Vector-Valued Function Calculator

Published on: July 21, 2026
Final Answer: Free Full Steps: Pro

This Vector-Valued Function Calculator differentiates \(\mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle\) component by component to find the velocity \(\mathbf{r}'(t)\) and the acceleration \(\mathbf{r}''(t)\).

Step-by-step method

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

Formula:

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

Example 1:

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

Step 1 - Set up the components of r(t).

In this problem: The vector function is \(\mathbf{r}(t) = \langle t^{2},\sin{\left(t \right)},e^{t} \rangle\).

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

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

In this problem: Differentiating each component gives \(\mathbf{r}'(t) = \langle 2 t,\cos{\left(t \right)},e^{t} \rangle\).

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

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

In this problem: Differentiating again gives \(\mathbf{r}''(t) = \langle 2,- \sin{\left(t \right)},e^{t} \rangle\).

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

Final answer:

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

Example 2:

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

Step 1 - Set up the components of r(t).

In this problem: The vector function is \(\mathbf{r}(t) = \langle \cos{\left(t \right)},\sin{\left(t \right)},t \rangle\).

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

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

In this problem: Differentiating each component gives \(\mathbf{r}'(t) = \langle - \sin{\left(t \right)},\cos{\left(t \right)},1 \rangle\).

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

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

In this problem: Differentiating again gives \(\mathbf{r}''(t) = \langle - \cos{\left(t \right)},- \sin{\left(t \right)},0 \rangle\).

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

Final answer:

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