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 or tap 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 $${\left\| \mathbf{v} \right\|_{\max }} = \large{\frac{{3\sqrt 2 }}{2}}\normalsize.$$