3.2 Calculus of vector-valued functions (Page 3/11)

Calculus volume 3 Page 3 / 11

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Now for some examples using these properties.

Using the properties of derivatives of vector-valued functions

Given the vector-valued functions

r (t) = (6 t + 8) i + (4 t^{2} + 2 t - 3) j + 5 t k

and

u (t) = (t^{2} - 3) i + (2 t + 4) j + (t^{3} - 3 t) k,

calculate each of the following derivatives using the properties of the derivative of vector-valued functions.

$\frac{d}{d t} [r (t) \cdot u (t)]$
$\frac{d}{d t} [u (t) \times u^{'} (t)]$

We have $r^{'} (t) = 6 i + (8 t + 2) j + 5 k$ and $u^{'} (t) = 2 t i + 2 j + (3 t^{2} - 3) k .$ Therefore, according to property iv.:

$\begin{matrix} \frac{d}{d t} [r (t) \cdot u (t)] & = r^{'} (t) \cdot u (t) + r (t) \cdot u^{'} (t) \\ = (6 i + (8 t + 2) j + 5 k) \cdot ((t^{2} - 3) i + (2 t + 4) j + (t^{3} - 3 t) k) \\ + ((6 t + 8) i + (4 t^{2} + 2 t - 3) j + 5 t k) \cdot (2 t i + 2 j + (3 t^{2} - 3) k) \\ = 6 (t^{2} - 3) + (8 t + 2) (2 t + 4) + 5 (t^{3} - 3 t) \\ + 2 t (6 t + 8) + 2 (4 t^{2} + 2 t - 3) + 5 t (3 t^{2} - 3) \\ = 20 t^{3} + 42 t^{2} + 26 t - 16 . \end{matrix}$
First, we need to adapt property v. for this problem:

$\frac{d}{d t} [u (t) \times u^{'} (t)] = u^{'} (t) \times u^{'} (t) + u (t) \times u″ (t) .$

Recall that the cross product of any vector with itself is zero. Furthermore, $u″ (t)$ represents the second derivative of $u (t) :$

$u″ (t) = \frac{d}{d t} [u^{'} (t)] = \frac{d}{d t} [2 t i + 2 j + (3 t^{2} - 3) k] = 2 i + 6 t k .$

Therefore,

$\begin{matrix} \frac{d}{d t} [u (t) \times u^{'} (t)] & = 0 + ((t^{2} - 3) i + (2 t + 4) j + (t^{3} - 3 t) k) \times (2 i + 6 t k) \\ = | \begin{matrix} i & j & k \\ t^{2} - 3 & 2 t + 4 & t^{3} - 3 t \\ 2 & 0 & 6 t \end{matrix} | \\ = 6 t (2 t + 4) i - (6 t (t^{2} - 3) - 2 (t^{3} - 3 t)) j - 2 (2 t + 4) k \\ = (12 t^{2} + 24 t) i + (12 t - 4 t^{3}) j - (4 t + 8) k . \end{matrix}$

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Given the vector-valued functions $r (t) = cos t i + sin t j - e^{2 t} k$ and $u (t) = t i + sin t j + cos t k,$ calculate $\frac{d}{d t} [r (t) \cdot r^{'} (t)]$ and $\frac{d}{d t} [u (t) \times r (t)] .$

$\frac{d}{d t} [r (t) \cdot r^{'} (t)] = 8 e^{4 t}$

$\begin{array}{l} \frac{d}{d t} [u (t) \times r (t)] \\ = - (e^{2 t} (cos t + 2 sin t) + cos 2 t) i + (e^{2 t} (2 t + 1) - sin 2 t) j + (t cos t + sin t - cos 2 t) k \end{array}$

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Tangent vectors and unit tangent vectors

Recall from the Introduction to Derivatives that the derivative at a point can be interpreted as the slope of the tangent line to the graph at that point. In the case of a vector-valued function, the derivative provides a tangent vector to the curve represented by the function. Consider the vector-valued function $r (t) = cos t i + sin t j .$ The derivative of this function is $r^{'} (t) = - sin t i + cos t j .$ If we substitute the value $t = π / 6$ into both functions we get

r (\frac{π}{6}) = \frac{\sqrt{3}}{2} i + \frac{1}{2} j and r^{'} (\frac{π}{6}) = - \frac{1}{2} i + \frac{\sqrt{3}}{2} j .

The graph of this function appears in [link] , along with the vectors $r (\frac{π}{6})$ and $r^{'} (\frac{π}{6}) .$

This figure is the graph of a circle represented by the vector-valued function r(t) = cost i + sint j. It is a circle centered at the origin with radius of 1, and counter-clockwise orientation. It has a vector from the origin pointing to the curve and labeled r(pi/6). At the same point on the circle there is a tangent vector labeled “r’(pi/6)”. — The tangent line at a point is calculated from the derivative of the vector-valued function $r (t) .$

Notice that the vector $r^{'} (\frac{π}{6})$ is tangent to the circle at the point corresponding to $t = π / 6 .$ This is an example of a tangent vector to the plane curve defined by $r (t) = cos t i + sin t j .$

Definition

Let C be a curve defined by a vector-valued function r, and assume that $r^{'} (t)$ exists when $t = t_{0} .$ A tangent vector v at $t = t_{0}$ is any vector such that, when the tail of the vector is placed at point $r (t_{0})$ on the graph, vector v is tangent to curve C. Vector $r^{'} (t_{0})$ is an example of a tangent vector at point $t = t_{0} .$ Furthermore, assume that $r^{'} (t) \neq 0 .$ The principal unit tangent vector at t is defined to be

T (t) = \frac{r^{'} (t)}{‖ r^{'} (t) ‖},

provided $‖ r^{'} (t) ‖ \neq 0 .$

The unit tangent vector is exactly what it sounds like: a unit vector that is tangent to the curve. To calculate a unit tangent vector, first find the derivative $r^{'} (t) .$ Second, calculate the magnitude of the derivative. The third step is to divide the derivative by its magnitude.

Finding a unit tangent vector

Find the unit tangent vector for each of the following vector-valued functions:

$r (t) = cos t i + sin t j$
$u (t) = (3 t^{2} + 2 t) i + (2 - 4 t^{3}) j + (6 t + 5) k$

$\begin{matrix} First step: & r^{'} (t) & = & −sin t i + cos t j \\ Second step: & ‖ r^{'} (t) ‖ & = & \sqrt{{(− sin t)}^{2} + {(cos t)}^{2}} = 1 \\ Third step: & T (t) & = & \frac{r^{'} (t)}{‖ r^{'} (t) ‖} = \frac{−sin t i + cos t j}{1} = −sin t i + cos t j \end{matrix}$
$\begin{matrix} First step: & u^{'} (t) & = & (6 t + 2) i - 12 t^{2} j + 6 k \\ Second step: & ‖ u^{'} (t) ‖ & = & \sqrt{{(6 t + 2)}^{2} + {(−12 t^{2})}^{2} + 6^{2}} \\ = & \sqrt{144 t^{4} + 36 t^{2} + 24 t + 40} \\ = & 2 \sqrt{36 t^{4} + 9 t^{2} + 6 t + 10} \\ Third step: & T (t) & = & \frac{u^{'} (t)}{‖ u^{'} (t) ‖} = \frac{(6 t + 2) i - 12 t^{2} j + 6 k}{2 \sqrt{36 t^{4} + 9 t^{2} + 6 t + 10}} \\ = & \frac{3 t + 1}{\sqrt{36 t^{4} + 9 t^{2} + 6 t + 10}} i - \frac{6 t^{2}}{\sqrt{36 t^{4} + 9 t^{2} + 6 t + 10}} j + \frac{3}{\sqrt{36 t^{4} + 9 t^{2} + 6 t + 10}} k \end{matrix}$

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Source: OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2

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