Vector-Valued Function Notes

Published on: September 29, 2026

This Vector-Valued Function Calculator differentiates a curve given by three component functions of a parameter, and shows the working line by line. Each component is differentiated on its own, because the axes are independent of one another. The resulting vector is tangent to the curve and points the way the curve is travelling. Following the same pattern every time keeps each component matched to its own axis.

Step-by-step method

  1. Set up the problem.
  2. Differentiate each component to get the velocity r'(t).
  3. Differentiate again to get the acceleration r''(t).

Formula:

\(\mathbf{r}'(t) = \langle x'(t), y'(t), z'(t) \rangle\)
See Example 1 Hide Example 1

Problem

\(\mathbf{r}(t) = \langle t^{2},\sin{\left(t \right)},e^{t} \rangle\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Differentiate each component to get the velocity r'(t).
  3. Differentiate again to get the acceleration r''(t).

Step 1

Step 1 - Set up the problem.

In this problem:

\(\mathbf{r}(t) = \langle t^{2},\sin{\left(t \right)},e^{t} \rangle\)

Step 2

Step 2 - Differentiate each component to get the velocity r'(t).

In this problem:

\(\mathbf{r}'(t) = \langle 2 t,\cos{\left(t \right)},e^{t} \rangle\)

Step 3

Step 3 - Differentiate again to get the acceleration r''(t).

In this problem:

\(\mathbf{r}''(t) = \langle 2,- \sin{\left(t \right)},e^{t} \rangle\)

Final Answer

\(\begin{gathered} \mathbf{r}'(t) = \langle 2 t,\cos{\left(t \right)},e^{t} \rangle \\ \mathbf{r}''(t) = \langle 2,- \sin{\left(t \right)},e^{t} \rangle \end{gathered}\)
See Example 2 Hide Example 2

Problem

\(\mathbf{r}(t) = \langle \cos{\left(t \right)},\sin{\left(t \right)},t \rangle\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Differentiate each component to get the velocity r'(t).
  3. Differentiate again to get the acceleration r''(t).

Step 1

Step 1 - Set up the problem.

In this problem:

\(\mathbf{r}(t) = \langle \cos{\left(t \right)},\sin{\left(t \right)},t \rangle\)

Step 2

Step 2 - Differentiate each component to get the velocity r'(t).

In this problem:

\(\mathbf{r}'(t) = \langle - \sin{\left(t \right)},\cos{\left(t \right)},1 \rangle\)

Step 3

Step 3 - Differentiate again to get the acceleration r''(t).

In this problem:

\(\mathbf{r}''(t) = \langle - \cos{\left(t \right)},- \sin{\left(t \right)},0 \rangle\)

Final Answer

\(\begin{gathered} \mathbf{r}'(t) = \langle - \sin{\left(t \right)},\cos{\left(t \right)},1 \rangle \\ \mathbf{r}''(t) = \langle - \cos{\left(t \right)},- \sin{\left(t \right)},0 \rangle \end{gathered}\)
See Example 3 Hide Example 3

Problem

\(\mathbf{r}(t) = \langle t^{3},\cos{\left(t \right)},2 t \rangle\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Differentiate each component to get the velocity r'(t).
  3. Differentiate again to get the acceleration r''(t).

Step 1

Step 1 - Set up the problem.

In this problem:

\(\mathbf{r}(t) = \langle t^{3},\cos{\left(t \right)},2 t \rangle\)

Step 2

Step 2 - Differentiate each component to get the velocity r'(t).

In this problem:

\(\mathbf{r}'(t) = \langle 3 t^{2},- \sin{\left(t \right)},2 \rangle\)

Step 3

Step 3 - Differentiate again to get the acceleration r''(t).

In this problem:

\(\mathbf{r}''(t) = \langle 6 t,- \cos{\left(t \right)},0 \rangle\)

Final Answer

\(\begin{gathered} \mathbf{r}'(t) = \langle 3 t^{2},- \sin{\left(t \right)},2 \rangle \\ \mathbf{r}''(t) = \langle 6 t,- \cos{\left(t \right)},0 \rangle \end{gathered}\)
See Example 4 Hide Example 4

Problem

\(\mathbf{r}(t) = \langle e^{t},t^{2},\sin{\left(t \right)} \rangle\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Differentiate each component to get the velocity r'(t).
  3. Differentiate again to get the acceleration r''(t).

Step 1

Step 1 - Set up the problem.

In this problem:

\(\mathbf{r}(t) = \langle e^{t},t^{2},\sin{\left(t \right)} \rangle\)

Step 2

Step 2 - Differentiate each component to get the velocity r'(t).

In this problem:

\(\mathbf{r}'(t) = \langle e^{t},2 t,\cos{\left(t \right)} \rangle\)

Step 3

Step 3 - Differentiate again to get the acceleration r''(t).

In this problem:

\(\mathbf{r}''(t) = \langle e^{t},2,- \sin{\left(t \right)} \rangle\)

Final Answer

\(\begin{gathered} \mathbf{r}'(t) = \langle e^{t},2 t,\cos{\left(t \right)} \rangle \\ \mathbf{r}''(t) = \langle e^{t},2,- \sin{\left(t \right)} \rangle \end{gathered}\)