x = t - \sin t$ $0$ $0.08$ $0.57$ $1.65$ $\pi \approx 3.14$ $4.63$ $5.71$ $6.20$ $2\pi \approx 6.28$
$y = 1 - \cos t$ $0$ $0.29$ $1$ $1.71$ $2$ $1.71$ $1$ $0.29$ $0$
$(x,\,y)$ $(0,\,0)$ $(0.08,\,0.29)$ $(0.57,\,1)$ $(1.65,\,1.71)$ $(\pi,\,2)$ $(4.63,\,1.71)$ $(5.71,\,1)$ $(6.20,\,0.29)$ $(2\pi,\,0)$

Sample calculation at $t = \pi/4$: $x = \pi/4 - \sin(\pi/4) = 0.785 - 0.707 = 0.078 \approx 0.08$, and $y = 1 - \cos(\pi/4) = 1 - 0.707 = 0.293 \approx 0.29$. All other entries are computed similarly.

Step 2: Plot and connect. The nine points reveal the characteristic arch shape of the cycloid:

Step 3: Observe asymmetry of $x$ vs.\ $t$. Notice that $x$ grows very slowly near the cusps (when $t \approx 0$ or $t \approx 2\pi$) but grows faster in the middle (near $t = \pi$). This is because $dx/dt = 1 - \cos t$ equals 0 at the cusps and equals 2 at $t = \pi$. The curve is not symmetric about its midpoint in the horizontal direction — a key insight that the table reveals clearly.

Using Desmos to verify: Enter (t - sin(t), 1 - cos(t)) in Desmos with $0 \le t \le 2\pi$ to see the arch. Add a slider for $t_0$ and plot the point (t_0 - sin(t_0), 1 - cos(t_0)) to trace the arch interactively, matching each row of the table above. The interactive graph below does exactly this.

Worked Examples

Example 9.6.1 — Slope of the Cycloid

Problem: Find the slope of the cycloid with $r = 1$ at $t = \pi/3$.

Solution:

With $r = 1$, we have $x = t - \sin t$ and $y = 1 - \cos t$.

$$\frac{dx}{dt} = 1 - \cos t, \qquad \frac{dy}{dt} = \sin t$$ $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\sin t}{1 - \cos t}$$

At $t = \pi/3$: $\sin(\pi/3) = \sqrt{3}/2$ and $\cos(\pi/3) = 1/2$.

$$\frac{dy}{dx}\bigg|_{t=\pi/3} = \frac{\sqrt{3}/2}{1 - 1/2} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$$

The slope at $t = \pi/3$ is $\sqrt{3}$, corresponding to a $60°$ angle with the horizontal.

Alternative using the identity: $\dfrac{\sin t}{1 - \cos t} = \dfrac{2\sin(t/2)\cos(t/2)}{2\sin^2(t/2)} = \cot(t/2)$. At $t = \pi/3$: $\cot(\pi/6) = \sqrt{3}$. This confirms our answer.

Example 9.6.2 — Arc Length of One Arch

Problem: Compute the arc length of one complete arch of the cycloid $x = 3(t - \sin t)$, $y = 3(1 - \cos t)$ for $0 \le t \le 2\pi$.

Solution:

Here $r = 3$. Compute the derivatives:

$$\frac{dx}{dt} = 3(1 - \cos t), \qquad \frac{dy}{dt} = 3\sin t$$

The speed is:

$$\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} = \sqrt{9(1 - \cos t)^2 + 9\sin^2 t} = 3\sqrt{2(1 - \cos t)}$$

Applying $1 - \cos t = 2\sin^2(t/2)$:

$$= 3\sqrt{4\sin^2(t/2)} = 6\sin(t/2) \qquad (\text{since } 0 \le t/2 \le \pi)$$

Integrate:

$$L = \int_0^{2\pi} 6\sin(t/2)\,dt = 6\left[-2\cos(t/2)\right]_0^{2\pi} = -12[\cos\pi - \cos 0] = -12(-1 - 1) = 24$$

This confirms $L = 8r = 8(3) = 24$.

Example 9.6.3 — Area Under One Arch

Problem: Compute the area under one arch of the cycloid $x = 2(t - \sin t)$, $y = 2(1 - \cos t)$ for $0 \le t \le 2\pi$.

Solution:

Here $r = 2$. Using the parametric area formula:

$$A = \int_0^{2\pi} y\,\frac{dx}{dt}\,dt = \int_0^{2\pi} 2(1 - \cos t) \cdot 2(1 - \cos t)\,dt = 4\int_0^{2\pi}(1 - \cos t)^2\,dt$$

Expand:

$$(1 - \cos t)^2 = 1 - 2\cos t + \cos^2 t = \frac{3}{2} - 2\cos t + \frac{\cos 2t}{2}$$

Integrate:

$$4\int_0^{2\pi}\left(\frac{3}{2} - 2\cos t + \frac{\cos 2t}{2}\right)dt = 4\left[\frac{3}{2}t - 2\sin t + \frac{\sin 2t}{4}\right]_0^{2\pi} = 4(3\pi) = 12\pi$$

This confirms $A = 3\pi r^2 = 3\pi(4) = 12\pi$.

Cycloid Practice Problems

Problem: Horizontal and Vertical Tangent Lines

For the cycloid $x = r(t - \sin t)$, $y = r(1 - \cos t)$, find all values of $t$ in $[0, 2\pi]$ where the tangent line is horizontal, and determine the behavior at the cusps where $t = 0$ and $t = 2\pi$.

Show Solution

We have $\dfrac{dx}{dt} = r(1 - \cos t)$ and $\dfrac{dy}{dt} = r\sin t$.

Horizontal tangent occurs when $dy/dt = 0$ and $dx/dt \neq 0$.

$r\sin t = 0 \Rightarrow t = 0, \pi, 2\pi$.

At $t = \pi$: $dx/dt = r(1 - \cos\pi) = r(1 - (-1)) = 2r \neq 0$. So the tangent is horizontal at $t = \pi$, which corresponds to the point $(r\pi,\, 2r)$, the top of the arch.

Behavior at cusps ($t = 0$ and $t = 2\pi$): At these points, both $dx/dt = 0$ and $dy/dt = 0$ simultaneously, so $dy/dx$ takes the indeterminate form $0/0$. Using L'Hopital's Rule or the identity $\dfrac{dy}{dx} = \cot(t/2)$:

As $t \to 0^+$: $\cot(t/2) \to +\infty$, so the tangent line is vertical.

As $t \to 2\pi^-$: $\cot(t/2) = \cot(\pi^-) \to -\infty$, so the tangent line is also vertical.

Therefore, the cycloid has a horizontal tangent only at $t = \pi$ and vertical tangent lines at the cusps $t = 0$ and $t = 2\pi$.

Problem: Arc Length of Half an Arch

Compute the arc length of the cycloid $x = 5(t - \sin t)$, $y = 5(1 - \cos t)$ for $0 \le t \le \pi$ (the first half of the arch).

Show Solution

With $r = 5$:

$$\frac{dx}{dt} = 5(1 - \cos t), \qquad \frac{dy}{dt} = 5\sin t$$

The speed is:

$$\sqrt{25(1 - \cos t)^2 + 25\sin^2 t} = 5\sqrt{2(1 - \cos t)} = 5\sqrt{4\sin^2(t/2)} = 10\sin(t/2)$$

(valid since $\sin(t/2) \ge 0$ for $0 \le t \le \pi$).

Integrate:

$$L = \int_0^{\pi} 10\sin(t/2)\,dt = 10\left[-2\cos(t/2)\right]_0^{\pi} = -20\left[\cos(\pi/2) - \cos(0)\right]$$ $$= -20(0 - 1) = 20$$

The arc length of the first half of the arch is $20$. Note that this is exactly half of $8r = 40$, which makes sense by the symmetry of the cycloid about $t = \pi$.

Problem: Area with Specific Radius

A wheel of radius $r = 4$ rolls along a flat road. Find the area between the cycloid traced by a point on the rim and the road for one complete revolution of the wheel.

Show Solution

The cycloid has parametric equations $x = 4(t - \sin t)$, $y = 4(1 - \cos t)$ with $0 \le t \le 2\pi$.

Using the area formula:

$$A = \int_0^{2\pi} y\,\frac{dx}{dt}\,dt = \int_0^{2\pi} 4(1 - \cos t) \cdot 4(1 - \cos t)\,dt = 16\int_0^{2\pi}(1 - \cos t)^2\,dt$$

We already know $\int_0^{2\pi}(1 - \cos t)^2\,dt = 3\pi$, so:

$$A = 16 \cdot 3\pi = 48\pi$$

Alternatively, by the formula $A = 3\pi r^2 = 3\pi(16) = 48\pi$.

For comparison, the area of the wheel itself is $\pi r^2 = 16\pi$, so the area under the arch is exactly 3 times the area of the generating circle.

Cycloid curve traced by a point on a rolling circle. Adjust the radius $r$ and watch the generating circle roll. The blue curve is the cycloid, and the red dot shows the tracing point.

9.7 Practice Problems

Test your understanding of the concepts from this chapter. Each problem is followed by a detailed solution that you can reveal after attempting it on your own.

Problem 1

A curve is defined parametrically by $x = t + \sin t$ and $y = 1 - \cos t$ for $0 \le t \le 2\pi$. Find $dy/dx$ at $t = \pi/2$ and determine where the curve has horizontal tangent lines.

Show Solution

$\dfrac{dx}{dt} = 1 + \cos t$ and $\dfrac{dy}{dt} = \sin t$.

$\dfrac{dy}{dx} = \dfrac{\sin t}{1 + \cos t}$.

At $t = \pi/2$: $\dfrac{dy}{dx} = \dfrac{1}{1 + 0} = 1$.

Horizontal tangent when $\sin t = 0$ and $1 + \cos t \neq 0$: $t = 0$ or $t = 2\pi$ (at both, $1 + \cos t = 2 \neq 0$). At $t = \pi$, $\sin\pi = 0$ but $1 + \cos\pi = 0$, so $t = \pi$ is not a horizontal tangent (it is a cusp). Thus the horizontal tangent occurs at $(0, 0)$ and $(2\pi, 0)$ (the same point).

Problem 2

Find $\dfrac{d^2y}{dx^2}$ for the parametric curve $x = t - t^2$ and $y = t^3$ at $t = 1$.

Show Solution

$\dfrac{dx}{dt} = 1 - 2t$, $\dfrac{dy}{dt} = 3t^2$.

$\dfrac{dy}{dx} = \dfrac{3t^2}{1 - 2t}$.

Differentiate with respect to $t$ using the quotient rule:

$\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right) = \dfrac{6t(1 - 2t) - 3t^2(-2)}{(1 - 2t)^2} = \dfrac{6t - 12t^2 + 6t^2}{(1 - 2t)^2} = \dfrac{6t - 6t^2}{(1 - 2t)^2}$

$\dfrac{d^2y}{dx^2} = \dfrac{6t - 6t^2}{(1 - 2t)^2} \cdot \dfrac{1}{1 - 2t} = \dfrac{6t(1 - t)}{(1 - 2t)^3}$

At $t = 1$: $\dfrac{d^2y}{dx^2} = \dfrac{6(1)(0)}{(1 - 2)^3} = \dfrac{0}{-1} = 0$.

Problem 3

Find the arc length of the curve $x = e^t\cos t$, $y = e^t\sin t$ for $0 \le t \le \pi$.

Show Solution

$\dfrac{dx}{dt} = e^t\cos t - e^t\sin t = e^t(\cos t - \sin t)$

$\dfrac{dy}{dt} = e^t\sin t + e^t\cos t = e^t(\sin t + \cos t)$

$\left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac{dy}{dt}\right)^2 = e^{2t}(\cos t - \sin t)^2 + e^{2t}(\sin t + \cos t)^2$

$= e^{2t}[\cos^2 t - 2\sin t\cos t + \sin^2 t + \sin^2 t + 2\sin t\cos t + \cos^2 t]$

$= e^{2t}[1 + 1] = 2e^{2t}$

$L = \int_0^\pi \sqrt{2e^{2t}}\,dt = \sqrt{2}\int_0^\pi e^t\,dt = \sqrt{2}(e^\pi - 1)$

Problem 4

Convert $r = 2\sec\theta$ to Cartesian form and identify the curve.

Show Solution

$r = \dfrac{2}{\cos\theta}$, so $r\cos\theta = 2$. Since $x = r\cos\theta$, we get $x = 2$. This is a vertical line.

Problem 5

Find the total area enclosed by the rose $r = 2\sin(2\theta)$.

Show Solution

Since $n = 2$ is even, the rose has $2n = 4$ petals. One petal (in the first quadrant) is traced for $0 \le \theta \le \pi/2$.

Area of one petal:

$A_1 = \dfrac{1}{2}\int_0^{\pi/2} [2\sin(2\theta)]^2\,d\theta = \dfrac{1}{2}\int_0^{\pi/2} 4\sin^2(2\theta)\,d\theta = 2\int_0^{\pi/2}\dfrac{1 - \cos(4\theta)}{2}\,d\theta$

$= \int_0^{\pi/2}[1 - \cos(4\theta)]\,d\theta = \left[\theta - \dfrac{\sin(4\theta)}{4}\right]_0^{\pi/2} = \dfrac{\pi}{2}$

Total area $= 4 \cdot \dfrac{\pi}{2} = 2\pi$.

Problem 6

Find the area of the region inside both $r = \sin\theta$ and $r = \cos\theta$.

Show Solution

The curves intersect when $\sin\theta = \cos\theta$, i.e., $\theta = \pi/4$. For $0 \le \theta \le \pi/4$, $\sin\theta \le \cos\theta$, so $r = \sin\theta$ is the inner curve. For $\pi/4 \le \theta \le \pi/2$, $\cos\theta \le \sin\theta$, so $r = \cos\theta$ is the inner curve.

By symmetry about $\theta = \pi/4$:

$A = 2 \cdot \dfrac{1}{2}\int_0^{\pi/4}\sin^2\theta\,d\theta = \int_0^{\pi/4}\dfrac{1 - \cos(2\theta)}{2}\,d\theta$

$= \dfrac{1}{2}\left[\theta - \dfrac{\sin(2\theta)}{2}\right]_0^{\pi/4} = \dfrac{1}{2}\left(\dfrac{\pi}{4} - \dfrac{1}{2}\right) = \dfrac{\pi}{8} - \dfrac{1}{4}$

Problem 7

A particle moves with velocity $\vec{v}(t) = \langle 2t,\; 3t^2 - 1 \rangle$ and has position $\vec{r}(0) = \langle 1, -2 \rangle$. Find the position at $t = 2$ and the speed at $t = 1$.

Show Solution

Position: Integrate the velocity.

$x(t) = \int 2t\,dt = t^2 + C_1$. $x(0) = 1 \Rightarrow C_1 = 1$. So $x(t) = t^2 + 1$.

$y(t) = \int (3t^2 - 1)\,dt = t^3 - t + C_2$. $y(0) = -2 \Rightarrow C_2 = -2$. So $y(t) = t^3 - t - 2$.

At $t = 2$: $\vec{r}(2) = \langle 5, 4 \rangle$.

Speed at $t = 1$:

$\vec{v}(1) = \langle 2, 2 \rangle$, so $|\vec{v}(1)| = \sqrt{4 + 4} = 2\sqrt{2}$.

Problem 8

Find the total distance traveled by a particle with position $\vec{r}(t) = \langle 4\cos t,\; 3\sin t \rangle$ over $0 \le t \le 2\pi$. (This is the perimeter of an ellipse—set up the integral.)

Show Solution

$\vec{v}(t) = \langle -4\sin t, 3\cos t \rangle$.

$|\vec{v}(t)| = \sqrt{16\sin^2 t + 9\cos^2 t} = \sqrt{9 + 7\sin^2 t}$

$\text{distance} = \int_0^{2\pi}\sqrt{9 + 7\sin^2 t}\,dt$

This is an elliptic integral and cannot be evaluated in closed form. Using a calculator: $\text{distance} \approx 22.103$.

Problem 9

For the polar curve $r = 2 + \cos\theta$, find the slope $dy/dx$ at $\theta = \pi/2$.

Show Solution

$r = 2 + \cos\theta$, $dr/d\theta = -\sin\theta$.

$\dfrac{dy}{dx} = \dfrac{(-\sin\theta)\sin\theta + (2 + \cos\theta)\cos\theta}{(-\sin\theta)\cos\theta - (2 + \cos\theta)\sin\theta}$

At $\theta = \pi/2$: $\sin(\pi/2) = 1$, $\cos(\pi/2) = 0$, $r = 2$.

Numerator: $(-1)(1) + (2)(0) = -1$.

Denominator: $(-1)(0) - (2)(1) = -2$.

$\dfrac{dy}{dx} = \dfrac{-1}{-2} = \dfrac{1}{2}$.

Problem 10

A particle has acceleration $\vec{a}(t) = \langle 6t,\; -2 \rangle$, initial velocity $\vec{v}(0) = \langle 1, 4 \rangle$, and initial position $\vec{r}(0) = \langle 0, 0 \rangle$. Find the position at $t = 3$.

Show Solution

Velocity: Integrate acceleration.

$\vec{v}(t) = \langle 3t^2 + C_1,\; -2t + C_2 \rangle$.

$\vec{v}(0) = \langle 1, 4 \rangle \Rightarrow C_1 = 1,\; C_2 = 4$.

$\vec{v}(t) = \langle 3t^2 + 1,\; -2t + 4 \rangle$.

Position: Integrate velocity.

$\vec{r}(t) = \langle t^3 + t + D_1,\; -t^2 + 4t + D_2 \rangle$.

$\vec{r}(0) = \langle 0, 0 \rangle \Rightarrow D_1 = 0,\; D_2 = 0$.

$\vec{r}(t) = \langle t^3 + t,\; -t^2 + 4t \rangle$.

At $t = 3$: $\vec{r}(3) = \langle 27 + 3,\; -9 + 12 \rangle = \langle 30, 3 \rangle$.

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