Question

Find \(\vec{r}(t),\) given that \(\vec{r}^{\prime}(t)=\langle 1 /(t+1), \tan t\rangle\) and \(\vec{r}(0)=\langle 1,2\rangle\)Find \(\vec{r}(t),\) given that \(\vec{r}^{\prime \prime}(t)=\left\langle t^{2}, t, 1\right\rangle\) \(\vec{r}^{\prime}(0)=\langle 1,2,3\rangle\) and \(\vec{r}(0)=\langle 4,5,6\rangle\)

Short answer

\(\vec{r}(t) = \left\langle \frac{t^4}{12} + t + 4, \frac{t^3}{6} + 2t + 5, \frac{t^2}{2} + 3t + 6 \right\rangle\).

Step-by-step solution

Integrate the Second Derivative

First, we need to find the first derivative, \(\vec{r}^{\prime}(t)\), by integrating \(\vec{r}^{\prime\prime}(t) = \langle t^2, t, 1 \rangle\) with respect to \(t\). The integral of each component is:- \(\int t^2 \, dt = \frac{t^3}{3} + C_1\)- \(\int t \, dt = \frac{t^2}{2} + C_2\)- \(\int 1 \, dt = t + C_3\)Thus, \(\vec{r}^{\prime}(t) = \left\langle \frac{t^3}{3} + C_1, \frac{t^2}{2} + C_2, t + C_3 \right\rangle.\)

Apply the First Initial Condition

Use the initial condition \(\vec{r}^{\prime}(0) = \langle 1, 2, 3 \rangle\) to find the constants \(C_1, C_2,\) and \(C_3\).Substitute \(t = 0\) into \(\vec{r}^{\prime}(t) = \left\langle \frac{t^3}{3} + C_1, \frac{t^2}{2} + C_2, t + C_3 \right\rangle\):\[ \vec{r}^{\prime}(0) = \left\langle 0 + C_1, 0 + C_2, 0 + C_3 \right\rangle = \langle 1, 2, 3 \rangle \]Thus, \(C_1 = 1\), \(C_2 = 2\), \(C_3 = 3\).We find \(\vec{r}^{\prime}(t) = \left\langle \frac{t^3}{3} + 1, \frac{t^2}{2} + 2, t + 3 \right\rangle.\)

Integrate the First Derivative

Now, integrate \(\vec{r}^{\prime}(t) = \left\langle \frac{t^3}{3} + 1, \frac{t^2}{2} + 2, t + 3 \right\rangle\) to find \(\vec{r}(t)\).- \(\int (\frac{t^3}{3} + 1) \, dt = \frac{t^4}{12} + t + D_1\)- \(\int (\frac{t^2}{2} + 2) \, dt = \frac{t^3}{6} + 2t + D_2\)- \(\int (t + 3) \, dt = \frac{t^2}{2} + 3t + D_3\)Thus, \(\vec{r}(t) = \left\langle \frac{t^4}{12} + t + D_1, \frac{t^3}{6} + 2t + D_2, \frac{t^2}{2} + 3t + D_3 \right\rangle.\)

Apply the Second Initial Condition

Use the initial condition \(\vec{r}(0) = \langle 4, 5, 6 \rangle\) to find the new constant terms \(D_1, D_2, D_3\).Substitute \(t = 0\) into \(\vec{r}(t) = \left\langle \frac{t^4}{12} + t + D_1, \frac{t^3}{6} + 2t + D_2, \frac{t^2}{2} + 3t + D_3 \right\rangle\):- \(\langle 0 + D_1, 0 + D_2, 0 + D_3 \rangle = \langle 4, 5, 6 \rangle\)Thus, \(D_1 = 4\), \(D_2 = 5\), \(D_3 = 6\).We find \(\vec{r}(t) = \left\langle \frac{t^4}{12} + t + 4, \frac{t^3}{6} + 2t + 5, \frac{t^2}{2} + 3t + 6 \right\rangle.\)

Key concepts

Integration

In vector calculus, integration plays a vital role in solving problems related to velocity and position vectors. To understand integration in this context, let's think of it as the reverse process of differentiation. When you integrate a vector function, you are essentially finding an original function whose derivative is the given vector function. In the problem we are addressing, we start with the second derivative of the position vector, often related to acceleration, \[\vec{r}^{\prime\prime}(t) = \left\langle t^2, t, 1 \right\rangle.\]
Our first task is to integrate this expression to get the first derivative or the velocity vector, \(\vec{r}^{\prime}(t)\). We integrate each component separately:

• \(\int t^2 \, dt = \frac{t^3}{3} + C_1\)
• \(\int t \, dt = \frac{t^2}{2} + C_2\)
• \(\int 1 \, dt = t + C_3\)

These sorts of integrations help to build up from what is known back to what we are seeking. The constants \(C_1, C_2,\) and \(C_3\) are then determined using the given initial conditions, allowing us to exactly find \(\vec{r}^{\prime}(t)\). Following this, we integrate the resulting velocity vector to eventually derive the position vector \(\vec{r}(t)\).

Initial Conditions

Initial conditions are essential for finding the specific solution to problems in vector calculus, particularly when dealing with differential equations. Initial conditions act like clues that help us figure out the integration constants that appear when performing integration. In our context, we have two initial conditions:

• \(\vec{r}^{\prime}(0) = \langle 1, 2, 3 \rangle\)
• \(\vec{r}(0) = \langle 4, 5, 6 \rangle\)

These initial conditions guide us to apply particular values of \(t\) to the integrated functions:- For \(\vec{r}^{\prime}(t)\), substituting \(t = 0\) allows us to solve for the constants \(C_1, C_2,\) and \(C_3\). This ensures that the function meets the velocity condition at \(t = 0\).- Similarly, for \(\vec{r}(t)\), using \(t = 0\) helps us in determining \(D_1, D_2,\) and \(D_3\), aligning the function with the given position at \(t = 0\).In essence, the initial conditions enable us to tailor a general solution to fit the precise specifics of a given problem.

Velocity and Position Vectors

Velocity and position vectors are foundational in describing motion within vector calculus. The velocity vector \(\vec{r}^{\prime}(t)\) represents the rate of change of the position vector \(\vec{r}(t)\) over time. By integrating the acceleration vector \(\vec{r}^{\prime\prime}(t)\), we derive this velocity vector. In the problem given, once we integrated from the acceleration to get:\[\vec{r}^{\prime}(t) = \left\langle \frac{t^3}{3} + 1, \frac{t^2}{2} + 2, t + 3 \right\rangle,\]we then focused on integrating this resultant to find the position vector \(\vec{r}(t)\).
The position vector, based on the velocity, can be found by:

• \(\int (\frac{t^3}{3} + 1) \, dt = \frac{t^4}{12} + t + D_1\)
• \(\int (\frac{t^2}{2} + 2) \, dt = \frac{t^3}{6} + 2t + D_2\)
• \(\int (t + 3) \, dt = \frac{t^2}{2} + 3t + D_3\)

The position vector \(\vec{r}(t)\) shows the exact path or location at any given point in time.
These relationships between the vectors illustrate the systematic nature by which calculus can describe dynamic physical systems.