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Next: Surfaces. Up: Curves in the plane Previous: Limits and Continuity.

Derivatives.

Definition 2.11   The function $\overrightarrow{F(t)} $ is differentiable at t0 if there exists a vector $\overrightarrow{V} $ such that

\begin{displaymath}\underset{t \rightarrow t_0}{\lim}\frac {1}{t-t_0}
(\overrightarrow{F(t)} - \overrightarrow{F(t_0)} ) = \overrightarrow{V} .
\end{displaymath}

We denote:

\begin{displaymath}\overrightarrow{F'(t_0)} =\frac {dF}{dt} \Bigg\vert_{t_0} = \overrightarrow{V} .
\end{displaymath}

Proposition 2.12   The function $\overrightarrow{F(t)} =f_1(t)\overrightarrow{i} + f_2(t)\overrightarrow{j} +f_3(t)\overrightarrow{k} $ is differentiable at t0 if, and only if each component fi is differentiable at t0. We have:

\begin{displaymath}\frac {dF}{dt} \Bigg\vert_{t_0} =
\frac {df_1}{dt} \Bigg\vert...
...ow{j} +
\frac {df_3}{dt} \Bigg\vert_{t_0} \overrightarrow{k} .
\end{displaymath}

Example 2.13   If $\overrightarrow{F(t)} = \cos 2t \overrightarrow{i} + \sin 3t \overrightarrow{j} +
t \overrightarrow{k} $, then

$\overrightarrow{F'(t)} = -2 \sin 2t \overrightarrow{i} + 3 \cos 3t \overrightarrow{j} +
\overrightarrow{k} $.

Definition 2.14   With the notations of  refdef derivation. Let the function $\overrightarrow{F(t)} $ describe the motion of a particle.
1.
The vector $\frac {dF}{dt} \Bigg\vert_{t_0}$ is the velocity vector of the particle at time t0.
2.
The number $\begin{vmatrix}V(t_0) \end{vmatrix}$ is the speed of the particle at t0.

Definition 2.15   With the same notations. The acceleration at t0 is the derivative of $\overrightarrow{V} (t)$ at t0. we denote it by $\overrightarrow{\gamma} (t_0)$.

Example 2.16   Let $\overrightarrow{F(t)} = \cos t \overrightarrow{i} + \sin t \overrightarrow{j} +
t \overrightarrow{k} $. Then:

Please check that, in this example, $\overrightarrow{V(t)$\space and $\overrightarrow{\gamma}(t)} $ are orthogonal for every value of t.



Noah Dana-Picard
2001-05-30