Integrals of Vector-Valued Functions
Solved Problems
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Example 7
Find R (t) if \[\mathbf{R}^\prime\left( t \right) = \left\langle {\sin \frac{t}{3},\cos \frac{t}{2}} \right\rangle \] and R (π) = 〈1/2, 1/2〉.
Example 8
Compute the integral \[\int\limits_0^1 {\left\langle {\frac{{2t}}{{1 + {t^2}}},\frac{2}{{1 + {t^2}}}} \right\rangle dt}.\]
Example 9
Compute the integral \[\int\limits_0^{\frac{\pi }{6}} {\left\langle {2\cos t,\sin 2t} \right\rangle dt}.\]
Example 10
A particle starts moving from the origin with the velocity \[\mathbf{v}\left( t \right) = \left\langle {4t,3{t^2} - 1} \right\rangle,\] where \(t\) is measured in seconds and the components are in meters per second. Determine the displacement \(\left| \mathbf{r} \right|\) of the particle in \(t = 2\) seconds.
Example 11
A particle starts moving from the origin with the acceleration \[\mathbf{a}\left( t \right) = \left\langle {6t,4} \right\rangle,\] where \(t\) is measured in seconds and the components are in meters per seconds squared. Determine the displacement \(\left| \mathbf{r} \right|\) of the particle in \(t = 1\) seconds if \(\mathbf{v}\left( 0 \right) = \left\langle {2,2} \right\rangle .\)
Example 7.
Find \(\mathbf{R}\left( t \right)\) if \[\mathbf{R}^\prime\left( t \right) = \left\langle {\sin \frac{t}{3},\cos \frac{t}{2}} \right\rangle \] and \(\mathbf{R}\left( \pi \right) = \left\langle {\frac{1}{2},\frac{1}{2}} \right\rangle .\)
Solution.
Integrating the vector function yields:
We find the vector \(\mathbf{C} = \left\langle {{C_1},{C_2}} \right\rangle \) from the initial condition \(\mathbf{R}\left( \pi \right) = \left\langle {\frac{1}{2},\frac{1}{2}} \right\rangle :\)
Then
The final answer is
Example 8.
Compute the integral \[\int\limits_0^1 {\left\langle {\frac{{2t}}{{1 + {t^2}}},\frac{2}{{1 + {t^2}}}} \right\rangle dt}.\]
Solution.
Integrating component-by-component, we can write:
We evaluate each of the integrals separately. To find the first integral, we make the substitution \(u = 1 + {t^2},\) \(du = 2tdt.\) Then
and
Calculate the second integral:
Thus, the initial integral is represented by the vector
Example 9.
Compute the integral \[\int\limits_0^{\frac{\pi }{6}} {\left\langle {2\cos t,\sin 2t} \right\rangle dt}.\]
Solution.
By integrating componentwise, we obtain:
Example 10.
A particle starts moving from the origin with the velocity \[\mathbf{v}\left( t \right) = \left\langle {4t,3{t^2} - 1} \right\rangle,\] where \(t\) is measured in seconds and the components are in meters per second. Determine the displacement \(\left| \mathbf{r} \right|\) of the particle in \(t = 2\) seconds.
Solution.
Determine the position of the particle in \(t = 2\) seconds by integrating the velocity vector:
Hence, the displacement of the particle is equal
Example 11.
A particle starts moving from the origin with the acceleration \[\mathbf{a}\left( t \right) = \left\langle {6t,4} \right\rangle,\] where \(t\) is measured in seconds and the components are in meters per seconds squared. Determine the displacement \(\left| \mathbf{r} \right|\) of the particle in \(t = 1\) seconds if \(\mathbf{v}\left( 0 \right) = \left\langle {2,2} \right\rangle .\)
Solution.
First we integrate the acceleration vector to obtain the velocity vector:
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 \(\mathbf{v}\left( 0 \right) = \left\langle {2,2} \right\rangle :\)
Hence, the particle's velocity is defined by the vector
Now we integrate the velocity vector to get the position vector:
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
Calculate the coordinates of the particle at \(t = 1:\)
Then the displacement of the particle is equal