This lesson covers sections 14.4 (Length of Curves) and 14.5 (Curvature) from the textbook.
Next: Functions of several variables
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Given vectors \(\vec{u}\), \(\vec{v}\), and \(\vec{w}\) as shown in the diagram, find \(\vec{w} - \vec{u} = \vec{v}\).
Diagram Description: A coordinate system showing three vectors. Vector \(\vec{u}\) points from the origin in one direction, vector \(\vec{v}\) forms another direction, and vector \(\vec{w}\) is shown as the resultant such that \(\vec{w} - \vec{u} = \vec{v}\).
Given \(F(x) = \int_a^x f(t)\,dt\).
By the Fundamental Theorem of Calculus:
\[F'(x) = f(x)\]
For a function \(y = f(x)\) with \(a \leq x \leq b\), the arc length formula is:
\[\text{Arc Length} = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx\]Solution:
First, compute \(\frac{dy}{dx} = 2x\)
Then:
\[\text{Arc length} = \int_0^1 \sqrt{1 + (2x)^2}\,dx = \int_0^1 \sqrt{1 + 4x^2}\,dx\]Graph Description: A coordinate plane showing the parabola \(y = x^2\) from \(x = 0\) to \(x = 1\). The curve starts at the origin and rises to the point (1, 1).
For \(y = x^2\) with \(0 \leq x \leq 1\), we found:
\[\text{Arc length} = \int_0^1 \sqrt{1+4x^2}\,dx\]We can parametrize this curve as:
\[\vec{r}(t) = \langle t, t^2 \rangle, \quad 0 \leq t \leq 1\]Then \(\vec{r}'(t) = \langle 1, 2t \rangle\)
And \(|\vec{r}'(t)| = \sqrt{1 + 4t^2}\)
Therefore, the arc length becomes:
\[\text{Arc length} = \int_0^1 |\vec{r}'(t)|\,dt = \int_0^1 \sqrt{1+4t^2}\,dt\]
Arc Length Formula:
For a parametrized curve in 3D space:
\[\vec{r}(t) = \langle f(t), g(t), h(t) \rangle\]The arc length of the curve from \(t = a\) to \(t = b\) is:
\[\int_a^b |\vec{r}'(t)|\,dt\]
We can interpret the arc length formula in two ways:
As a Vector:
\[\vec{r}(b) - \vec{r}(a)\]This represents the displacement vector from the starting point \(\vec{r}(a)\) to the ending point \(\vec{r}(b)\).
As a Scalar (Length):
\[\int_a^b |\vec{r}'(t)|\,dt\]This represents the actual length of the curve traveled from \(t = a\) to \(t = b\).
Diagram Description: Two curved paths are shown. The first (in red) shows a vector arc with endpoints labeled, demonstrating \(\vec{r}(a)\) and \(\vec{r}(b)\) with notation "(0,0,0)" at the origin. The second (in purple) shows the length of the curve from \(t = a\) to \(t = b\), emphasizing the scalar nature of arc length.
For a curve \(\vec{r}(t) = \langle f(t), g(t), h(t) \rangle\) in 3D space:
Displacement Vector:
\[\int_a^b \vec{r}'(t)\,dt\]This represents the vector from \(\vec{r}(a)\) to \(\vec{r}(b)\).
Arc Length:
\[\int_a^b |\vec{r}'(t)|\,dt\]This represents the length of the curve from \(t = a\) to \(t = b\).
Given \(\vec{r}(t) = \langle \cos t, \sin t \rangle\), compute \(\int_0^{\pi/2} |\vec{r}'(t)|\,dt\).
Solution:
First, find the derivative:
\[\vec{r}'(t) = \langle -\sin t, \cos t \rangle\]Evaluate at \(t = \frac{\pi}{2}\):
\[\vec{r}\left(\frac{\pi}{2}\right) - \vec{r}(0) = \langle 0,1 \rangle - \langle 1,0 \rangle = \langle -1, 1 \rangle\]Compute the magnitude:
\[|\vec{r}'(t)| = \sqrt{\sin^2 t + \cos^2 t} = 1\]Therefore:
\[\int_0^{\pi/2} |\vec{r}'(t)|\,dt = \int_0^{\pi/2} 1\,dt = \frac{\pi}{2}\]Graph Description: A unit circle centered at the origin with coordinate axes shown. The curve traces from \((1,0)\) to \((0,1)\), representing the first quarter of the circle (from \(t=0\) to \(t=\pi/2\)). The position vectors \(\vec{r}(0)\) and \(\vec{r}(\pi/2)\) are labeled.
Interpretation: The length of the curve from \(t = 0\) to \(t = \frac{\pi}{2}\) is \(\frac{\pi}{2}\), which is exactly one quarter of the circle's circumference.
Given \(\vec{r}(t) = \langle \cos t, \sin t, 2t \rangle\), compute the arc length.
Solution:
Find the derivative:
\[\vec{r}'(t) = \langle -\sin t, \cos t, 2 \rangle\]Compute the magnitude:
\[|\vec{r}'(t)| = \sqrt{\sin^2 t + \cos^2 t + 4} = \sqrt{1 + 4} = \sqrt{5}\]Calculate the arc length:
\[\text{Arc length} = \int_0^{2\pi} \sqrt{5}\,dt = \sqrt{5} \cdot 2\pi = 2\pi\sqrt{5}\]
Given \(\vec{r}(t) = \langle \cos t, \sin t, 2t \rangle\)
Solution:
From the previous example, we know \(|\vec{r}'(u)| = \sqrt{5}\)
The arc length from 0 to \(t\) is:
\[s(t) = \int_0^t |\vec{r}'(u)|\,du = \int_0^t \sqrt{5}\,du = \sqrt{5}t\]
Definition: For a curve \(\vec{r}(t) = \langle f(t), g(t), h(t) \rangle\), we define \(s(t)\) as the arc length function:
Input: \(t\)
Output: Arc length from \(a\) to \(t\)
\[s(t) = \int_a^t |\vec{r}'(u)|\,du\]Note: By the Fundamental Theorem of Calculus:
\[s'(t) = |\vec{r}'(t)|\]
For \(\vec{r}(t) = \langle \cos t, \sin t, 2t \rangle\):
We found \(s(t) = \int_0^t |\vec{r}'(u)|\,du = \sqrt{5}t\)
This gives us the arc length measured from \(t = 0\).
Using this relationship, we can reparametrize \(\vec{r}\) with \(s\) as the new parameter.
Since \(s = \sqrt{5}t\), we have \(t = \frac{s}{\sqrt{5}}\)
Substituting this into \(\vec{r}(t)\):
\[\vec{r}(s) = \left\langle \cos\left(\frac{s}{\sqrt{5}}\right), \sin\left(\frac{s}{\sqrt{5}}\right), \frac{2s}{\sqrt{5}} \right\rangle\]Arc Length Parametrization: A parametrization where the parameter represents the arc length along the curve measured from a starting point.
| Regular Parametrization | Arc Length Parametrization |
|---|---|
| Parameter = "time" | Parameter = length of curve |
| Given \(t\), \(\vec{r}(t)\) represents position on the curve at time \(t\) | Given \(s\), \(\vec{r}(s)\) represents position on curve when you travel \(s\) units from start |
| Speed = \(|\vec{r}'(t)|\) (varies) | Speed = \(|\vec{r}'(s)| = 1\) |
Key Property: In arc length parametrization, the speed is always constant and equal to 1.
For \(\vec{r}(s) = \left\langle \cos\left(\frac{s}{\sqrt{5}}\right), \sin\left(\frac{s}{\sqrt{5}}\right), \frac{2s}{\sqrt{5}} \right\rangle\)
Solution:
Find the derivative with respect to \(s\):
\[\vec{r}'(s) = \left\langle -\frac{1}{\sqrt{5}}\sin\left(\frac{s}{\sqrt{5}}\right), \frac{1}{\sqrt{5}}\cos\left(\frac{s}{\sqrt{5}}\right), \frac{2}{\sqrt{5}} \right\rangle\]Compute the magnitude:
\[|\vec{r}'(s)| = \sqrt{\frac{1}{5}\sin^2\left(\frac{s}{\sqrt{5}}\right) + \frac{1}{5}\cos^2\left(\frac{s}{\sqrt{5}}\right) + \frac{4}{5}}\] \[= \sqrt{\frac{1}{5}\left(\sin^2\left(\frac{s}{\sqrt{5}}\right) + \cos^2\left(\frac{s}{\sqrt{5}}\right)\right) + \frac{4}{5}}\] \[= \sqrt{\frac{1}{5} + \frac{4}{5}} = \sqrt{1} = 1\]
Reparametrize \(\vec{r}(t) = \langle \cos t, \sin t, 2t \rangle\) with respect to arc length as measured from \(\langle -1, 0, 2\pi \rangle\) in the \(t\) direction (\(t\) increasing).
Note: At \(t = \pi\), we have \(\vec{r}(\pi) = \langle -1, 0, 2\pi \rangle\), which is the starting point.
Solution:
Compute \(\vec{r}'(t) = \langle -\sin t, \cos t, 2 \rangle\)
Find the magnitude:
\[|\vec{r}'(t)| = \sqrt{\sin^2 t + \cos^2 t + 4} = \sqrt{5}\]The arc length function from \(\pi\) is:
\[s(t) = \int_\pi^t \sqrt{5}\,du = \sqrt{5}(t - \pi)\]From \(s = \sqrt{5}(t - \pi)\), we get:
\[t = \frac{s}{\sqrt{5}} + \pi\]
Diagram Description: Three curves are shown to illustrate different amounts of bending:
Curvature \(\kappa\): The magnitude of the rate of change of the unit tangent vector.
\[\kappa = \left|\frac{d\vec{T}}{ds}\right| = |\vec{T}'(s)|\]where \(\vec{T}\) is the unit tangent vector and \(s\) is the arc length parameter.
Curvature with arc length parameter:
\[\kappa = |\vec{T}'(s)| = |\vec{r}''(s)|\]where \(s\) is the arc length parameter.
Curvature with general parameter \(t\):
\[\kappa = \frac{|\vec{T}'(t)|}{|\vec{r}'(t)|}\]Using the chain rule: \(\frac{d\vec{T}}{dt} = \frac{d\vec{T}}{ds} \cdot \frac{ds}{dt}\)
Since \(\frac{ds}{dt} = |\vec{r}'(t)|\), we get:
\[\left|\frac{d\vec{T}}{dt}\right| = \left|\frac{d\vec{T}}{ds}\right| \cdot \left|\frac{ds}{dt}\right|\]Practical formula (most useful for computation):
\[\kappa = \frac{|\vec{r}'(t) \times \vec{r}''(t)|}{|\vec{r}'(t)|^3}\]Proof of this formula is in the textbook.
Given \(\vec{r}(t) = \langle 3, -4t, 5t^2 \rangle\), find the curvature at \(t = 0\).
Solution:
Find the first derivative:
\[\vec{r}'(t) = \langle 0, -4, 10t \rangle\]At \(t = 0\): \(\vec{r}'(0) = \langle 0, -4, 0 \rangle\)
Find the second derivative:
\[\vec{r}''(t) = \langle 0, 0, 10 \rangle\]At \(t = 0\): \(\vec{r}''(0) = \langle 0, 0, 10 \rangle\)
Compute the cross product:
\[\vec{r}'(0) \times \vec{r}''(0) = \langle 0, -4, 0 \rangle \times \langle 0, 0, 10 \rangle = \langle -40, 0, 0 \rangle\]Find the magnitudes:
\[|\vec{r}'(0) \times \vec{r}''(0)| = \sqrt{1600} = 40\] \[|\vec{r}'(0)| = \sqrt{16} = 4\] \[|\vec{r}'(0)|^3 = 64\]Calculate the curvature:
\[\kappa = \frac{|\vec{r}' \times \vec{r}''|}{|\vec{r}'|^3} = \frac{40}{64} = \frac{5}{8}\]Therefore, \(\kappa\) at \(t = 0\) is \(\frac{5}{8}\).