AP Calculus BC Flashcards: Arc Length Of Smooth Planar Curve

What this deck covers

This deck focuses on Arc Length Of Smooth Planar Curve, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus BC.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: Calculate the total distance traveled by $x(t) = t, y(t) = t^2$ from $t = 0$ to $t = 3$.

Answer: $D = 5.196$ (approximately). Integrates $\sqrt{1 + 4t^2}$ from 0 to 3 to get the total path length.

Flashcard 2: Compute the radius of curvature for $y = x^3$ at $x = 0$.

Answer: $R = \infty$. At $x = 0$: $f''(0) = 0$, so curvature is 0 and radius is infinite.

Flashcard 3: What is the relationship between curvature and radius of curvature?

Answer: $R = \frac{1}{\kappa}$. Curvature and radius of curvature are reciprocals of each other.

Flashcard 4: How do you express arc length in polar coordinates from $\theta = a$ to $\theta = b$?

Answer: $L = \int_{a}^{b} \sqrt{(\frac{dr}{d\theta})^2 + r^2} \, d\theta$. Combines radial and tangential components: $\frac{dr}{d\theta}$ and $r$ respectively.

Flashcard 5: What is the curvature of the parametric curve $x(t) = t, y(t) = t^2$?

Answer: $\kappa = \frac{2}{(1 + 4t^2)^{3/2}}$. Applies the parametric curvature formula with $x'(t) = 1, y'(t) = 2t, y''(t) = 2$.

Flashcard 6: Find the total distance traveled by a particle with velocity $v(t) = 3t$ from $t = 0$ to $t = 2$.

Answer: $D = 6$. Distance equals $\int_0^2 3t \, dt = \frac{3t^2}{2}|_0^2 = 6$.

Flashcard 7: Identify the expression for the arc length of a curve $x = g(y)$ from $y = c$ to $y = d$.

Answer: $L = \int_{c}^{d} \sqrt{1 + (g'(y))^2} \, dy$. Similar to $y = f(x)$ formula but with $x$ as a function of $y$.

Flashcard 8: Determine the arc length of $r = 1 + \sin(\theta)$ from $\theta = 0$ to $\theta = \pi$.

Answer: $L = 5.333$ (approximately). The cardioid $r = 1 + \sin(\theta)$ has a complex arc length integral.

Flashcard 9: Find the radius of curvature of $y = \sin(x)$ at $x = 0$.

Answer: $R = 1$. At $x = 0$: $f'(0) = 1, f''(0) = -1$, giving $\kappa = 1$.

Flashcard 10: What is the curvature of a circle with radius $r$?

Answer: $\kappa = \frac{1}{r}$. For a circle, curvature is the reciprocal of the radius.

Flashcard 11: How do you express arc length in polar coordinates from $\theta = a$ to $\theta = b$?

Answer: $L = \int_{a}^{b} \sqrt{(\frac{dr}{d\theta})^2 + r^2} \, d\theta$. Combines radial and tangential components: $\frac{dr}{d\theta}$ and $r$ respectively.

Flashcard 12: What is the differential arc length $ds$ in polar coordinates $(r, \theta)$?

Answer: $ds = \sqrt{(dr)^2 + (r \, d\theta)^2}$. In polar coordinates, $r \, d\theta$ represents the tangential component of arc length.

Flashcard 13: What is the curvature of the parametric curve $x(t) = t, y(t) = t^2$?

Answer: $\kappa = \frac{2}{(1 + 4t^2)^{3/2}}$. Applies the parametric curvature formula with $x'(t) = 1, y'(t) = 2t, y''(t) = 2$.

Flashcard 14: Determine the arc length of $x(t) = t^2, y(t) = t^3$ from $t = 0$ to $t = 2$.

Answer: $L = \frac{1}{3} (17\sqrt{17} - 1)$. Uses parametric formula with $x'(t) = 2t, y'(t) = 3t^2$, then integrates.

Flashcard 15: Determine the arc length of $x(t) = t^2, y(t) = t^3$ from $t = 0$ to $t = 2$.

Answer: $L = \frac{1}{3} (17\sqrt{17} - 1)$. Uses parametric formula with $x'(t) = 2t, y'(t) = 3t^2$, then integrates.

Flashcard 16: Find the arc length for $r = 2\cos(\theta)$ from $\theta = 0$ to $\theta = \frac{\pi}{2}$.

Answer: $L = 2$. For $r = 2\cos(\theta)$, this represents a semicircle with diameter 2.

Flashcard 17: What is the formula for the speed of a particle in terms of its parametric derivatives?

Answer: $v = \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2}$. Speed is the magnitude of the velocity vector in parametric form.

Flashcard 18: Find the total distance traveled by a particle with velocity $v(t) = 3t$ from $t = 0$ to $t = 2$.

Answer: $D = 6$. Distance equals $\int_0^2 3t \, dt = \frac{3t^2}{2}|_0^2 = 6$.

Flashcard 19: Find the arc length of $y = 3x$ from $x = 0$ to $x = 4$.

Answer: $L = 4 \sqrt{10}$. For $y = 3x$, $f'(x) = 3$, so $\sqrt{1 + 9} = \sqrt{10}$ over length 4.

Flashcard 20: What is the formula for the total distance traveled by a particle with position $s(t)$?

Answer: $D = \int_{a}^{b} |v(t)| \, dt$. Integrates the absolute value of velocity to account for direction changes.

Flashcard 21: Compute the radius of curvature for $y = x^3$ at $x = 0$.

Answer: $R = \infty$. At $x = 0$: $f''(0) = 0$, so curvature is 0 and radius is infinite.

Flashcard 22: Find the arc length for $r = 2\cos(\theta)$ from $\theta = 0$ to $\theta = \frac{\pi}{2}$.

Answer: $L = 2$. For $r = 2\cos(\theta)$, this represents a semicircle with diameter 2.

Flashcard 23: Identify the expression for the arc length of a curve $x = g(y)$ from $y = c$ to $y = d$.

Answer: $L = \int_{c}^{d} \sqrt{1 + (g'(y))^2} \, dy$. Similar to $y = f(x)$ formula but with $x$ as a function of $y$.

Flashcard 24: Calculate the arc length of $y = x^2$ from $x = 0$ to $x = 1$.

Answer: $L = \frac{\sqrt{5} + \ln(1 + \sqrt{5})}{2}$. Uses $f'(x) = 2x$ in the arc length formula and evaluates the resulting integral.

Flashcard 25: What is the formula for the speed of a particle in terms of its parametric derivatives?

Answer: $v = \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2}$. Speed is the magnitude of the velocity vector in parametric form.

Flashcard 26: Calculate the arc length of $y = x^2$ from $x = 0$ to $x = 1$.

Answer: $L = \frac{\sqrt{5} + \ln(1 + \sqrt{5})}{2}$. Uses $f'(x) = 2x$ in the arc length formula and evaluates the resulting integral.

Flashcard 27: What is the arc length differential $ds$ in terms of $dx$ and $dy$?

Answer: $ds = \sqrt{(dx)^2 + (dy)^2}$. Represents the infinitesimal arc length element using the Pythagorean theorem.

Flashcard 28: Find the total distance traveled by $x(t) = 2t, y(t) = 3t$ from $t = 0$ to $t = 2$.

Answer: $D = 4\sqrt{13}$. Linear motion with constant speed $\sqrt{4 + 9} = \sqrt{13}$ over time interval 2.

Flashcard 29: State the formula for the arc length of a curve $y = f(x)$ from $x = a$ to $x = b$.

Answer: $L = \int_{a}^{b} \sqrt{1 + (f'(x))^2} \, dx$. Uses Pythagorean theorem on small segments with $1 + (f'(x))^2$ under the square root.

Flashcard 30: What is the expression for the curvature $\kappa$ of a curve $y = f(x)$?

Answer: $\kappa = \frac{|f''(x)|}{(1 + (f'(x))^2)^{3/2}}$. Measures how quickly the curve deviates from its tangent line.

Flashcard 31: Determine the arc length of $r = 1 + \sin(\theta)$ from $\theta = 0$ to $\theta = \pi$.

Answer: $L = 5.333$ (approximately). The cardioid $r = 1 + \sin(\theta)$ has a complex arc length integral.

Flashcard 32: What is the arc length for $y = \sin(x)$ from $x = 0$ to $x = \frac{\pi}{2}$?

Answer: $L = 1.910$ (approximately). The arc length integral for $\sin(x)$ cannot be expressed in elementary functions.

Flashcard 33: Calculate the total distance traveled by $x(t) = t, y(t) = t^2$ from $t = 0$ to $t = 3$.

Answer: $D = 5.196$ (approximately). Integrates $\sqrt{1 + 4t^2}$ from 0 to 3 to get the total path length.

Flashcard 34: Find the arc length of $y = 3x$ from $x = 0$ to $x = 4$.

Answer: $L = 4 \sqrt{10}$. For $y = 3x$, $f'(x) = 3$, so $\sqrt{1 + 9} = \sqrt{10}$ over length 4.

Flashcard 35: What is the differential arc length $ds$ in polar coordinates $(r, \theta)$?

Answer: $ds = \sqrt{(dr)^2 + (r \, d\theta)^2}$. In polar coordinates, $r \, d\theta$ represents the tangential component of arc length.

Flashcard 36: Find the total distance traveled by $x(t) = 2t, y(t) = 3t$ from $t = 0$ to $t = 2$.

Answer: $D = 4\sqrt{13}$. Linear motion with constant speed $\sqrt{4 + 9} = \sqrt{13}$ over time interval 2.

Flashcard 37: Calculate the curvature of $r(\theta) = 1 + \cos(\theta)$ at $\theta = 0$.

Answer: $\kappa = \frac{3}{4}$. For cardioid at $\theta = 0$: $r = 2, r' = 0, r'' = -1$.

Flashcard 38: What is the formula for the arc length of a parametric curve $x(t), y(t)$ from $t = a$ to $t = b$?

Answer: $L = \int_{a}^{b} \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \, dt$. Combines $x$ and $y$ velocity components using the distance formula in parametric form.

Flashcard 39: Find the radius of curvature of $y = \sin(x)$ at $x = 0$.

Answer: $R = 1$. At $x = 0$: $f'(0) = 1, f''(0) = -1$, giving $\kappa = 1$.

Flashcard 40: What is the arc length for $y = \sin(x)$ from $x = 0$ to $x = \frac{\pi}{2}$?

Answer: $L = 1.910$ (approximately). The arc length integral for $\sin(x)$ cannot be expressed in elementary functions.

Flashcard 41: What is the formula for the radius of curvature $R$ of a curve $y = f(x)$?

Answer: $R = \frac{1}{\kappa} = \frac{(1 + (f'(x))^2)^{3/2}}{|f''(x)|}$. Radius of curvature is the reciprocal of curvature.

Flashcard 42: What is the curvature of a circle with radius $r$?

Answer: $\kappa = \frac{1}{r}$. For a circle, curvature is the reciprocal of the radius.

Flashcard 43: What is the formula for the arc length of a parametric curve $x(t), y(t)$ from $t = a$ to $t = b$?

Answer: $L = \int_{a}^{b} \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \, dt$. Combines $x$ and $y$ velocity components using the distance formula in parametric form.

Flashcard 44: Determine the curvature of $y = x^2$ at $x = 1$.

Answer: $\kappa = \frac{2}{(1 + 4)^{3/2}}$. At $x = 1$: $f'(1) = 2, f''(1) = 2$, so $\kappa = \frac{2}{5^{3/2}}$.

Flashcard 45: What is the relationship between curvature and radius of curvature?

Answer: $R = \frac{1}{\kappa}$. Curvature and radius of curvature are reciprocals of each other.

Flashcard 46: What is the expression for the curvature $\kappa$ of a curve $y = f(x)$?

Answer: $\kappa = \frac{|f''(x)|}{(1 + (f'(x))^2)^{3/2}}$. Measures how quickly the curve deviates from its tangent line.

Flashcard 47: Determine the curvature of $y = x^2$ at $x = 1$.

Answer: $\kappa = \frac{2}{(1 + 4)^{3/2}}$. At $x = 1$: $f'(1) = 2, f''(1) = 2$, so $\kappa = \frac{2}{5^{3/2}}$.

Flashcard 48: What is the formula for the total distance traveled by a particle with position $s(t)$?

Answer: $D = \int_{a}^{b} |v(t)| \, dt$. Integrates the absolute value of velocity to account for direction changes.

Flashcard 49: State the formula for the arc length of a curve $y = f(x)$ from $x = a$ to $x = b$.

Answer: $L = \int_{a}^{b} \sqrt{1 + (f'(x))^2} \, dx$. Uses Pythagorean theorem on small segments with $1 + (f'(x))^2$ under the square root.

Flashcard 50: State the formula for curvature $\kappa$ in parametric form.

Answer: $\kappa = \frac{|x'(t)y''(t) - y'(t)x''(t)|}{((x'(t))^2 + (y'(t))^2)^{3/2}}$. Uses the cross product of velocity and acceleration vectors in the numerator.

Flashcard 51: Calculate the curvature of $r(\theta) = 1 + \cos(\theta)$ at $\theta = 0$.

Answer: $\kappa = \frac{3}{4}$. For cardioid at $\theta = 0$: $r = 2, r' = 0, r'' = -1$.

Flashcard 52: What is the formula for the radius of curvature $R$ of a curve $y = f(x)$?

Answer: $R = \frac{1}{\kappa} = \frac{(1 + (f'(x))^2)^{3/2}}{|f''(x)|}$. Radius of curvature is the reciprocal of curvature.

Flashcard 53: What is the arc length differential $ds$ in terms of $dx$ and $dy$?

Answer: $ds = \sqrt{(dx)^2 + (dy)^2}$. Represents the infinitesimal arc length element using the Pythagorean theorem.

Flashcard 54: State the formula for curvature $\kappa$ in parametric form.

Answer: $\kappa = \frac{|x'(t)y''(t) - y'(t)x''(t)|}{((x'(t))^2 + (y'(t))^2)^{3/2}}$. Uses the cross product of velocity and acceleration vectors in the numerator.