Question

In Exercises 35 and \(36,\) use the properties of the derivative to find the following. (a) \(\mathbf{r}^{\prime}(t)\) (b) \(\mathbf{r}^{\prime \prime}(t)\) (c) \(D_{t}[\mathbf{r}(t) \cdot \mathbf{u}(t)]\) (d) \(D_{t}[3 \mathbf{r}(t)-\mathbf{u}(t)]\) (e) \(D_{t}[\mathbf{r}(t) \times \mathbf{u}(t)]\) (f) \(D_{t}[\|\mathbf{r}(t)\|], \quad t>0\) $$ \mathbf{r}(t)=t \mathbf{i}+3 t \mathbf{j}+t^{2} \mathbf{k}, \quad \mathbf{u}(t)=4 t \mathbf{i}+t^{2} \mathbf{j}+t^{3} \mathbf{k} $$

Short answer

The derivatives are: (a) \(\mathbf{r}^{\prime}(t) = \mathbf{i} + 3\mathbf{j} + 2t\mathbf{k}\) (b) \(\mathbf{r}^{\prime \prime}(t) = 2\mathbf{k}\) (c) \(D_t[\mathbf{r}(t) \cdot \mathbf{u}(t)] = 8t + 9t^2 + 5t^4\) (d) \(D_{t}[3 \mathbf{r}(t)-\mathbf{u}(t)] = -\mathbf{i}+(9-2t)\mathbf{j}+(6t-3t^2)\mathbf{k}\) (e) \(D_{t}[\mathbf{r}(t) \times \mathbf{u}(t)] = -4t\mathbf{i} + (3t^2-8t)\mathbf{j} + (2t - 12)\mathbf{k}\) (f) \(D_{t}[\|\mathbf{r}(t)\|] = \frac{9t+2t^3}{\sqrt{1+9t^2 + t^4}}\)

Step-by-step solution

Calculation of \(\mathbf{r}^{\prime}(t)\)

To find the derivative of \(\mathbf{r}(t)=t \mathbf{i}+3 t \mathbf{j}+t^{2} \mathbf{k}\), derive each component's function. \(\mathbf{r}^{\prime}(t) = \mathbf{i} + 3\mathbf{j} + 2t\mathbf{k}\)

Calculation of \(\mathbf{r}^{\prime \prime}(t)\)

The second derivative of \(\mathbf{r}(t)\) is obtained by deriving \(\mathbf{r}^{\prime}(t)\) or each component's function of \(\mathbf{r}(t)\). \(\mathbf{r}^{\prime \prime}(t) = 2\mathbf{k}\)

Calculation of \(D_{t}[\mathbf{r}(t) \cdot \mathbf{u}(t)]\)

This requires the product rule for derivatives. First calculate the dot product and then derive that result. \(\mathbf{r}(t) \cdot \mathbf{u}(t) = t*4t + 3t*t^2 + t^2*t^3 = 4t^2+3t^3+t^5\). Derivative of this is \(D_t[\mathbf{r}(t) \cdot \mathbf{u}(t)] = 8t + 9t^2 + 5t^4\)

Calculation of \(D_{t}[3 \mathbf{r}(t)-\mathbf{u}(t)]\)

First calculate \(3 \mathbf{r}(t)-\mathbf{u}(t)\) and then derive each component of the resultant vector. \(3 \mathbf{r}(t)-\mathbf{u}(t) = (3t - 4t)\mathbf{i} + (9t - t^2)\mathbf{j} + (3t^2 - t^3)\mathbf{k} = (-t)\mathbf{i} + (9t - t^2)\mathbf{j} + (3t^2 - t^3)\mathbf{k}\). Now the derivative is: \(D_{t}[3 \mathbf{r}(t)-\mathbf{u}(t)] = -\mathbf{i}+(9-2t)\mathbf{j}+(6t-3t^2)\mathbf{k}\)

Calculation of \(D_{t}[\mathbf{r}(t) \times \mathbf{u}(t)]\)

First compute the cross product and then take its derivative. The cross product: \(\mathbf{r}(t) \times \mathbf{u}(t)=((-3t^2)+t^2)\mathbf{i} + ((t^3) - 4t^2)\mathbf{j} + ((t^2) -12t)\mathbf{k} = (-2t^2)\mathbf{i} + (t^3-4t^2)\mathbf{j} + (t^2 -12t)\mathbf{k}\). Now the derivative is : \(D_{t}[\mathbf{r}(t) \times \mathbf{u}(t)] = -4t\mathbf{i} + (3t^2-8t)\mathbf{j} + (2t - 12)\mathbf{k}\)

Calculation of \(D_{t}[\|\mathbf{r}(t)\|], \quad t>0\)

First compute the magnitude of \(\mathbf{r}(t)\) and then take its derivative. The magnitude of \(\mathbf{r}(t)\) is \(\sqrt{t^2 + (3t)^2 + t^4} = \sqrt{1+9t^2 + t^4}\). The derivative is \(D_{t}[\|\mathbf{r}(t)\|] = \frac{18t + 4t^3}{2\sqrt{1+9t^2 + t^4}} = \frac{9t+2t^3}{\sqrt{1+9t^2 + t^4}}\)