Definition of Vector-Valued Functions

A function of the form

\[{\mathbf{r}\left( t \right) = f\left( t \right)\mathbf{i} + g\left( t \right)\mathbf{j} + h\left( t \right)\mathbf{k}\;\;\;}\kern0pt{\text{or}\;\;\;\mathbf{r}\left( t \right) = \left\langle {f\left( t \right),g\left( t \right),h\left( t \right)} \right\rangle }\]

is called a vector-valued function in \(3D\) space, where \(f\left( t \right),\) \(g\left( t \right),\) \(h\left( t \right)\) are the component functions depending on the parameter \(t.\)

We can likewise define a vector-valued function in \(2D\) space (in plane):

\[{\mathbf{r}\left( t \right) = f\left( t \right)\mathbf{i} + g\left( t \right)\mathbf{j}\;\;\;}\kern0pt{\text{or}\;\;\;\mathbf{r}\left( t \right) = \left\langle {f\left( t \right),g\left( t \right)} \right\rangle .}\]

Antiderivatives of Vector-Valued Functions

The vector-valued function \(\mathbf{R}\left( t \right)\) is called an antiderivative of the vector-valued function \(\mathbf{r}\left( t \right)\) whenever

\[{\mathbf{R}^\prime\left( t \right) = \mathbf{r}\left( t \right).}\]

In component form, if \(\mathbf{R}\left( t \right) = \left\langle {F\left( t \right),G\left( t \right),H\left( t \right)} \right\rangle \) and \(\mathbf{r}\left( t \right) = \left\langle {f\left( t \right),g\left( t \right),h\left( t \right)} \right\rangle,\) then

\[{\left\langle {F^\prime\left( t \right),G^\prime\left( t \right),H^\prime\left( t \right)} \right\rangle }={ \left\langle {f\left( t \right),g\left( t \right),h\left( t \right)} \right\rangle .}\]

Note that the vector function

\[{\left\langle {F\left( t \right) + {C_1},\,}\kern0pt{G\left( t \right) + {C_2},\,}\kern0pt{H\left( t \right) + {C_3}} \right\rangle }\]

is also an antiderivative of \(\mathbf{r}\left( t \right)\).

The \(3\) scalar constants \({C_1},{C_2},{C_3}\) produce one vector constant, so the most general antiderivative of \(\mathbf{r}\left( t \right)\) has the form

\[{{\mathbf{R}\left( t \right)} + \mathbf{C},}\]

where \(\mathbf{C} = \left\langle {{C_1},{C_2},{C_3}} \right\rangle .\)

Indefinite Integral of a Vector-Valued Function

If \(\mathbf{R}\left( t \right)\) is an antiderivative of \(\mathbf{r}\left( t \right),\) the indefinite integral of \(\mathbf{r}\left( t \right)\) is

\[{\int {\mathbf{r}\left( t \right)dt} }= {\mathbf{R}\left( t \right) + \mathbf{C},}\]

where \(\mathbf{C}\) is an arbitrary constant vector.

In component form, the indefinite integral is given by

\[{\int {\mathbf{r}\left( t \right)dt} \text{ = }}\kern0pt{ \int {\left\langle {f\left( t \right),g\left( t \right),h\left( t \right)} \right\rangle dt} \text{ = }}\kern0pt{ \left\langle {\int {f\left( t \right)dt} ,\int {g\left( t \right)dt} ,\int {h\left( t \right)dt} } \right\rangle}\]

Definite Integral of a Vector-Valued Function

The definite integral of \(\mathbf{r}\left( t \right)\) on the interval \(\left[ {a,b} \right]\) is defined by

\[{\int\limits_a^b {\mathbf{r}\left( t \right)dt} \text{ =}}\kern0pt{ \int\limits_a^b {\left\langle {f\left( t \right),g\left( t \right),h\left( t \right)} \right\rangle dt} \text{ = }}\kern0pt{ \left\langle {\int\limits_a^b {f\left( t \right)dt} ,\int\limits_a^b {g\left( t \right)dt} ,\int\limits_a^b {h\left( t \right)dt} } \right\rangle}\]

We can extend the Fundamental Theorem of Calculus to vector-valued functions.

If \(\mathbf{r}\left( t \right)\) is continuous on \(\left( {a,b} \right),\) then

\[{\int\limits_a^b {\mathbf{r}\left( t \right)dt} }={ \mathbf{R}\left( b \right) – \mathbf{R}\left( a \right),}\]

where \(\mathbf{R}\left( t \right)\) is any antiderivative of \(\mathbf{r}\left( t \right).\)

Vector-valued integrals obey the same linearity rules as scalar-valued integrals.

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Solved Problems

Click or tap a problem to see the solution.

Solution.

By integrating componentwise, we have

\[{\int\limits_0^{\frac{\pi }{2}} {\left\langle {\sin t,2\cos t,1} \right\rangle dt} }={ \left\langle {{\int\limits_0^{\frac{\pi }{2}} {\sin tdt}} ,{\int\limits_0^{\frac{\pi }{2}} {2\cos tdt}} ,{\int\limits_0^{\frac{\pi }{2}} {1dt}} } \right\rangle }={ \left\langle {\left. { – \cos t} \right|_0^{\frac{\pi }{2}},\left. {2\sin t} \right|_0^{\frac{\pi }{2}},\left. t \right|_0^{\frac{\pi }{2}}} \right\rangle }={ \left\langle {0 + 1,2 – 0,\frac{\pi }{2} – 0} \right\rangle }={ \left\langle {{1},{2},{\frac{\pi }{2}}} \right\rangle .}\]

Solution.

We integrate on a component-by-component basis:

\[{I = \int {\left( {{{\sec }^2}t\mathbf{i} + \ln t\mathbf{j}} \right)dt} }={ \left( {\int {{{\sec }^2}tdt} } \right)\mathbf{i} }+{ \left( {\int {\ln td} t} \right)\mathbf{j}.}\]

The first integral is given by

\[\int {{{\sec }^2}tdt} = \tan t.\]

The second integral can be computed using integration by parts:

\[{\int {\ln td} t }={ \left[ {\begin{array}{*{20}{l}} {u = \ln t}\\ {dv = dt}\\ {du = \frac{1}{t}dt}\\ {v = t} \end{array}} \right] }={ t\ln t – \int {t \cdot \frac{1}{t}dt} }={ t\ln t – \int {dt} }={ t\ln t – t }={ t\left( {\ln t – 1} \right).}\]

Thus, the given integral is equal to

\[{I }={ \tan t\mathbf{i} + t\left( {\ln t – 1} \right)\mathbf{j} }+{ \mathbf{C},}\]

where \(\mathbf{C} = {C_1}\mathbf{i} + {C_2}\mathbf{j}\) is an arbitrary constant vector.

Solution.

Integrating on a component-by-component basis yields:

\[{\int {\left( {\frac{1}{{{t^2}}}\mathbf{i} + \frac{1}{{{t^3}}}\mathbf{j} + t\mathbf{k}} \right)dt} }={ \left( {\int {\frac{{dt}}{{{t^2}}}} } \right)\mathbf{i} }+{ \left( {\int {\frac{{dt}}{{{t^3}}}} } \right)\mathbf{j} }+{ \left( {\int {tdt} } \right)\mathbf{k} }={ \left( {\int {{t^{ – 2}}dt} } \right)\mathbf{i} }+{ \left( {\int {{t^{ – 3}}dt} } \right)\mathbf{j} }+{ \left( {\int {tdt} } \right)\mathbf{k} }={ \frac{{{t^{ – 1}}}}{{\left( { – 1} \right)}}\mathbf{i} }+{ \frac{{{t^{ – 2}}}}{{\left( { – 2} \right)}}\mathbf{j} }+{ \frac{{{t^2}}}{2}\mathbf{k} + \mathbf{C} }={ – \frac{1}{t}\mathbf{i} }-{ \frac{1}{{2{t^2}}}\mathbf{j} }+{ \frac{{{t^2}}}{2}\mathbf{k} }+{ \mathbf{C},}\]

where \(\mathbf{C} = {C_1}\mathbf{i} + {C_2}\mathbf{j}\) is a constant vector.

Solution.

Integrating componentwise yields:

\[{I }={ \int {\left\langle {4\cos 2t,4t{e^{{t^2}}},2t + 3{t^2}} \right\rangle dt} \text{ = }}\kern0pt{ \left\langle {\int {4\cos 2tdt} ,}\kern0pt{\int {4t{e^{{t^2}}}dt} ,}\kern0pt{\int {\left( {2t + 3{t^2}} \right)dt} } \right\rangle }\]

We evaluate each integral separately.

The first integral is given by

\[{\int {4\cos 2tdt} }={ 4 \cdot \frac{{\sin 2t}}{2} + {C_1} }={ 2\sin 2t + {C_1}.}\]

To compute the second integral, we make the substitution \(u = {t^2},\) \(du = 2tdt.\) Then

\[{\int {4t{e^{{t^2}}}dt} }={ 2\int {{e^u}du} }={ 2{e^u} + {C_2} }={ 2{e^{{t^2}}} + {C_2}.}\]

The third integral is pretty straightforward:

\[{\int {\left( {2t + 3{t^2}} \right)dt} }={ {t^2} + {t^3} + {C_3}.}\]

Thus, the initial integral is equal

\[{I }={ \left\langle {2\sin 2t + {C_1},\,}\kern0pt{2{e^{{t^2}}} + {C_2},\,}\kern0pt{{t^2} + {t^3} + {C_3}} \right\rangle }={ \left\langle {2\sin 2t,2{e^{{t^2}}},{t^2} + {t^3}} \right\rangle }+{ \left\langle {{C_1},{C_2},{C_3}} \right\rangle }={ \left\langle {2\sin 2t,2{e^{{t^2}}},{t^2} + {t^3}} \right\rangle }+{ \mathbf{C},}\]

where \(\mathbf{C} = \left\langle {{C_1},{C_2},{C_3}} \right\rangle \) is an arbitrary constant vector.

Solution.

We integrate component-by-component:

\[{\int {\left\langle {\frac{1}{t},4{t^3},\sqrt t } \right\rangle dt} }={ \left\langle {\int {\frac{{dt}}{t}} ,\int {4{t^3}dt} ,\int {\sqrt t dt} } \right\rangle }={ \left\langle {\ln t,{t^4},\frac{{2\sqrt {{t^3}} }}{3}} \right\rangle }+{ \left\langle {{C_1},{C_2},{C_3}} \right\rangle }={ \left\langle {\ln t,3{t^4},\frac{{3\sqrt {{t^3}} }}{2}} \right\rangle }+{ \mathbf{C},}\]

where \(\mathbf{C} = \left\langle {{C_1},{C_2},{C_3}} \right\rangle \) is any number vector.

Solution.

First we integrate the vector-valued function:

\[{\mathbf{R}\left( t \right) }={ \int {\left\langle {1 + 2t,2{e^{2t}}} \right\rangle dt} }={ \left\langle {\int {\left( {1 + 2t} \right)dt} ,\int {2{e^{2t}}dt} } \right\rangle }={ \left\langle {t + {t^2},{e^{2t}}} \right\rangle }+{ \left\langle {{C_1},{C_2}} \right\rangle }={ \left\langle {t + {t^2},{e^{2t}}} \right\rangle + \mathbf{C}.}\]

We determine the vector \(\mathbf{C}\) from the initial condition \(\mathbf{R}\left( 0 \right) = \left\langle {1,3} \right\rangle :\)

\[{\mathbf{R}\left( 0 \right) }={ \left\langle {0 + {0^2},{e^0}} \right\rangle + \mathbf{C} }={ \left\langle {0,1} \right\rangle + \mathbf{C} }={ \left\langle {1,3} \right\rangle .}\]

Hence

\[{\mathbf{C} }={ \left\langle {1,3} \right\rangle – \left\langle {0,1} \right\rangle }={ \left\langle {1,2} \right\rangle .}\]

The answer is given by

\[{\mathbf{R}\left( t \right) }={ \left\langle {t + {t^2},{e^{2t}}} \right\rangle }+{ \left\langle {1,2} \right\rangle .}\]

Solution.

Integrating the vector function yields:

\[{\mathbf{R}\left( t \right) }={ \int {\left\langle {\sin \frac{t}{3},\cos \frac{t}{2}} \right\rangle dt} }={ \left\langle {\int {\sin \frac{t}{3}dt} ,\int {\cos \frac{t}{2}dt} } \right\rangle }={ \left\langle { – 3\cos \frac{t}{3},2\sin \frac{t}{2}} \right\rangle }+{ \mathbf{C}.}\]

We find the vector \(\mathbf{C} = \left\langle {{C_1},{C_2}} \right\rangle \) from the initial condition \(\mathbf{R}\left( \pi \right) = \left\langle {\large{\frac{1}{2}}\normalsize,\large{\frac{1}{2}}\normalsize} \right\rangle :\)

\[{\mathbf{R}\left( \pi \right) }={ \left\langle { – 3\cos \frac{\pi }{3},2\sin \frac{\pi }{2}} \right\rangle + \mathbf{C} }={ \left\langle { – 3 \cdot \frac{1}{2},2 \cdot 1} \right\rangle + \mathbf{C} }={ \left\langle { – \frac{3}{2},2} \right\rangle + \mathbf{C} }={ \left\langle {\frac{1}{2},\frac{1}{2}} \right\rangle .}\]

Then

\[{\mathbf{C} }={ \left\langle {\frac{1}{2},\frac{1}{2}} \right\rangle – \left\langle { – \frac{3}{2},2} \right\rangle }={ \left\langle {2, – \frac{3}{2}} \right\rangle .}\]

The final answer is

\[{\mathbf{R}\left( t \right) }={ \left\langle { – 3\cos \frac{t}{3},2\sin \frac{t}{2}} \right\rangle }+{ \left\langle {2, – \frac{3}{2}} \right\rangle .}\]

Solution.

Integrating component-by-component, we can write:

\[{\int\limits_0^1 {\left\langle {\frac{{2t}}{{1 + {t^2}}},\frac{2}{{1 + {t^2}}}} \right\rangle dt} }={ \left\langle {\int\limits_0^1 {\frac{{2tdt}}{{1 + {t^2}}}} ,\int\limits_0^1 {\frac{{2dt}}{{1 + {t^2}}}} } \right\rangle .}\]

We evaluate each of the integrals separately. To find the first integral, we make the substitution \(u = 1 + {t^2},\) \(du = 2tdt.\) Then

\[{\int {\frac{{2tdt}}{{1 + {t^2}}}} }={ \int {\frac{{du}}{u}} }={ \ln \left| u \right| }={ \ln \left( {1 + {t^2}} \right),}\]

and

\[{\int\limits_0^1 {\frac{{2tdt}}{{1 + {t^2}}}} }={ \left. {\ln \left( {1 + {t^2}} \right)} \right|_0^1 }={ \ln 2 – \ln 0 }={ \ln 2.}\]

Calculate the second integral:

\[{\int\limits_0^1 {\frac{{2dt}}{{1 + {t^2}}}} }={ \left. {2\arctan t} \right|_0^1 }={ 2\left( {\arctan 1 – \arctan 0} \right) }={ 2\left( {\frac{\pi }{4} – 0} \right) }={ \frac{\pi }{2}.}\]

Thus, the initial integral is represented by the vector

\[{\int\limits_0^1 {\left\langle {\frac{{2t}}{{1 + {t^2}}},\frac{2}{{1 + {t^2}}}} \right\rangle dt} }={ \left\langle {\ln 2,\pi } \right\rangle .}\]

Solution.

By integrating componentwise, we obtain:

\[{\int\limits_0^{\frac{\pi }{6}} {\left\langle {2\cos t,\sin 2t} \right\rangle dt} }={ \left\langle {\int\limits_0^{\frac{\pi }{6}} {2\cos tdt} ,\int\limits_0^{\frac{\pi }{6}} {\sin 2tdt} } \right\rangle }={ \left\langle {\left. {2\sin t} \right|_0^{\frac{\pi }{6}}, – \left. {\frac{{\cos 2t}}{2}} \right|_0^{\frac{\pi }{6}}} \right\rangle }={ \left\langle {2\left( {\frac{1}{2} – 0} \right), – \frac{1}{2}\left( {\frac{1}{2} – 1} \right)} \right\rangle }={ \left\langle {1,\frac{1}{4}} \right\rangle .}\]

Solution.

Determine the position of the particle in \(t = 2\) seconds by integrating the velocity vector:

\[{\mathbf{r}\left( t \right) = \int\limits_0^2 {\mathbf{v}\left( t \right)dt} }={ \int\limits_0^2 {\left\langle {4t,3{t^2} – 1} \right\rangle dt} }={ \left\langle {\int\limits_0^2 {4tdt} ,\int\limits_0^2 {\left( {3{t^2} – 1} \right)dt} } \right\rangle }={ \left\langle {\left. {2{t^2}} \right|_0^2,\left. {{t^3} – t} \right|_0^2} \right\rangle }={ \left\langle {8,8 – 2} \right\rangle }={ \left\langle {8,6} \right\rangle .}\]

Hence, the displacement of the particle is equal

\[{\left| {\mathbf{r}\left( t \right)} \right| }={ \sqrt {{8^2} + {6^2}} }={ 10\,\text{m}.}\]

Solution.

First we integrate the acceleration vector to obtain the velocity vector:

\[{\mathbf{v}\left( t \right) }={ \int {\mathbf{a}\left( t \right)dt} }={ \int {\left\langle {6t,4} \right\rangle dt} }={ \left\langle {3{t^2} + {k_1},4t + {k_2}} \right\rangle ,}\]

where \(\mathbf{K} = \left\langle {{k_1},{k_2}} \right\rangle \) is a constant vector depending on the initial conditions. We determine this vector from the initial condition \(\mathbf{v}\left( 0 \right) = \left\langle {2,2} \right\rangle :\)

\[{\mathbf{v}\left( 0 \right) }={ \left\langle {3{t^2} + {k_1},4t + {k_2}} \right\rangle }={ \left\langle {0 + {k_1},0 + {k_2}} \right\rangle }={ \left\langle {{k_1},{k_2}} \right\rangle }={ \left\langle {2,2} \right\rangle .}\]

Hence, the particle’s velocity is defined by the vector

\[{\mathbf{v}\left( t \right) }={ \left\langle {3{t^2} + 2,4t + 2} \right\rangle .}\]

Now we integrate the velocity vector to get the position vector:

\[{\mathbf{r}\left( t \right) }={ \int {\mathbf{v}\left( t \right)dt} }={ \int {\left\langle {3{t^2} + 2,4t + 2} \right\rangle dt} }={ \left\langle {{t^3} + 2t + {m_1},\,}\kern0pt{2{t^2} + 2t + {m_2}} \right\rangle .}\]

The constant vector \(\mathbf{M} = \left\langle {{m_1},{m_2}} \right\rangle \) also depends on the initial condition. As \(\mathbf{r}\left( 0 \right) = \left\langle {0,0} \right\rangle ,\) we have \(\mathbf{M} = \left\langle {{0},{0}} \right\rangle .\)

So the position vector is given by

\[{\mathbf{r}\left( t \right) }={ \left\langle {{t^3} + 2t,2{t^2} + 2t} \right\rangle .}\]

Calculate the coordinates of the particle at \(t = 1:\)

\[{\mathbf{r}\left( {t = 1} \right) }={ \left\langle {{1^3} + 2 \cdot 1,2 \cdot {1^2} + 2 \cdot 1} \right\rangle }={ \left\langle {3,4} \right\rangle .}\]

Then the displacement of the particle is equal

\[{\left| {\mathbf{r}\left( {t = 1} \right)} \right| }={ \sqrt {{3^2} + {4^2}} }={ 5\,\text{m}.}\]