Question

Find \(\mathbf{r}(t)\) for the given conditions. $$ \mathbf{r}^{\prime \prime}(t)=-4 \cos t \mathbf{j}-3 \sin t \mathbf{k}, \quad \mathbf{r}^{\prime}(0)=3 \mathbf{k}, \quad \mathbf{r}(0)=4 \mathbf{j} $$

Short answer

\(\mathbf{r}(t) = 4 (\cos t + 1) \mathbf{j} +3 \sin t \mathbf{k}\)

Step-by-step solution

Integrate the second derivative to find the first derivative

The process begins with integrating the given second derivative \(\mathbf{r}^{\prime \prime}(t)=-4 \cos t \mathbf{j}-3 \sin t \mathbf{k}\) to find the first derivative. \(\mathbf{r}^{\prime}(t) = -4 \sin t \mathbf{j} + 3 \cos t \mathbf{k} + C_1\), where \(C_1\) is the vector constant of integration.

Apply the initial condition to find the constant of integration \(C_1\)

Substitute the given initial condition for the first derivative, \(\mathbf{r}^{\prime}(0)= 3\mathbf{k}\), into the equation from Step 1. This should give \(C_1 = 3\mathbf{k}\).

Integrate the first derivative to find the vector function \(\mathbf{r}(t)\)

next step is to integrate the found first derivative to get \(\mathbf{r}(t)\). This gives \(\mathbf{r}(t) = 4 \cos t \mathbf{j} +3 \sin t \mathbf{k} + C_2\), with \(C_2\) as another vector constant of integration.

Apply the second initial condition to find \(C_2\)

Substitute the given initial condition for the vector function into the equation from Step 3, \(\mathbf{r}(0) = 4\mathbf{j}\), to find \(C_2 = 4\mathbf{j}\).

Write the final solution

The final step is to write the form of the vector function with the found constants, providing the complete solution to the exercise which is: \(\mathbf{r}(t) = 4 \cos t \mathbf{j} +3 \sin t \mathbf{k} + 4\mathbf{j}\).