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Lessons for Multivariable Calculus

Mark Fitch

Section 1.5 Space Curve Properties

In Section 1.4 we defined a derivative on a space curve (function \(f:\R \to \R^n\) for \(n \ge 2\)). In this section we will use derivatives to analyze properties of space curves.

Subsection 1.5.1 Tangents for Space Curves

From previous calculus we know that the derivative of a function gives us the direction of the tangent. This may be interpreted as the direction that a particle traveling the curve is going at that point. The same is true for space curves.

Example 1.5.1.

Consider the curve
\begin{equation*} C(t)=(\cos(t),\sin(t),t) \end{equation*}
If we are travelling along the curve with \(t\) increasing, which way are we facing at \(t=\pi/2\text{?}\) First we calculate the derivative.
\begin{equation*} C^\prime(t) = (-\sin(t),\cos(t),1) \end{equation*}
At \(t=\pi/2\) the point is
\begin{equation*} C(\pi/2)=(0,1,\pi/2) \end{equation*}
and the tangent vector is
\begin{equation*} C^\prime(\pi/2)=(-1,0,1) \end{equation*}
If we are travelling along the curve with \(t\) increasing, which way are we facing at \(t=\pi\text{?}\) At \(t=\pi\) the point is
\begin{equation*} C(\pi)=(-1,0,\pi) \end{equation*}
and the tangent vector is
\begin{equation*} C^\prime(\pi/2)=(0,-1,1) \end{equation*}
Notice that \(C_z^\prime(t)=1\text{.}\) This implies that no matter where we are we are looking up with a slope of 1. Only the x and y components of the tangent are changing.
Figure 1.5.2. Space Curve with Tangents

Checkpoint 1.5.3.

Calculate the tangent vector for the curve \(C(t)=(t,t^2,e^t)\) at \(t=5\)
While the direction of the tangent vector is the direction the curve is heading at that input, the magnitude of the tangent vector is the speed at which it is changing.

Example 1.5.4.

Consider the curve
\begin{equation*} C(t)=(\cos(t),\sin(t),t) \end{equation*}
What is the speed at which the curve is changing at \(t=\pi/2\text{?}\) First we calculate the derivative.
\begin{equation*} C^\prime(t) = (-\sin(t),\cos(t),1) \end{equation*}
At \(t=\pi/2\) the tangent vector is
\begin{equation*} C^\prime(\pi/2)=(-1,0,1)\text{.} \end{equation*}
The magnitude is
\begin{equation*} \|(-1,0,1)\|=\sqrt{(-1)^2+0^2+1^2}=\sqrt{2}. \end{equation*}
Note we can calculate the magnitude of the tangent vector using the derivative.
\begin{align*} \|C^\prime(t)|= & \\ \|(-\sin(t),\cos(t),1)\|= & \sqrt{(-\sin(t))^2+(\cos(t))^2+1^2}=\sqrt{2}. \end{align*}
For this curve the speed of change is constant.

Checkpoint 1.5.5.

Calculate the magnitude of the tangent vector for the curve \(C(t)=(t,t^2,e^t)\) at \(t=5\)

Subsection 1.5.2 Unit Tangent Vector for a Space Curve

We know that the parameterization of a curve affects results. Here we define a tangent that removes the effect of parameterization.

Definition 1.5.6. Unit Tangent Vector.

The unit vector in the direction of the tangent vector of a space curve is
\begin{equation*} \vec{T}=\frac{C^\prime(t)}{\|C^\prime(t)\|} \end{equation*}

Example 1.5.7.

Consider the curve
\begin{equation*} C(t)=(\cos(t),\sin(t),t) \end{equation*}
Calculate the unit tangent for this curve.
First we calculate the derivative.
\begin{equation*} C^\prime(t) = (-\sin(t),\cos(t),1) \end{equation*}
Next we calculate the norm of the derivative.
\begin{equation*} \|C^\prime(t)\|=\sqrt{(-\sin(t))^2+(\cos(t))^2+1^2}=\sqrt{2}\text{.} \end{equation*}
Thus the unit tangent is
\begin{equation*} \vec{T}(t)=\frac{1}{\sqrt{2}}(-\sin(t),\cos(t),1) \end{equation*}

Subsection 1.5.3 Normal Vector for a Space Curve

The unit tangent gives the direction of change of the curve. We may also desire the change of unit tangent of a curve at a point. This is a type of second derivative.
The following theorem allows us to define a useful vector for the change of the tangent. For an explanation of the theorem check your textbook.
Now we note that because we defined \(\vec{T}(t)\) as a unit vector the above theorem applies. This means that \(\vec{T}^\prime(t)\) is orthogonal to \(\vec{T}(t)\text{.}\) Furthermore, because it is the derivative of the unit tangent it gives the change of the unit tangent. For future reasons we desire a unit vector version of this derivative.

Definition 1.5.9. Unit Normal Vector.

The unit normal vector of a space curve is
\begin{equation*} \vec{N}(t)=\frac{\vec{T}^\prime(t)}{\|\vec{T}^\prime(t)\|} \end{equation*}

Example 1.5.10.

Consider the curve
\begin{equation*} C(t)=(\cos(t),\sin(t),t) \end{equation*}
Calculate the unit binormal for this curve.
First we calculate the unit tangent of this curve.
\begin{equation*} \vec{T}(t)=\frac{1}{\sqrt{2}}(-\sin(t),\cos(t),1) \end{equation*}
Next we need to calculate the derivative of the unit tangent.
\begin{equation*} \vec{T}^\prime(t)=\frac{1}{\sqrt{2}}(-\cos(t),-\sin(t),0)) \end{equation*}
We also need the magnitude of this derivative.
\begin{align*} \|\vec{T}^\prime(t)\| & =\sqrt{\left(\frac{-\cos(t)}{\sqrt{2}}\right)^2+\left(\frac{-\sin(t)}{\sqrt{2}}\right)^2+0^2}\\ & = \sqrt{\frac{\cos^2(t)}{2}+\frac{\sin^2(t)}{2}}\\ & = \frac{1}{\sqrt{2}}. \end{align*}
Thus the unit normal is
\begin{equation*} \vec{N}(t)=(-\cos(t),-\sin(t),0)) \end{equation*}

Subsection 1.5.4 Binormal Vector for a Space Curve

The unit tangent and unit normal vectors can be thought of as defining an x and y coordinate system for that curve at the point. To extend this to 3D we define the following.

Definition 1.5.11. Binormal.

The unit binormal vector of a space curve is
\begin{equation*} \vec{B}=\frac{\vec{T}(t) \times \vec{N}(t)}{\|\vec{T}(t) \times \vec{N}(t)\|} \end{equation*}

Example 1.5.12.

Consider the curve
\begin{equation*} C(t)=(\cos(t),\sin(t),t) \end{equation*}
Calculate the unit binormal for this curve.
First we calculate the unit tangent and unit normal vectors.
\begin{equation*} \vec{T}(t)=\frac{1}{\sqrt{2}}(-\sin(t),\cos(t),1) \end{equation*}
\begin{equation*} \vec{N}(t)=(-\cos(t),-\sin(t),0)) \end{equation*}
Next we need to calculate the cross product.
\begin{align*} \vec{T}(t) \times \vec{N}(t) = & \left\langle \frac{\cos(t)}{\sqrt{2}} \cdot 0 - \frac{1}{\sqrt{2}} \cdot -\sin(t) \right.\\ & -\frac{\sin(t)}{\sqrt{2}} \cdot 0 - \frac{1}{\sqrt{2}} \cdot -\cos(t) \\ & \left. -\frac{\sin(t)}{\sqrt{2}} \cdot -\sin(t)-\frac{\cos(t)}{\sqrt{2}} \cdot -\frac{\cos(t)}{\sqrt{2}} \right\rangle \\ = & \left\langle -\frac{\sin(t)}{\sqrt{2}}, -\frac{\cos(t)}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right\rangle \end{align*}
The norm of this is
\begin{equation*} \|\vec{T}(t) \times \vec{N}(t)\| = \frac{1}{2} \end{equation*}
Thus the unit binormal is
\begin{equation*} \left\langle -\frac{\sin(t)}{4\sqrt{2}}, -\frac{\cos(t)}{4\sqrt{2}}, \frac{1}{4\sqrt{2}} \right\rangle \end{equation*}
Figure 1.5.13. Space Curve with Tangent, Normal, Binormal

Subsection 1.5.5 Curvature

The unit tangent and normal tell us about how a curve is changing. Here we develop a parameter that measures how curved it is.

Definition 1.5.14. Curvature.

The curvature of a curve is
\begin{equation*} \kappa = \left| \frac{d\vec{T}}{ds} \right| \end{equation*}
which is the derivative of the unit tangent with respect to arclength.
To make sense of curvature we will first look at two simple cases.

Example 1.5.16.

First, we look at the curvature of a line. Consider \(\ell(t)=(8,4,-4)+\langle 1,-3,1 \rangle t = \langle t+8,-3t+4,t-4 \rangle\text{.}\)
\begin{align*} \vec{T}(t) = & \frac{\ell^\prime(t)}{|\ell^\prime(t)|}\\ = & \frac{\langle 1,-3,1 \rangle}{\sqrt{1^2+(-3)^2+1^2}}\\ = & \left\langle \frac{1}{\sqrt{11}}, -\frac{3}{\sqrt{11}},\frac{1}{\sqrt{11}} \right\rangle\\ \vec{T}^\prime(t) = & \langle 0,0,0 \rangle. \end{align*}
Thus the curvature will be 0. This makes sense: a line does not curve.

Example 1.5.17.

Second, we look at the curvature of a circle. Consider \(C=(4\cos\theta,4\sin\theta,17)\text{.}\)
\begin{align*} C^\prime(t) = & (-4\sin\theta,4\cos\theta,0)\\ C^{\prime \prime}(t) = & (-4\cos\theta,-4\sin\theta,0)\\ C^\prime(t) \times C^{\prime \prime}(t) = & (0,0,4^2\sin^2 \theta + 4^2\cos^2 \theta)\\ = & (0,0,4)\\ |C^\prime(t) \times C^{\prime \prime}(t)| = & 4^2\\ |C^\prime(t)| = & \sqrt{(-4\sin\theta)^2+(4\cos\theta)^2+0^2}\\ = & 4\\ \kappa(\theta) = & \frac{4^2}{4^3}\\ = & \frac{1}{4} \end{align*}
Notice that the curvature of a circle is dependent solely on its radius and not on the particular point. Also consider that the larger the radius, the smaller the curvature.
The next example illustrates the more general use of curvature.

Example 1.5.18.

Consider the curve \(P=(t,t^2,t^3)\text{.}\)
\begin{align*} P^\prime(t) = & (1,2t,3t^2).\\ P^{\prime \prime}(t) = & (0,2,6t).\\ P^\prime(t) \times P^{\prime \prime}(t) = & (6t^2,-6t,2).\\ |P^\prime(t) \times P^{\prime \prime}(t)| = & 2\sqrt{1+9t^2+9t^4}.\\ |P^\prime(t)|^3 = & (1+4t^2+9t^4)^{3/2}.\\ \kappa(t) = & \frac{2\sqrt{1+9t^2+9t^4}}{(1+4t^2+9t^4)^{3/2}}. \end{align*}
We can now calculate the curvature at any point. For example at \(t=1\) which is the point \(P=(1,1,1)\) the curvature is \(\kappa(1)=\frac{2\sqrt{19}}{(14)^{3/2}} \approx 0.1664\text{.}\)

Checkpoint 1.5.19.

Calculate the curvature of \(E=(5\cos\theta,3\sin\theta,1)\) at \(\theta=\pi/2\text{.}\)
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