3.2 Calculus of vector-valued functions (Page 4/11)

Calculus volume 3 Page 4 / 11

Find the unit tangent vector for the vector-valued function

r (t) = (t^{2} - 3) i + (2 t + 1) j + (t - 2) k .

$T (t) = \frac{2 t}{\sqrt{4 t^{2} + 5}} i + \frac{2}{\sqrt{4 t^{2} + 5}} j + \frac{1}{\sqrt{4 t^{2} + 5}} k$

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Integrals of vector-valued functions

We introduced antiderivatives of real-valued functions in Antiderivatives and definite integrals of real-valued functions in The Definite Integral . Each of these concepts can be extended to vector-valued functions. Also, just as we can calculate the derivative of a vector-valued function by differentiating the component functions separately, we can calculate the antiderivative in the same manner. Furthermore, the Fundamental Theorem of Calculus applies to vector-valued functions as well.

The antiderivative of a vector-valued function appears in applications. For example, if a vector-valued function represents the velocity of an object at time t , then its antiderivative represents position. Or, if the function represents the acceleration of the object at a given time, then the antiderivative represents its velocity.

Definition

Let f, g, and h be integrable real-valued functions over the closed interval $[a, b] .$

The indefinite integral of a vector-valued function $r (t) = f (t) i + g (t) j$ is

$\int [f (t) i + g (t) j] d t = [\int f (t) d t] i + [\int g (t) d t] j .$

The definite integral of a vector-valued function is

$\int_{a}^{b} [f (t) i + g (t) j] d t = [\int_{a}^{b} f (t) d t] i + [\int_{a}^{b} g (t) d t] j .$
The indefinite integral of a vector-valued function $r (t) = f (t) i + g (t) j + h (t) k$ is

$\int [f (t) i + g (t) j + h (t) k] d t = [\int f (t) d t] i + [\int g (t) d t] j + [\int h (t) d t] k .$

The definite integral of the vector-valued function is

$\int_{a}^{b} [f (t) i + g (t) j + h (t) k] d t = [\int_{a}^{b} f (t) d t] i + [\int_{a}^{b} g (t) d t] j + [\int_{a}^{b} h (t) d t] k .$

Since the indefinite integral of a vector-valued function involves indefinite integrals of the component functions, each of these component integrals contains an integration constant. They can all be different. For example, in the two-dimensional case, we can have

\int f (t) d t = F (t) + C_{1} and \int g (t) d t = G (t) + C_{2},

where F and G are antiderivatives of f and g, respectively. Then

\begin{matrix} \int [f (t) i + g (t) j] d t & = [\int f (t) d t] i + [\int g (t) d t] j \\ = (F (t) + C_{1}) i + (G (t) + C_{2}) j \\ = F (t) i + G (t) j + C_{1} i + C_{2} j \\ = F (t) i + G (t) j + C, \end{matrix}

where $C = C_{1} i + C_{2} j .$ Therefore, the integration constant becomes a constant vector.

Integrating vector-valued functions

Calculate each of the following integrals:

$\int [(3 t^{2} + 2 t) i + (3 t - 6) j + (6 t^{3} + 5 t^{2} - 4) k] d t$
$\int [⟨ t, t^{2}, t^{3} ⟩ \times ⟨ t^{3}, t^{2}, t ⟩] d t$
$\int_{0}^{π / 3} [sin 2 t i + tan t j + e^{−2 t} k] d t$

We use the first part of the definition of the integral of a space curve:

$\begin{matrix} \int [(3 t^{2} + 2 t) i + (3 t - 6) j + (6 t^{3} + 5 t^{2} - 4) k] d t \\ = [\int 3 t^{2} + 2 t d t] i + [\int 3 t - 6 d t] j + [\int 6 t^{3} + 5 t^{2} - 4 d t] k \\ = (t^{3} + t^{2}) i + (\frac{3}{2} t^{2} - 6 t) j + (\frac{3}{2} t^{4} + \frac{5}{3} t^{3} - 4 t) k + C . \end{matrix}$
First calculate $⟨ t, t^{2}, t^{3} ⟩ \times ⟨ t^{3}, t^{2}, t ⟩ :$

$\begin{matrix} ⟨ t, t^{2}, t^{3} ⟩ \times ⟨ t^{3}, t^{2}, t ⟩ & = | \begin{matrix} i & j & k \\ t & t^{2} & t^{3} \\ t^{3} & t^{2} & t \end{matrix} | \\ = (t^{2} (t) - t^{3} (t^{2})) i - (t^{2} - t^{3} (t^{3})) j + (t (t^{2}) - t^{2} (t^{3})) k \\ = (t^{3} - t^{5}) i + (t^{6} - t^{2}) j + (t^{3} - t^{5}) k . \end{matrix}$

Next, substitute this back into the integral and integrate:

$\begin{matrix} \int [⟨ t, t^{2}, t^{3} ⟩ \times ⟨ t^{3}, t^{2}, t ⟩] d t & = \int (t^{3} - t^{5}) i + (t^{6} - t^{2}) j + (t^{3} - t^{5}) k d t \\ = (\frac{t^{4}}{4} - \frac{t^{6}}{6}) i + (\frac{t^{7}}{7} - \frac{t^{3}}{3}) j + (\frac{t^{4}}{4} - \frac{t^{6}}{6}) k + C . \end{matrix}$
Use the second part of the definition of the integral of a space curve:

$\begin{matrix} \int_{0}^{π / 3} [sin 2 t i + tan t j + e^{−2 t} k] d t \\ = [\int_{0}^{π / 3} sin 2 t d t] i + [\int_{0}^{π / 3} tan t d t] j + [\int_{0}^{π / 3} e^{−2 t} d t] k \\ = {(- \frac{1}{2} cos 2 t) |}_{0}^{π / 3} i - {(ln (cos t)) |}_{0}^{π / 3} j - {(\frac{1}{2} e^{−2 t}) |}_{0}^{π / 3} k \\ = (- \frac{1}{2} cos \frac{2 π}{3} + \frac{1}{2} cos 0) i - (ln (cos \frac{π}{3}) - ln (cos 0)) j - (\frac{1}{2} e^{−2 π / 3} - \frac{1}{2} e^{−2 (0)}) k \\ = (\frac{1}{4} + \frac{1}{2}) i - (− ln 2) j - (\frac{1}{2} e^{−2 π / 3} - \frac{1}{2}) k \\ = \frac{3}{4} i + (ln 2) j + (\frac{1}{2} - \frac{1}{2} e^{−2 π / 3}) k . \end{matrix}$

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Calculate the following integral:

\int_{1}^{3} [(2 t + 4) i + (3 t^{2} - 4 t) j] d t .

$\int_{1}^{3} [(2 t + 4) i + (3 t^{2} - 4 t) j] d t = 16 i + 10 j$

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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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