Calculus

Differentiation of Functions

Differentiation of Functions Logo

Derivatives of Vector-Valued Functions

  • Definition of Vector-Valued Functions

    A vector-valued function of one variable in Cartesian \(3D\) space has the form

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

    where \(f\left( t \right),\) \(g\left( t \right),\) \(h\left( t \right)\) are called the component functions.

    Similarly, a vector-valued function in in Cartesian \(2D\) space is given by

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

    Limits and Continuity of Vector-Valued Functions

    Suppose that \(\mathbf{r}\left( t \right) = \left\langle {f\left( t \right),g\left( t \right),h\left( t \right)} \right\rangle.\) The limit of \(\mathbf{r}\left( t \right)\) as \(t\) approaches \(a\) is given by

    \[{\mathop {\lim }\limits_{t \to a} \mathbf{r}\left( t \right) }={ \mathop {\lim }\limits_{t \to a} \left\langle {f\left( t \right),g\left( t \right),h\left( t \right)} \right\rangle }={ \left\langle {\mathop {\lim }\limits_{t \to a} f\left( t \right),\mathop {\lim }\limits_{t \to a} g\left( t \right),\mathop {\lim }\limits_{t \to a} h\left( t \right)} \right\rangle ,}\]

    provided the limit of the component functions exist.

    The vector-valued function \(\mathbf{r}\left( t \right)\) is continuous at \(t = a\) if

    \[{\mathop {\lim }\limits_{t \to a} \mathbf{r}\left( t \right) = \mathbf{r}\left( a \right).}\]

    Derivative of a Vector-Valued Function

    The derivative \(\mathbf{r}^\prime\left( t \right)\) of the vector-valued function \(\mathbf{r}\left( t \right)\) is defined by

    \[{\frac{{d\mathbf{r}}}{{dt}} = \mathbf{r}^\prime\left( t \right) }={ \mathop {\lim }\limits_{\Delta t \to 0} \frac{{\mathbf{r}\left( {t + \Delta t} \right) – \mathbf{r}\left( t \right)}}{{\Delta t}}}\]

    for any values of \(t\) for which the limit exists.

    The vector \(\mathbf{r}^\prime\left( t \right)\) is called the tangent vector to the curve defined by \(\mathbf{r}.\)

    If \(\mathbf{r}\left( t \right) = \left\langle {f\left( t \right),g\left( t \right),h\left( t \right)} \right\rangle\) where \(f, g\) and \(h\) are differentiable functions, then

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

    Thus, we can differentiate vector-valued functions by differentiating their component functions.

    Physical Interpretation

    If \(\mathbf{r}\left( t \right)\) represents the position of a particle, then the derivative is the velocity of the particle:

    \[{\mathbf{v}\left( t \right) = \frac{{d\mathbf{r}}}{{dt}} = \mathbf{r}^\prime\left( t \right).}\]

    The speed of the particle is the magnitude of the velocity vector:

    \[{\left\| {\mathbf{v}\left( t \right)} \right\| \text{ = }}\kern0pt{ \sqrt {{{\left( {f^\prime\left( t \right)} \right)}^2} + {{\left( {g^\prime\left( t \right)} \right)}^2} + {{\left( {h^\prime\left( t \right)} \right)}^2}.} }\]

    In a similar way, the derivative of the velocity is the acceleration:

    \[{\mathbf{a}\left( t \right) = \frac{{d\mathbf{v}}}{{dt}} = \mathbf{v}^\prime\left( t \right) = \mathbf{r}^{\prime\prime}\left( t \right).}\]


  • Solved Problems

    Click a problem to see the solution.

    Example 1

    A particle moves along the curve \[{\mathbf{r}\left( t \right) = \left( {{t^2} – 1} \right)\mathbf{i} + \left( {2{t^3} + 3} \right)\mathbf{j} }+{ \left( {3t – 4} \right)\mathbf{k}.}\] Find the speed of the particle at \(t = 1.\)

    Example 2

    Compute the derivative of the vector-valued function \(\mathbf{r}\left( t \right) = \left\langle {\sin 2t,{e^{{t^2}}}} \right\rangle .\)

    Example 3

    Evaluate the derivative of the function \(\mathbf{r}\left( t \right) = \left\langle {\ln t,3t + 1,{t^2}} \right\rangle \) at \(t = 1.\)

    Example 4

    Differentiate the vector-valued function \(\mathbf{r}\left( t \right) = \left\langle {\sin \left( {{t^2}} \right),4{t^2} – t} \right\rangle .\)

    Example 5

    Evaluate the derivative of the function \(\mathbf{r}\left( t \right) = \left\langle {\sin 2t,\cos t,2\sec t} \right\rangle \) at \(t = \large{\frac{\pi }{6}}\normalsize.\)

    Example 6

    Two straight lines are given by the vector functions \({\mathbf{r}_1}\left( t \right) = \left\langle {t,t – 7} \right\rangle \) and \({\mathbf{r}_2}\left( t \right) = \left\langle {u – 1,2 – u} \right\rangle .\) Find the angle \(\alpha\) between the lines.

    Example 7

    A particle moves along the curve\[\mathbf{r}\left( t \right) = \left\langle {\frac{{2t}}{{1 + {t^2}}},\frac{{1 – {t^2}}}{{1 + {t^2}}}} \right\rangle .\]Find the speed of the particle at \(t = 1.\)

    Example 8

    A particle moves along the curve \(\mathbf{r}\left( t \right) = \left\langle {\arctan t,\sqrt {t + 1} } \right\rangle .\) Find the speed of the particle at \(t = 3.\)

    Example 9

    Two plane curves are defined by the vector functions \({\mathbf{r}_1}\left( t \right) = \left\langle {t,2t – 4} \right\rangle \) and \({\mathbf{r}_2}\left( u \right) = \left\langle {2u,{u^2}} \right\rangle .\) Find the angle \(\alpha\) between the curves at the point of their intersection.

    Example 10

    A particle moves along the curve \(\mathbf{r}\left( t \right) = \left\langle {{t^2},{t^2} – 4t,t} \right\rangle .\) Find the minimum speed of the particle.

    Example 11

    The position of a particle is defined by the vector \(\mathbf{r}\left( t \right) = \left\langle {\sin t,\cos 2t} \right\rangle .\) Find the maximum speed of the particle.

    Example 1.

    A particle moves along the curve \[{\mathbf{r}\left( t \right) = \left( {{t^2} – 1} \right)\mathbf{i} }+{ \left( {2{t^3} + 3} \right)\mathbf{j} + \left( {3t – 4} \right)\mathbf{k}.}\] Find the speed of the particle at \(t = 1.\)

    Solution.

    We differentiate the vector function \(\mathbf{r}\left( t \right)\) to find the velocity:

    \[{{\mathbf{r}^\prime\left( t \right) = \mathbf{v}\left( t \right) }={ 2t\mathbf{i} + 6{t^2}\mathbf{j} + 3\mathbf{k}.}}\]

    Compute the speed of the particle at \(t = 1:\)

    \[{\left\| {\mathbf{v}\left( t \right)} \right\| }={ \sqrt {{2^2} + {6^2} + {3^2}} }={ \sqrt {49} }={ 7.}\]

    Example 2.

    Compute the derivative of the vector-valued function \(\mathbf{r}\left( t \right) = \left\langle {\sin 2t,{e^{{t^2}}}} \right\rangle .\)

    Solution.

    We differentiate the vector function on a component-by-component basis using the chain rule. This yields:

    \[{\left( {\sin 2t} \right)^\prime = \cos 2t \cdot \left( {2t} \right)^\prime }={ 2\cos 2t;}\]

    \[\left( {{e^{{t^2}}}} \right)^\prime = {e^{{t^2}}} \cdot \left( {{t^2}} \right)^\prime = 2t{e^{{t^2}}}.\]

    So the derivative is given by

    \[\mathbf{r}^\prime\left( t \right) = \left\langle {2\cos 2t,2t{e^{{t^2}}}} \right\rangle .\]

    Example 3.

    Evaluate the derivative of the function \(\mathbf{r}\left( t \right) = \left\langle {\ln t,3t + 1,{t^2}} \right\rangle \) at \(t = 1.\)

    Solution.

    We differentiate each component of the vector function. This yields:

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

    Substitute the parameter value \(t = 1:\)

    \[{\mathbf{r}^\prime\left( 1 \right) = \left\langle {1,3,2} \right\rangle }={ \mathbf{i} + 3\mathbf{j} + 2\mathbf{k}.}\]

    Example 4.

    Differentiate the vector-valued function \(\mathbf{r}\left( t \right) = \left\langle {\sin \left( {{t^2}} \right),4{t^2} – t} \right\rangle .\)

    Solution.

    We take the derivative of each component of the function:

    \[{\left( {\sin \left( {{t^2}} \right)} \right)^\prime = \cos \left( {{t^2}} \right) \cdot \left( {{t^2}} \right)^\prime }={ 2t\cos \left( {{t^2}} \right);}\]

    \[\left( {4{t^2} – t} \right)^\prime = 8t – 1.\]

    Hence, the derivative of the vector function is given by

    \[\mathbf{r}^\prime\left( t \right) = \left\langle {2t\cos \left( {{t^2}} \right),8t – 1} \right\rangle .\]

    Example 5.

    Evaluate the derivative of the function \(\mathbf{r}\left( t \right) = \left\langle {\sin 2t,\cos t,2\sec t} \right\rangle \) at \(t = \large{\frac{\pi }{6}}\normalsize.\)

    Solution.

    We differentiate each component of the vector function:

    \[\left( {\sin 2t} \right)^\prime = 2\cos 2t;\]

    \[\left( {\cos t} \right)^\prime = – \sin t;\]

    \[{\left( {2\sec t} \right)^\prime = \left( {\frac{2}{{\cos t}}} \right)^\prime }={ – \frac{2}{{{{\cos }^2}t}} \cdot \left( { – \sin t} \right) }={ \frac{{2\sin t}}{{{{\cos }^2}t}}.}\]

    So the derivative is written as

    \[\mathbf{r}^\prime\left( t \right) = \left\langle {2\cos 2t, – \sin t,\frac{{2\sin t}}{{{{\cos }^2}t}}} \right\rangle .\]

    When \(t = \large{\frac{\pi }{6}}\normalsize,\) we obtain:

    \[{\mathbf{r}^\prime\left( {\frac{\pi }{6}} \right) \text{ = }}\kern0pt{\left\langle {2\cos \frac{\pi }{3}, – \sin \frac{\pi }{6},\frac{{2\sin \frac{\pi }{6}}}{{{{\cos }^2}\frac{\pi }{6}}}} \right\rangle }={ \left\langle {2 \cdot \frac{1}{2}, – \frac{1}{2},\frac{{2 \cdot \frac{1}{2}}}{{{{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}}} \right\rangle }={ \left\langle {1, – \frac{1}{2},\frac{1}{{\frac{3}{4}}}} \right\rangle }={ \left\langle {1, – \frac{1}{2},\frac{4}{3}} \right\rangle .}\]

    Example 6.

    Two straight lines are given by the vector functions \({\mathbf{r}_1}\left( t \right) = \left\langle {t,t – 7} \right\rangle \) and \({\mathbf{r}_2}\left( t \right) = \left\langle {u – 1,2 – u} \right\rangle .\) Find the angle \(\alpha\) between the lines.

    Solution.

    To determine the direction of the straight lines, we compute their tangent vectors:

    \[{{\mathbf{r}_1^\prime}\left( t \right) = \left\langle {t^\prime,\left( {t – 7} \right)^\prime} \right\rangle }={ \left\langle {1,1} \right\rangle ;}\]

    \[{{\mathbf{r}_2}\left( u \right) = \left\langle {\left( {u – 1} \right)^\prime,\left( {2 – u} \right)^\prime} \right\rangle }={ \left\langle {1, – 1} \right\rangle .}\]

    The angle \(\alpha\) between the lines is the angle between the corresponding tangent vectors. Using the scalar (dot) product, we have

    \[{\cos \alpha = \frac{{{x_1}{x_2} + {y_1}{y_2}}}{{\sqrt {{x_1} + {y_1}} \sqrt {{x_2} + {y_2}} }} }={ \frac{{1 \cdot 1 + 1 \cdot \left( { – 1} \right)}}{{\sqrt {{1^2} + {1^2}} \sqrt {{1^2} + {{\left( { – 1} \right)}^2}} }} }={ \frac{0}{{\sqrt 2 \sqrt 2 }} }={ 0.}\]

    Hence, \(\alpha = \large{\frac{\pi }{2}}\normalsize,\) that is the lines are perpendicular to each other.

    Example 7.

    A particle moves along the curve \[\mathbf{r}\left( t \right) = \left\langle {\frac{{2t}}{{1 + {t^2}}},\frac{{1 – {t^2}}}{{1 + {t^2}}}} \right\rangle .\] Find the speed of the particle at \(t = 1.\)

    Solution.

    First we need to find the velocity vector. Differentiate each component of the position vector \(\mathbf{r}\left( t \right):\)

    \[{\left( {\frac{{2t}}{{1 + {t^2}}}} \right)^\prime = \frac{{2 \cdot \left( {1 + {t^2}} \right) – 2t \cdot 2t}}{{{{\left( {1 + {t^2}} \right)}^2}}} }={ \frac{{2 + 2{t^2} – 4{t^2}}}{{{{\left( {1 + {t^2}} \right)}^2}}} }={ \frac{{2\left( {1 – {t^2}} \right)}}{{{{\left( {1 + {t^2}} \right)}^2}}};}\]

    \[\require{cancel}{\left( {\frac{{1 – {t^2}}}{{1 + {t^2}}}} \right)^\prime \text{ = }}\kern0pt{\frac{{\left( { – 2t} \right) \cdot \left( {1 + {t^2}} \right) – \left( {1 – {t^2}} \right) \cdot 2t}}{{{{\left( {1 + {t^2}} \right)}^2}}} }={ \frac{{ – 2t – \cancel{2{t^3}} – 2t + \cancel{2{t^3}}}}{{{{\left( {1 + {t^2}} \right)}^2}}} }={ \frac{{ – 4t}}{{{{\left( {1 + {t^2}} \right)}^2}}}.}\]

    Hence, the velocity vector is

    \[{\mathbf{v}\left( t \right) \text{ = }}\kern0pt{\left\langle {\frac{{2\left( {1 – {t^2}} \right)}}{{{{\left( {1 + {t^2}} \right)}^2}}},\frac{{ – 4t}}{{{{\left( {1 + {t^2}} \right)}^2}}}} \right\rangle}\]

    or

    \[\mathbf{v}\left( t \right) = \frac{{2\left( {1 – {t^2}} \right)}}{{{{\left( {1 + {t^2}} \right)}^2}}}\mathbf{i} – \frac{{4t}}{{{{\left( {1 + {t^2}} \right)}^2}}}\mathbf{j}.\]

    At \(t = 1,\) we obtain

    \[{\mathbf{v}\left( 1 \right) = \frac{{2\left( {1 – {1^2}} \right)}}{{{{\left( {1 + {1^2}} \right)}^2}}}\mathbf{i} – \frac{{4 \cdot 1}}{{{{\left( {1 + {1^2}} \right)}^2}}}\mathbf{j} }={ 0 \cdot \mathbf{i} – 1 \cdot \mathbf{j} }={ – \mathbf{j}.}\]

    The speed of the particle is the magnitude of the velocity vector. Then

    \[\left| {\mathbf{v}\left( 1 \right)} \right| = \sqrt {{0^2} + {{\left( { – 1} \right)}^2}} = 1.\]

    Example 8.

    A particle moves along the curve \(\mathbf{r}\left( t \right) = \left\langle {\arctan t,\sqrt {t + 1} } \right\rangle .\) Find the speed of the particle at \(t = 3.\)

    Solution.

    We differentiate the function \(\mathbf{r}\left( t \right)\) to find the velocity vector:

    \[{\mathbf{v}\left( t \right) = \mathbf{r}^\prime\left( t \right) = \left\langle {\left( {\arctan t} \right)^\prime,\left( {\sqrt {t + 1} } \right)^\prime} \right\rangle }={ \left\langle {\frac{1}{{1 + {t^2}}},\frac{1}{{2\sqrt {t + 1} }}} \right\rangle .}\]

    When \(t = 3,\) the velocity is given by

    \[{\mathbf{v}\left( 3 \right) = \left\langle {\frac{1}{{1 + {3^2}}},\frac{1}{{2\sqrt {3 + 1} }}} \right\rangle }={ \left\langle {\frac{1}{{10}},\frac{1}{4}} \right\rangle .}\]

    Calculate the speed of the particle at that moment:

    \[{\left| {\mathbf{v}\left( 3 \right)} \right| = \sqrt {{{\left( {\frac{1}{{10}}} \right)}^2} + {{\left( {\frac{1}{4}} \right)}^2}} }={ \sqrt {\frac{1}{{100}} + \frac{1}{{16}}} }={ \sqrt {\frac{4}{{400}} + \frac{{25}}{{400}}} }={ \frac{{\sqrt {29} }}{{20}}.}\]

    Example 9.

    Two plane curves are defined by the vector functions \({\mathbf{r}_1}\left( t \right) = \left\langle {t,2t – 4} \right\rangle \) and \({\mathbf{r}_2}\left( u \right) = \left\langle {2u,{u^2}} \right\rangle .\) Find the angle \(\alpha\) between the curves at the point of their intersection.

    Solution.

    We first determine the coordinates of the point of intersection of these curves by solving the following system of equations:

    \[{\left\{ \begin{array}{l} t = 2u\\ 2t – 4 = {u^2} \end{array} \right.,}\;\; \Rightarrow {\left\{ \begin{array}{l} t = 2u\\ 4u – 4 = {u^2} \end{array} \right.,}\;\; \Rightarrow {\left\{ \begin{array}{l} {u^2} – 4u + 4 = 0\\ t = 2u \end{array} \right.,}\;\; \Rightarrow {\left\{ \begin{array}{l} {\left( {u – 2} \right)^2} = 0\\ t = 2u \end{array} \right.,}\;\; \Rightarrow {\left\{ \begin{array}{l} u = 2\\ t = 4 \end{array} \right..}\]

    Now we differentiate both these functions and find the tangent vectors at the point of intersection.

    \[{{\mathbf{r}_1^\prime}\left( t \right) = \left\langle {t^\prime,\left( {2t – 4} \right)^\prime} \right\rangle }={ \left\langle {1,2} \right\rangle ;}\]

    \[{{\mathbf{r}_2^\prime}\left( u \right) = \left\langle {\left( {2u} \right)^\prime,\left( {{u^2}} \right)^\prime} \right\rangle }={ \left\langle {2,2u} \right\rangle .}\]

    At the point of intersection where \(t = 4\) and \(u = 2\), the tangent vectors are equal to

    \[{{\mathbf{r}_1^\prime}\left( t \right) = \left\langle {1,2} \right\rangle ,\;\;}\kern0pt{{\mathbf{r}_2^\prime}\left( u \right) = \left\langle {2,4} \right\rangle .}\]

    The angle \(\alpha\) between the curves is the angle between the corresponding tangent lines. Using the scalar (dot) product, we get

    \[{\cos \alpha = \frac{{{x_1}{x_2} + {y_1}{y_2}}}{{\sqrt {{x_1^2} + {y_1^2}} \sqrt {{x_2^2} + {y_2^2}} }} }={ \frac{{1 \cdot 2 + 2 \cdot 4}}{{\sqrt {{1^2} + {2^2}} \sqrt {{2^2} + {4^2}} }} }={ \frac{{10}}{{\sqrt 5 \sqrt {20} }} }={ \frac{{10}}{{\sqrt {100} }} = 1.}\]

    Hence, \(\alpha = 0\,\text{rad}.\)

    Example 10.

    A particle moves along the curve \(\mathbf{r}\left( t \right) = \left\langle {{t^2},{t^2} – 4t,t} \right\rangle .\) Find the minimum speed of the particle.

    Solution.

    Let’s first find the velocity vector:

    \[{\mathbf{v}\left( t \right) = \mathbf{r}^\prime\left( t \right) }={ \left\langle {\left( {{t^2}} \right)^\prime,\left( {{t^2} – 4t} \right)^\prime,t^\prime} \right\rangle }={ \left\langle {2t,2t – 4,1} \right\rangle .}\]

    Determine the speed of the particle:

    \[{\left| {\mathbf{v}\left( t \right)} \right| = \sqrt {{{\left( {2t} \right)}^2} + {{\left( {2t – 4} \right)}^2} + {1^2}} }={ \sqrt {4{t^2} + 4{t^2} – 16t + 16 + 1} }={ \sqrt {8{t^2} – 16t + 17} .}\]

    Now we explore the speed for extreme values. We denote it as \(F\left( t \right)\) and differentiate with respect to time \(t:\)

    \[{F^\prime\left( t \right) = \left( {\sqrt {8{t^2} – 16t + 17} } \right)^\prime }={ \frac{{\left( {8{t^2} – 16t + 17} \right)^\prime}}{{2\sqrt {8{t^2} – 16t + 17} }} }={ \frac{{16t – 16}}{{2\sqrt {8{t^2} – 16t + 17} }} }={ \frac{{8t – 8}}{{2\sqrt {8{t^2} – 16t + 17} }}.}\]

    It is clear that \(F^\prime\left( t \right) = 0\) when \(t = 1\). This value is a point of minimum as the derivative \(F^\prime\left( t \right)\) changes its sign from negative to positive when passing through this point.

    Now we can calculate the minimum value of the speed:

    \[{{\left| \mathbf{v} \right|_{\min }} = \left| {\mathbf{v}\left( 1 \right)} \right| }={ \sqrt {8 \cdot {1^2} – 16 \cdot 1 + 17} }={ \sqrt 9 }={ 3.}\]

    Example 11.

    The position of a particle is defined by the vector \(\mathbf{r}\left( t \right) = \left\langle {\sin t,\cos 2t} \right\rangle .\) Find the maximum speed of the particle.

    Solution.

    The velocity vector is written is the form

    \[{\mathbf{v}\left( t \right) = \mathbf{r}^\prime\left( t \right) }={ \left\langle {\left( {\sin t} \right)^\prime,\left( {\cos 2t} \right)^\prime} \right\rangle }={ \left\langle {\cos t, – 2\sin 2t} \right\rangle .}\]

    Hence, the speed of the particle is expressed by the function:

    \[\left| {\mathbf{v}\left( t \right)} \right| = \sqrt {{{\cos }^2}t + 4{{\sin }^2}2t} .\]

    Let’s find the extreme values of the function. It’s enough to explore the extreme values of the radicand (the expression under the square root sign). Be denoting the radicand as \(F\left( t \right),\) we have

    \[{F^\prime\left( t \right) = \left( {{{\cos }^2}t + 4{{\sin }^2}2t} \right)^\prime }={ – 2\cos t\sin t + 16\sin 2t\cos 2t }={ 15\sin 2t\cos 2t }={ \frac{{15}}{2} \cdot 2\sin 2t\cos 2t }={ \frac{{15}}{2}\sin 4t.}\]

    Equate the derivative to zero and determine the critical points:

    \[{F^\prime\left( t \right) = 0,}\;\; \Rightarrow {\frac{{15}}{2}\sin 4t = 0,}\;\; \Rightarrow {\sin 4t = 0,}\;\; \Rightarrow {4t = \pi n,}\;\; \Rightarrow {\frac{{\pi n}}{4},n \in Z.}\]

    So the function \(F\left( t \right)\) has the following critical points in the interval \(\left[ {0,\pi } \right]:\)

    \[t = 0,\frac{\pi }{4},\frac{\pi }{2},\frac{{3\pi }}{4},\pi .\]

    For each of these points, we calculate the value of the speed of the particle:

    \[{\left| {\mathbf{v}\left( 0 \right)} \right| }={ \sqrt {{{\cos }^2}0 + 4{{\sin }^2}0} }={ \sqrt {1 + 4 \cdot 0} }={ 1;}\]

    \[{\left| {\mathbf{v}\left( {\frac{\pi }{4}} \right)} \right| }={ \sqrt {{{\cos }^2}\frac{\pi }{4} + 4{{\sin }^2}\frac{\pi }{2}} }={ \sqrt {\frac{1}{2} + 4 \cdot 1} }={ \sqrt {\frac{9}{2}} }={ \frac{3}{{\sqrt 2 }} }={ \frac{{3\sqrt 2 }}{2} }\approx {2.12;}\]

    \[{\left| {\mathbf{v}\left( {\frac{\pi }{2}} \right)} \right| }={ \sqrt {{{\cos }^2}\frac{\pi }{2} + 4{{\sin }^2}\pi } }={ \sqrt {0 + 4 \cdot 0} }={ 0;}\]

    \[{\left| {\mathbf{v}\left( {\frac{3\pi }{4}} \right)} \right| }={ \sqrt {{{\cos }^2}\frac{{3\pi }}{4} + 4{{\sin }^2}\frac{{3\pi }}{2}} }={ \sqrt {\frac{1}{2} + 4 \cdot 1} }={ \sqrt {\frac{9}{2}} }={ \frac{3}{{\sqrt 2 }} }={ \frac{{3\sqrt 2 }}{2} }\approx{ 2.12;}\]

    \[{\left| {\mathbf{v}\left( \pi \right)} \right| }={ \sqrt {{{\cos }^2}\pi + 4{{\sin }^2}2\pi } }={ \sqrt {1 + 4 \cdot 0} }={ 1.}\]

    Thus, the maximum speed is \({\left\| \mathbf{v} \right\|_{\max }} = \large{\frac{{3\sqrt 2 }}{2}}\normalsize.\)