Question

Compute \(\mathbf{r}^{\prime \prime}(t)\) when \(\mathbf{r}(t)=\left\langle t^{10}, 8 t, \cos t\right\rangle\)

Short answer

Answer: The second derivative of the vector function is \(\mathbf{r}''(t) = \left\langle 90t^{8}, 0, -\cos t \right\rangle\).

Step-by-step solution

Differentiate \(\mathbf{r}(t)\) once to get \(\mathbf{r}'(t)\)

First, we need to find the first derivative of \(\mathbf{r}(t)\). To do this, differentiate each component with respect to \(t\). Let's compute the first derivative of \(\mathbf{r}(t)\): - For the first component: \(\frac{d(t^{10})}{dt} = 10 t^{9}\) - For the second component: \(\frac{d(8t)}{dt} = 8\) - For the third component: \(\frac{d(\cos t)}{dt} = -\sin t\) Therefore, \(\mathbf{r}'(t) = \left\langle 10t^{9}, 8, -\sin t \right\rangle\).

Differentiate \(\mathbf{r}'(t)\) once more to get \(\mathbf{r}''(t)\)

Now, we need to find the second derivative of \(\mathbf{r}(t)\). To do this, differentiate each component of \(\mathbf{r}'(t)\) with respect to \(t\). Let's compute the second derivative of \(\mathbf{r}(t)\): - For the first component: \(\frac{d(10t^{9})}{dt} = 90 t^{8}\) - For the second component: \(\frac{d(8)}{dt} = 0\) - For the third component: \(\frac{d(-\sin t)}{dt} = -\cos t\) So, \(\mathbf{r}''(t) = \left\langle 90t^{8}, 0, -\cos t \right\rangle\). Our final answer is \(\mathbf{r}''(t) = \left\langle 90t^{8}, 0, -\cos t \right\rangle\).