AP Calculus BC 8.13 The Arc Length of a Smooth, Planar Curve and Distance Traveled Study Notes - New Syllabus
AP Calculus BC 8.13 The Arc Length of a Smooth, Planar Curve and Distance Traveled bc only Study Notes- New syllabus
AP Calculus BC 8.13 The Arc Length of a Smooth, Planar Curve and Distance Traveled bc only Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- Definite integrals allow us to solve problems involving the accumulation of change over an interval.
Key Concepts:
- The Arc Length of a Smooth, Planar Curve and Distance Traveled
The Arc Length of a Smooth, Planar Curve and Distance Traveled
Arc Length of a Smooth, Planar Curve
The arc length of a smooth curve measures the total distance along the curve between two points. It is found by summing (integrating) the lengths of infinitesimally small segments of the curve.
Derivation of the Arc Length Formula
By the Pythagorean theorem, the length of the line segment is \( \sqrt{(\Delta x)^2 + (\Delta y_i)^2} \).
We can also write this as \( \Delta x \sqrt{1 + \Big(\dfrac{\Delta y_i}{\Delta x}\Big)^2} \).
Now, by the Mean Value Theorem, there is a point \(x_i^* \in [x_{i-1}, x_i]\) such that \( f'(x_i^*) = \dfrac{\Delta y_i}{\Delta x} \). Then the length of the line segment is given by \( \Delta x \sqrt{1 + [f'(x_i^*)]^2} \).
Adding up the lengths of all the line segments, we get \( \text{Arc Length} \approx \sum_{i=1}^n \sqrt{1 + [f'(x_i^*)]^2}\,\Delta x. \)
This is a Riemann sum. Taking the limit as \( n \to \infty \), we have
\( = \lim_{n \to \infty} \sum_{i=1}^n \sqrt{1 + [f'(x_i^*)]^2}\,\Delta x \)
\( = \displaystyle \int_a^b \sqrt{1 + [f'(x)]^2}\, dx. \)
Formula when the curve is expressed as \( y = f(x) \):
\( L = \displaystyle \int_{a}^{b} \sqrt{1 + \left( \dfrac{dy}{dx} \right)^2} \, dx \)
Formula when the curve is expressed as \( x = g(y) \):
\( L = \displaystyle \int_{c}^{d} \sqrt{1 + \left( \dfrac{dx}{dy} \right)^2} \, dy \)
Formula for parametric curves \( x = x(t),\ y = y(t) \):
\( L = \displaystyle \int_{t_1}^{t_2} \sqrt{\left( \dfrac{dx}{dt} \right)^2 + \left( \dfrac{dy}{dt} \right)^2} \, dt \)
Important Notes:
- The derivative term inside the square root accounts for the slope of the curve at each point.
- The function must be smooth and differentiable in the given interval.
- Arc length problems often require simplification before integrating.
Example:
Arc Length of \( y = \dfrac{x^{3/2}}{3} \) on \( [0, 4] \)
▶️ Answer/Explanation
We have \( y = \dfrac{x^{3/2}}{3} \). First, find the derivative:
\( \dfrac{dy}{dx} = \dfrac{\dfrac{3}{2}x^{1/2}}{3} = \dfrac{\sqrt{x}}{2} \)
Arc length: \( L = \displaystyle \int_{0}^{4} \sqrt{1 + \left( \dfrac{\sqrt{x}}{2} \right)^2} \, dx \)
\( L = \displaystyle \int_{0}^{4} \sqrt{1 + \dfrac{x}{4}} \, dx \)
Let \( u = 1 + \dfrac{x}{4} \), so \( du = \dfrac{dx}{4} \), \( dx = 4\,du \).
When \( x = 0 \), \( u = 1 \); when \( x = 4 \), \( u = 2 \).
\( L = \displaystyle \int_{1}^{2} \sqrt{u} \cdot 4 \, du \)
\( L = 4 \cdot \left[ \dfrac{2}{3} u^{3/2} \right]_{1}^{2} \)
\( L = \dfrac{8}{3} \left( 2^{3/2} – 1 \right) \)
Example :
Arc Length for Parametric Equations \( x = t^2, \ y = \dfrac{t^3}{3} \), \( 0 \leq t \leq 2 \)
▶️ Answer/Explanation
We have: \( \dfrac{dx}{dt} = 2t, \quad \dfrac{dy}{dt} = t^2 \)
Arc length formula for parametric curves:
\( L = \displaystyle \int_{0}^{2} \sqrt{(2t)^2 + (t^2)^2} \, dt \)
\( L = \displaystyle \int_{0}^{2} \sqrt{4t^2 + t^4} \, dt \)
Factor \( t^2 \): \( L = \displaystyle \int_{0}^{2} t \sqrt{4 + t^2} \, dt \)
Let \( u = 4 + t^2 \), \( du = 2t \, dt \), so \( t\,dt = \dfrac{du}{2} \).
When \( t = 0 \), \( u = 4 \); when \( t = 2 \), \( u = 8 \).
\( L = \dfrac{1}{2} \displaystyle \int_{4}^{8} \sqrt{u} \, du \)
\( L = \dfrac{1}{2} \cdot \left[ \dfrac{2}{3} u^{3/2} \right]_{4}^{8} \)
\( L = \dfrac{1}{3} \left( 8^{3/2} – 4^{3/2} \right) = \dfrac{1}{3} (16\sqrt{2} – 8) \)
Distance Traveled
The distance traveled by a particle is the total length of the path it moves along during a given time interval. It is different from displacement, which only measures the net change in position.
Distance traveled is obtained by integrating the speed (the absolute value of velocity) over the given time interval.
Formula:
\(\text{Distance} = \displaystyle \int_{t_1}^{t_2} |v(t)| \, dt\)
Where:
- \( v(t) \) is the velocity function of the particle.
- \( |v(t)| \) ensures that all movement is counted as positive distance, even if the particle changes direction.
Important Notes:
- If velocity changes sign during the interval, the integration must be split at the points where \( v(t) = 0 \).
- For motion along a curve described parametrically (\( x(t), y(t) \)), distance is the arc length of the curve:
\(\text{Distance} = \displaystyle \int_{t_1}^{t_2} \sqrt{ \left( \dfrac{dx}{dt} \right)^2 + \left( \dfrac{dy}{dt} \right)^2 } \, dt\)
Example:
Find the Distance Traveled for \( v(t) = t^2 – 4t + 3 \) on \( [0, 4] \)
▶️ Answer/Explanation
First, find where velocity changes sign by solving:
\(t^2 – 4t + 3 = 0 \quad \Rightarrow \quad (t – 1)(t – 3) = 0\)
So \( t = 1 \) and \( t = 3 \) are the points where the velocity is zero.
We split the integral into three intervals:
\(\text{Distance} = \displaystyle \int_{0}^{1} (t^2 – 4t + 3) \, dt – \int_{1}^{3} (t^2 – 4t + 3) \, dt + \int_{3}^{4} (t^2 – 4t + 3) \, dt\)
Note: We put a minus sign on the second integral because \( v(t) < 0 \) there, and we want absolute value.
Computing each:
\(\int (t^2 – 4t + 3) \, dt = \dfrac{t^3}{3} – 2t^2 + 3t\)
From \( 0 \) to \( 1 \):
\(\left[ \dfrac{t^3}{3} – 2t^2 + 3t \right]_{0}^{1} = \left( \dfrac{1}{3} – 2 + 3 \right) – 0 = \dfrac{4}{3}\)
From \( 1 \) to \( 3 \):
\(\left[ \dfrac{t^3}{3} – 2t^2 + 3t \right]_{1}^{3} = \left( 9 – 18 + 9 \right) – \left( \dfrac{1}{3} – 2 + 3 \right) = 0 – \dfrac{4}{3} = -\dfrac{4}{3}\)
Absolute value gives \( \dfrac{4}{3} \).
From \( 3 \) to \( 4 \):
\(\left[ \dfrac{t^3}{3} – 2t^2 + 3t \right]_{3}^{4} = \left( \dfrac{64}{3} – 32 + 12 \right) – (9 – 18 + 9) = \left( \dfrac{64}{3} – 20 \right) – 0 = \dfrac{4}{3}\)
Total distance:
\(\text{Distance} = \dfrac{4}{3} + \dfrac{4}{3} + \dfrac{4}{3} = 4\)
Example:
Particle Moving Along a Curve \( x(t) = \cos t,\ y(t) = \sin t \) from \( t = 0 \) to \( t = \pi \) , Find the Distance Traveled.
▶️ Answer/Explanation
We have:
\(\dfrac{dx}{dt} = -\sin t, \quad \dfrac{dy}{dt} = \cos t\)
Speed:
\(\sqrt{(-\sin t)^2 + (\cos t)^2} = \sqrt{\sin^2 t + \cos^2 t} = 1\)
Distance traveled:
\(\text{Distance} = \displaystyle \int_{0}^{\pi} 1 \, dt = \pi\)
This makes sense because the particle moves along the upper half of the unit circle, whose length is \( \pi \).