Question

Compute \(\mathbf{r}^{\prime \prime}(t)\) and \(\mathbf{r}^{\prime \prime \prime}(t)\) for the following functions. $$\mathbf{r}(t)=\langle\cos 3 t, \sin 4 t, \cos 6 t\rangle$$

Short answer

Given the vector function \(\mathbf{r}(t) = \langle\cos 3t, \sin 4t, \cos 6t\rangle\), find its second derivative \(\mathbf{r}^{\prime\prime}(t)\) and third derivative \(\mathbf{r}^{\prime\prime\prime}(t)\). Solution: We found that \(\mathbf{r}^{\prime\prime}(t) = \langle -9\cos 3t, -16\sin 4t, -36\cos 6t\rangle\) and \(\mathbf{r}^{\prime\prime\prime}(t) = \langle 27\sin 3t, -64\cos 4t, 216\sin 6t \rangle\).

Step-by-step solution

Compute the first derivative \(\mathbf{r}^{\prime}(t)\)

To compute the first derivative of \(\mathbf{r}(t)\), we can differentiate each component separately: $$\frac{d}{dt}(\cos 3t) = -3\sin 3t,$$ $$\frac{d}{dt}(\sin 4t) = 4\cos 4t,$$ $$\frac{d}{dt}(\cos 6t) = -6\sin 6t.$$ Thus, \(\mathbf{r}^{\prime}(t) = \langle -3\sin 3t, 4\cos 4t, -6\sin 6t\rangle\).

Compute the second derivative \(\mathbf{r}^{\prime\prime}(t)\)

Now, differentiate each component of \(\mathbf{r}^{\prime}(t)\) with respect to \(t\): $$\frac{d}{dt}(-3\sin 3t)=-9\cos 3t,$$ $$\frac{d}{dt}(4\cos 4t)=-16\sin 4t,$$ $$\frac{d}{dt}(-6\sin 6t)=-36\cos 6t.$$ Thus, \(\mathbf{r}^{\prime\prime}(t) = \langle -9\cos 3t, -16\sin 4t, -36\cos 6t\rangle\).

Compute the third derivative \(\mathbf{r}^{\prime\prime\prime}(t)\)

Finally, differentiate each component of \(\mathbf{r}^{\prime\prime}(t)\) with respect to \(t\): $$\frac{d}{dt}(-9\cos 3t)=27\sin 3t,$$ $$\frac{d}{dt}(-16\sin 4t)=-64\cos 4t,$$ $$\frac{d}{dt}(-36\cos 6t)=216\sin 6t.$$ Thus, \(\mathbf{r}^{\prime\prime\prime}(t) = \langle 27\sin 3t, -64\cos 4t, 216\sin 6t\rangle\). Now, we have calculated the second derivative \(\mathbf{r}^{\prime\prime}(t)\) and the third derivative \(\mathbf{r}^{\prime\prime\prime}(t)\).