Calculus III 12.04 Tangent Vectors and Normal Vectors

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12.4 Tangent Vectors and Normal Vectors

• Find a unit tangent vector and a principal unit normal vector at a point on a space curve.
• Find the tangential and normal components of acceleration.

Tangent Vectors and Normal Vectors

Definition 12.4.1 Unit Tangent Vector

Let \(C\) be any smooth curve represented by \( \textbf{r} \) on an open interval \(I\). The unit tangent vector \( \textbf{T}(t) \) at \(t\) is defined as

$$ \textbf{T}(t) = \frac{ \textbf{r}^{\prime}(t)}{ \| \textbf{r}^{\prime}(t) \|}, \: \textbf{r}^{\prime}(t) \ne \textbf(0) $$

Recall that a curve is smooth on an interval when is continuous and nonzero on the interval. The term “smoothness” is sufficient to guarantee that a curve has a unit tangent vector.

Example 12.4.1 Finding the Unit Tangent Vector

The tangent vector's direction depends on the curve's orientation. Figure 12.4.1

Find the unit tangent vector to the curve given by \( \textbf{r} = t\textbf{i} + t^{2}\textbf{j} \) where \(t=1\). Solution Take the first derivative and apply Definition 12.4.1. \( \textbf{r}^{\prime}(t) \) \(= \textbf{i} + 2t\textbf{j} \) \( \textbf{T}(t) \) $$= \frac{ \textbf{r}^{\prime}(t)}{ \| \textbf{r}^{\prime}(t) \|}$$ $$= \frac{1}{ \sqrt{ 1+4t^{2}}} ( \textbf{i} + 2t\textbf{j} ) \:\:\:\: $$ Substitute in \( \textbf{r}^{\prime}(t) \) . When \(t=1\), the unit tangent vector is $$ \textbf{T}(1) = \frac{1}{ \sqrt{ 5 }} ( \textbf{i} + 2\textbf{j} ) $$ as shown in Figure 12.4.1. The unit tangent vector's direction depends on the curve's orientation. For the parabola described by \( \textbf{r} = -(t-2)\textbf{i} + (t-2)^{2}\textbf{j} \) \( \textbf{T}(1) \) would still represent the unit tangent vector at the point (1,1), but it would point in the opposite direction.

Example 12.4.2 Finding the Tangetnt Line at a Point on a Curve

The tangent line to a curve at a point is determined by the unit tangent vector at the point. Figure 12.4.2

Find \( \textbf{T}(t) \) and then find parametric equations for the tangent line to the helix given by \( \textbf{r}(t) = 2 \cos t\textbf{i} + 2 \sin t\textbf{j} + t\textbf{k} \) at the point $$ \left( \sqrt{2}, \sqrt{2}, \frac{\pi}{4} \right). $$ Solution For tangent vectors, the tangent line to a curve is a line that passes through the point and is parallel to the unit tangent vector, as shown in Figure 12.4.2. The derivative for \( \textbf{r}(t) \) is \( \textbf{r}^{\prime}(t) = -2 \sin t\textbf{i} + 2 \cos t\textbf{j} + \textbf{k} \) which produces $$ \| \textbf{r}^{\prime}(t) \| = \sqrt{ 4 \sin^{2}t + 4 \cos^{2}t + 1 } = \sqrt{5}. $$ The unit tangent vector is \( \textbf{T}(t) \) $$= \frac{ \textbf{r}^{\prime}(t) }{ \| \textbf{r}^{\prime}(t) \|} $$ $$= \frac{1}{ \sqrt{5} } \left( -2 \sin t\textbf{i} + 2 \cos t\textbf{j} + \textbf{k} \right) \:\:\:\: $$ Unit tangent vector At the point \( \left( \sqrt{2}, \sqrt{2}, \frac{\pi}{4} \right) \), \( t=\frac{\pi}{4} \) and the unit tangent vector is $$ \textbf{T}(\frac{\pi}{4}) $$ $$= \frac{1}{ \sqrt{5} } \left( -2 \frac{\sqrt{2}}{2} \textbf{i} + 2 \frac{\sqrt{2}}{2} \textbf{j} + \textbf{k} \right) $$ $$= \frac{1}{ \sqrt{5} } \left( - \sqrt{2} \textbf{i} + \sqrt{2} \textbf{j} + \textbf{k} \right) $$ Using the direction numbers, \(a=-\sqrt{2} \), \(b=\sqrt{2}\), and \(c=1\), and the point \((x,y,z)=( \sqrt{2}, \sqrt{2}, \frac{\pi}{4} ) \) the parametric equations (given with parameter \(s\) ) are listed below. \( x\) \(=x_{1}+as \) \(=\sqrt{2} - \sqrt{2}s \) \(y \) \(=y_{1}+bs \) \(= \sqrt{2} + \sqrt{2}s\) \( z\) \(=z_{1}+cs \) \(= \frac{\pi}{4} + s\) This tangent line is shown in Figure 12.4.2.

Definition 12.4.2 Principal Unit Normal Vector

Let \(C\) be a smooth curve represented by \( \textbf{r} \) on an open interval \(I\). If \(\textbf{T}^{\prime}(t) \ne \textbf{0} \), then the principle unit normal vector at \(t\) is defined as

$$ \textbf{N}(t) = \frac{ \textbf{T}^{\prime}(t)}{ \| \textbf{T}^{\prime}(t) \|} $$

Proof
There are infinitely many vectors that are orthogonal to the tangent vector \( \textbf{T}(t) \). One is the vector \( \textbf{T}^{\prime}(t) \). Applying Theorem 12.2.2 Property 7 produces

$$ \textbf{T}(t) \cdot \textbf{T}(t) = \| \textbf{T}(t) \|^{2} = 1 \rightarrow \textbf{T}(t) \cdot \textbf{T}^{\prime}(t) = 0.$$

This normalization proves Definition 12.4.2.

Example 12.4.3 Finding the Principal Unit Normal Vector in Two-Dimensions

The principal unit normal vector points toward the curve's concave side. Figure 12.4.3

Find \( \textbf{N}(t) \) and \( \textbf{N}(1) \) for the curve described by \( \textbf{r}(t) = 3t\textbf{i} + 2t^{2}\textbf{j} \). Solution Differentiating yields \( \textbf{r}^{\prime}(t) =3 \textbf{i} + 4t \textbf{j} \) and \( \| \textbf{r}^{\prime}(t) \| = \sqrt{ 9+ 16t^{2} }. \) The unit tangent vector is $$ \textbf{T}(t) $$ $$= \frac{ \textbf{r}^{\prime}(t)}{ \| \textbf{r}^{\prime}(t) \|} $$ $$= \frac{1}{ \sqrt{ 9+ 16t^{2} } } \left( 3 \textbf{i} + 4t \textbf{j} \right) \:\:\:\: $$ Unit Tangent Vector Apply Theorem 12.2.2 and differentiate \( \textbf{T}(t) \) with respect to \(t\) to produce \( \textbf{T}^{\prime}(t) \) $$= \frac{1}{ \sqrt{ 9+ 16t^{2} } } \left(4 \textbf{j} \right) - \frac{16t}{ ( 9+ 16t^{2})^{3/2} }\left(3\textbf{i} + 4t \textbf{j} \right) $$ $$= \frac{12}{ ( 9+ 16t^{2})^{3/2} }\left(-4t\textbf{i} + 3t \textbf{j} \right) $$ \( \| \textbf{T}^{\prime}(t) \| \) $$= 12 \sqrt{ \frac{ 9+ 16t^{2} }{(9+ 16t^{2} )^{3}}} = \frac{12}{ 9+ 16t^{2} }.$$ The principal unit normal vector is \( \textbf{N}(t) \) $$= \frac{ \textbf{T}^{\prime}(t) }{\| \textbf{T}^{\prime}(t) \| }$$ $$=\frac{1}{ \sqrt{ 9+ 16t^{2} } } \left(-4t\textbf{i} + 3t \textbf{j} \right) \:\:\:\:$$ Principal unit normal vector When \(t=1\), the principal unit normal vector is $$ \textbf{N}(1) = \frac{1}{5}\left(-4\textbf{i} + 3 \textbf{j} \right) $$ as shown in Figure 12.4.3.

Standard unit vectors i and j Figure 12.4.4

The principal unit normal vector can be difficult to evaluate algebraically. For two-dimensional curves the algebra can be simplified by finding \( \textbf{T}(t) = x(t)\textbf{i} + y(t)\textbf{j} \:\:\:\: \)Unit Tangent Vector and observing that \( \textbf{N}(t) \) must be either \( \textbf{N}_{1}(t) = y(t)\textbf{i} - x(t)\textbf{j} \:\: \text{ or } \:\: \textbf{N}_{2}(t) = -y(t)\textbf{i} + x(t)\textbf{j}. \) Because $$ \sqrt{ \left[ x(t) \right]^{2} + \left[ y(t) \right]^{2}} =1, $$ it follows that both \( \textbf{N}_{1}(t) \) and \( \textbf{N}_{2}(t) \) are unit normal vectors. The principle unit normal vector \( \textbf{N} \) is one that points toward the curve's concave side, as shown in Figure 12.4.3. This also holds for curves in three-dimensions. That is, for an object moving along a curve \(C\) in three-dimensions, the vector \( \textbf{N}(t) \) is orthogonal to \( \textbf{T}(t) \) and the points along which the object is moving, as shown in Figure 12.4.3.

Example 12.4.4 Finding the Principal Unit Normal Vector in Three-Dimensions

\( \textbf{N}(t) \) is horizontal and points toward the \(z\)-axis. Figure 12.4.5

Find the principal unit normal vector for the helix \( \textbf{r}(t) = 2 \cos t \textbf{i}+ 2 \sin t \textbf{j} + \textbf{k}\). Solution From Example 12.4.2, \( \textbf{T}(t) \) \(= \frac{1}{ \sqrt{5} } \left( -2 \sin t\textbf{i} + 2 \cos t\textbf{j} + \textbf{k} \right) \:\:\:\: \) Unit tangent vector \( \textbf{T}^{\prime}(t) \) \(= \frac{1}{ \sqrt{5} } \left( -2 \cos t\textbf{i} - 2 \sin t\textbf{j} \right) \) Because $$ \| \textbf{T}^{\prime}(t) \| = \frac{2}{ \sqrt{5} }, $$ it follows that the principal unit normal vector is \( \textbf{N}(t) \) $$= \frac{ \textbf{T}^{\prime}(t) }{\| \textbf{T}^{\prime}(t) \| }$$ $$=\frac{1}{2} \left(-2 \cos t\textbf{i} -2 \sin t \textbf{j} \right) \:\:\:\:$$ \(= - \cos t \textbf{i} - \sin t \textbf{j}.\) Principal unit normal vector Note that this vector is horizontal and points toward the \(z\)-axis, as shown in Figure 12.4.5.

Tangential and Normal Components of Acceleration

For an object traveling at a constant speed, the velocity and acceleration vectors are perpendicular. This seems reasonable, because the speed would not be constant if any acceleration were acting along the curve. This is verified by noting that

\( \textbf{r}^{\prime \prime}(t) \cdot \textbf{r}^{\prime }(t) = 0 \)

when \( \| \textbf{r}^{\prime }(t) \| \) is a constant, as shown in Theorem 12.2.2, Property 7.

For an object traveling along a curve at a variable speed the velocity and acceleration vectors are not necessarily perpendicular. For example, the acceleration vector for a projectile always points down, no matter where the curve goes.

In general, the acceleration, the tangential component, acts along the curve. While the normal component acts perpendicular to the curve. In order to determine these two components, the unit vectors \( \textbf{T}(t) \) and \( \textbf{N}(t) \), Use the unit vectors \( \textbf{T}(t) \) and \( \textbf{N}(t) \) to determine these components just as \( \textbf{i} \) \( \textbf{j} \) are used in representing vectors in two-dimensions. This is described in Theorem 12.4.1 below.

Theorem 12.4.1 Acceleration Vector

If \( \textbf{r}(t) \) is the position vector for a smooth curve \(C\) and \( \textbf{N}(t) \) exists, then the acceleration vector \( \textbf{a}(t) \) lies in the plane determined by \( \textbf{T}(t) \) and \( \textbf{N}(t) \).
Proof To simplify the notation, write \( \textbf{T} \) for \( \textbf{T}(t) \), \( \textbf{T}^{\prime } \) for \( \textbf{T}^{\prime }(t) \), and so on. Because \( \textbf{T} = \textbf{r}^{\prime } / \| \textbf{r}^{\prime } \| = \textbf{v} / \| \textbf{v} \| \), it follows that

\( \textbf{v} = \| \textbf{v} \| \textbf{T} \).

Differentiating produces

\( \textbf{a} \)	\(= \textbf{v}^{\prime }\)	Product Rule
	$$= \frac{d}{dt} \left[ \| \textbf{v} \| \right] \textbf{T} + \| \textbf{v} \| \textbf{T}^{\prime } $$
	$$= \frac{d}{dt} \left[ \| \textbf{v} \| \right] \textbf{T} + \| \textbf{v} \| \textbf{T}^{\prime} \left( \frac{\| \textbf{T}^{\prime} \|}{\| \textbf{T}^{\prime} \|}\right)$$
	$$= \frac{d}{dt} \left[ \| \textbf{v} \| \right] \textbf{T} + \| \textbf{v} \|\| \textbf{T}^{\prime} \| \textbf{N}. \:\:\:\: $$	\( \color{red}{ \textbf{N}=\textbf{T}^{\prime } / \| \textbf{T}^{\prime } \| }\)

Because \( \textbf{a} \) is written as a linear combination with \( \textbf{T} \) and \( \textbf{N} \), it follow that \( \textbf{a} \) lies in the plane determined by \( \textbf{T} \) and \( \textbf{N} \).

The coefficients for \( \textbf{T} \) and \( \textbf{N} \) are called the tangential and normal components of acceleration and are denoted by

$$ a_{\textbf{T}} = \frac{d}{dt} \left[ \| \textbf{v} \| \right] $$

and \( a_{\textbf{N}} = \| \textbf{v} \| \| \textbf{T}^{\prime} \| \). This can be written as

$$ \textbf{a}(t) = a_{\textbf{T}}\textbf{T}(t) + a_{\textbf{N}}\textbf{N}(t). $$

Theorem 12.4.2 describes some formulas for \( a_{\textbf{N}} \) and \( a_{\textbf{T}} \).

Theorem 12.4.2 Tangential and Normal Components for Acceleration

The tangential and normal components for acceleration are obtained by projecting \(\textbf{a}\) onto \( \textbf{T} \) and \( \textbf{N} \) Figure 12.4.6

If \( \textbf{r}(t) \) is the position vector for a smooth curve \(C\), where \(\textbf{N}(t)\) exists, then the tangential and normal components of acceleration are as follows. $$ a_{\textbf{T}} $$ $$= \frac{d}{dt} \left[ \| \textbf{v} \| \right] $$ $$ = \textbf{a} \cdot \textbf{T}$$ $$= \frac{\textbf{v} \cdot \textbf{a}}{\| \textbf{v} \| }$$ $$ a_{\textbf{N}} $$ $$= \| \textbf{v} \| \| \textbf{T}^{\prime} \| $$ $$ = \textbf{a} \cdot \textbf{N}$$ $$= \frac{ \| \textbf{v} \times \textbf{a} \|}{\| \textbf{v} \| }$$ $$= \sqrt{ \| \textbf{a} \|^{2} - (a_{\textbf{T}})^{2} } $$ Note that \( a_{\textbf{N}} \geqslant 0 \). The normal acceleration component is also called the centripetal component of acceleration. Proof Note that \(a\) lies in the plane \( \textbf{T} \) and \( \textbf{N} \). As shown in Figure 12.4.6, for any time \(t\), the projection components for the acceleration vector onto \( \textbf{T} \) and \( \textbf{N} \) are given by \( a_{\textbf{T}} = \textbf{a} \cdot \textbf{T} \) and \( a_{\textbf{N}} = \textbf{a} \cdot \textbf{N} \), respectively. Because \( \textbf{a} = \textbf{v}^{\prime }\) and \( \textbf{T} = \textbf{v}/\| \textbf{v} \|\) produces $$ a_{\textbf{T}}= \textbf{a} \cdot \textbf{T} = \textbf{T} \cdot \textbf{a}= \frac{\textbf{v} }{\| \textbf{v} \| }\cdot \textbf{a} = \frac{\textbf{v} \cdot \textbf{a}}{\| \textbf{v} \| }. $$

Example 12.4.5 Tangential and Normal Acceleration Components

Find the tangential and normal acceleration components for the position vector given by

\( \textbf{r}(t) = 3t \textbf{i} -t\textbf{j} + t^{2}\textbf{k}.\)

Solution Begin by finding the velocity, speed, and acceleration.

\( \textbf{v}(t) \)	\(= \textbf{r}^{\prime } = 3 \textbf{i} -\textbf{j} + 2t\textbf{k} \)	Velocity vector
\( \| \textbf{v}(t) \| \)	\(= \sqrt{9+1+4t^{2}} = \sqrt{10+4t^{2}} \:\:\:\: \)	Speed
\( \textbf{a}(t) \)	\(=\textbf{r}^{\prime \prime} = 2 \textbf{k} \)	Acceleration vector

By Theorem 12.4.2, the tangential acceleration component is

$$ a_{\textbf{T}} = \frac{\textbf{v} \cdot \textbf{a}}{\| \textbf{v} \| } = \frac{4t}{\sqrt{10+4t^{2}}} \:\:\:\: \color{red}{ \text{ Tangential acceleration component}}$$

and because

$$ \textbf{v} \times \textbf{a} = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \\ 3 & -1 & 2t\\ 0 & 0 & 2 \end{vmatrix} = -2\textbf{i}-6\textbf{j} $$

the normal acceleration component is

$$ a_{\textbf{N}} = \frac{\textbf{v} \times \textbf{a}}{\| \textbf{v} \| } = \frac{ \sqrt{4+36}}{\sqrt{10+4t^{2}}} = \frac{2 \sqrt{10}}{\sqrt{10+4t^{2}}}. \:\:\:\: \color{red}{ \text{ Normal acceleration component}} $$

The alternative formula for \( a_{\textbf{N}} \) also works.

$$ a_{\textbf{N}} = \sqrt{ \| \textbf{a} \|^{2} - (a_{\textbf{T}})^{2}} = \sqrt{ (2)^{2} - \frac{16t^{2}}{\sqrt{10+4t^{2}}} } = \frac{2 \sqrt{10}}{\sqrt{10+4t^{2}}} $$

Example 12.4.6 Finding \(a_{\textbf{T}}\) and \( a_{\textbf{N}} \) for a Circular Helix

The normal acceleration component is equal to the cylinder's radius the about which the helix is spiraling. Figure 12.4.7

Find the tangential and normal acceleration components for the helix given by \( \textbf{r}(t) = b \cos t \textbf{i} + b \sin t\textbf{j} + ct\textbf{k},\: b >0.\) Solution \( \textbf{v}(t) \) \(= \textbf{r}^{\prime}(t) = -b \sin t \textbf{i} + b \cos t\textbf{j} + c\textbf{k}\) Velocity vector \( \| \textbf{v}(t) \| \) \(= \sqrt{ b^{2} \sin^{2} t \textbf{i} + b^{2} \cos^{2} t\textbf{j} + c^{2} } = \sqrt{ b^{2}+c^{2}} \:\:\:\: \) Speed \( \textbf{a}(t) \) \(= \textbf{r}(t) = -b \cos t \textbf{i} - b \sin t\textbf{j} \) Acceleration vector By Theorem 12.4.2, the tangential acceleration component is $$ a_{\textbf{T}} = \frac{\textbf{v} \cdot \textbf{a}}{\| \textbf{v} \| } = \frac{ b^{2} \sin t \cos t - b^{2} \sin t \cos t + 0 }{ \sqrt{ b^{2}+c^{2}}} = 0. $$ In addition, because \( \| \textbf{a} \| = \sqrt{ b^{2} \cos^{2} t + b^{2} \sin^{2} t } = b \) the alternative formula for the normal acceleration component produces $$ a_{\textbf{N}} = \sqrt{ \| \textbf{a} \|^{2} - (a_{\textbf{T}})^{2}} = \sqrt{b^{2} - 0^{2}} = b$$ the normal acceleration component as well. Note that the normal acceleration component is equal to the acceleration magnitude. In other words, because the speed is constant, the acceleration is perpendicular to the velocity, as shown in Figure 12.4.6.

Example 12.4.7 Projectile Motion

Projectile path. Figure 12.4.8

The position vector for the projectile shown in Figure 12.4.8 is \( \textbf{r}(t) = (50 \sqrt{2}t)\textbf{i} + (50 \sqrt{2}t - 16t^{2})\textbf{j}. \:\:\:\: \)Position vector Find the tangential acceleration components when \(t=0, \: 1\), and \(25\sqrt{2}/16 \). Solution \( \textbf{v}(t) \) \(= (50 \sqrt{2}) \textbf{i} + (50 \sqrt{2} -32t)\textbf{j} \) Velocity vector \( \| \textbf{v}(t) \| \) \(= 2\sqrt{ 50^{2} - 16(50)\sqrt{2}t + 16^{2}t^{2} } \:\:\:\: \) Speed \( \textbf{a}(t) \) \(= -32\textbf{j} \) Acceleration vector The tangential acceleration component is $$ a_{\textbf{T}} = \frac{\textbf{v} \cdot \textbf{a}}{\| \textbf{v} \| } = \frac{ -32(50 \sqrt{2} -32t) }{2\sqrt{ 50^{2} - 16(50)\sqrt{2}t + 16^{2}t^{2} } } . $$ Plugging in the specified times yields $$ a_{\textbf{T}}(0) $$ $$= \frac{ -32(50^{2}\sqrt{2})}{100} = -16\sqrt{2} \approx -22.6 $$ $$ a_{\textbf{T}}(1) $$ $$= \frac{ -32(50 \sqrt{2} -32) }{2\sqrt{ 50^{2} - 16(50)\sqrt{2} + 16^{2} } } \approx -15.4 $$ $$ a_{\textbf{T}}\left( \frac{25\sqrt{2}}{16} \right) $$ $$= \frac{ -32(50 \sqrt{2} - 50 \sqrt{2}) }{50 \sqrt{2}} = 0 $$ Figure 12.4.8 shows the maximum height is reached when \(25\sqrt{2}/16 \) and the tangential component is 0. This is reasonable because the direction is horizontal at that point and the tangential acceleration component is equal to the horizontal acceleration component.

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Parent Article: Calculus III 12 Vector-Valued Functions