Question

Find the function \(\mathrm{r}\) that satisfies the given conditions. $$\mathbf{r}^{\prime}(t)=\langle 0,2,2 t\rangle ; \quad \mathbf{r}(1)=\langle 4,3,-5\rangle$$

Short answer

Question: Determine the function \(\mathbf{r}(t)\) if \(\mathbf{r}^{\prime}(t) = \langle 0, 2, 2t \rangle\) and \(\mathbf{r}(1) = \langle 4, 3, -5 \rangle\). Answer: \(\mathbf{r}(t) = \langle 4, 2t + 1, t^2 - 6 \rangle\)

Step-by-step solution

Integrate the derivative vector components

Since we are given the derivative vector, \(\mathbf{r}^{\prime}(t)=\langle 0,2,2t\rangle\), we need to integrate each component of the vector separately. $$ \int 0\,dt = C_x \\ \int 2\,dt = 2t + C_y \\ \int 2t\, dt = t^2 + C_z $$ So the integrated vector is given by \(\mathbf{r}(t) = \langle C_x, 2t + C_y, t^2 + C_z\rangle\).

Use the initial position to find the constant terms

We know that \(\mathbf{r}(1) = \langle 4,3,-5 \rangle\). Therefore, by plugging \(t=1\) into the integrated vector, we can find the constants \(C_x\), \(C_y\), and \(C_z\). $$ \mathbf{r}(1) = \langle C_x, 2(1) + C_y, (1)^2 + C_z \rangle \\ \langle 4,3,-5 \rangle = \langle C_x, 2 + C_y, 1 + C_z \rangle $$ Comparing the components, we find: $$ C_x = 4 \\ C_y = 1 \\ C_z = -6 $$

Write the final position vector function

Now that we have the constant values, we can write the final position vector function \(\mathbf{r}(t)\) by plugging the constants back into the integrated vector: $$ \mathbf{r}(t) = \langle 4, 2t + 1, t^2 - 6 \rangle $$ So the function that satisfies the given conditions is \(\mathbf{r}(t) = \langle 4, 2t + 1, t^2 - 6 \rangle\).

Key concepts

Integration

Integration is a fundamental concept in calculus, often described as the reverse process of differentiation. It allows us to find the original function given its derivative. In this exercise, you are given the derivative of a vector-valued function and tasked with finding the original function.
To solve the problem, we integrate each component of the vector function independently. This involves calculating an antiderivative for each component:

• The integration of the constant 0 gives us a constant of integration, which we represent as \(C_x\).
• Integrating 2 with respect to \(t\) results in \(2t + C_y\).
• Finally, integrating \(2t\) gives \(t^2 + C_z\).

By performing these integrations, we uncover the general form of the position vector \(\mathbf{r}(t) = \langle C_x, 2t + C_y, t^2 + C_z\rangle\). Employing integration here is crucial to solving such problems in vector calculus.

Initial Condition

The initial condition is an essential part of the problem-solving process, especially in calculus. It provides specific information allowing us to uniquely determine the constants of integration that arise during the integration process. In this exercise, the initial condition is given as \(\mathbf{r}(1) = \langle 4, 3, -5 \rangle\).
This condition provides a precise snapshot of the position vector at \(t = 1\). By substituting \(t = 1\) into the integrated form of the vector, you can match it to the given initial condition:

• Set \(C_x = 4\) because the x-component is constant and equals 4.
• The y-component equation, \(2 + C_y = 3\), leads to \(C_y = 1\).
• The z-component \(1 + C_z = -5\) results in \(C_z = -6\).

The presence of initial conditions in vector calculus problems is what pinpoints the exact solution from the infinite possibilities that arise due to integration constants.

Position Vector

A position vector is a vector representing the position of a point in space, relative to an origin. It is often used in physics and engineering to describe the trajectory of an object.
In this exercise, we find a position vector function \(\mathbf{r}(t)\), that not only fits the derivative conditions set by \(\mathbf{r}'(t)\) but also matches the initial characterization given by \(\mathbf{r}(1)\). This ultimately gives us a complete picture of how the position of a point changes over time.

• The resolved function \(\mathbf{r}(t) = \langle 4, 2t + 1, t^2 - 6 \rangle\) provides this evolving spatial information.
• Each particular component of the vector specifies how movement occurs in each dimension: constant in x, linear in y, and quadratic in z.

A clear understanding of position vectors is crucial for analyzing motion across different dimensions, providing insights into object paths and trajectories in various fields.