Question

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

Short answer

Answer: The original function \(\mathbf{r}(t)\) is given by \(\mathbf{r}(t) = \langle 4, 2t+1, t^2-6 \rangle\).

Step-by-step solution

Rewrite the given derivative and the initial condition

We are given the derivative and initial condition: $$\mathbf{r^{\prime}}(t)=\langle 0,2,2t \rangle \quad \text{and} \quad \mathbf{r}(1)=\langle 4,3,-5 \rangle$$

Integrate each component of \(\mathbf{r^{\prime}}(t)\)

To find the original function \(\mathbf{r}(t)\), integrate each component separately: $$\int 0 \ dt = c_1$$ $$\int 2 \ dt = 2t + c_2$$ $$\int 2t \ dt = t^2 + c_3$$

Rewrite the integrated function as a vector function

After integrating, rewrite the function in vector form: $$\mathbf{r}(t) = \langle c_1, 2t+c_2, t^2+c_3 \rangle$$

Apply the initial condition \(\mathbf{r}(1)=\langle 4,3,-5 \rangle\)

Replace \(t\) with \(1\) and set each component of the vector function equal to the corresponding component in the initial condition: $$\begin{cases} c_1 = 4 \\ 2(1) + c_2 = 3 \\ (1)^2 + c_3 = -5 \end{cases}$$

Solve for the integration constants

From the above equations, we get the constants: $$\begin{cases} c_1 = 4 \\ c_2 = 1 \\ c_3 = -6 \end{cases}$$

Write the final function \(\mathbf{r}(t)\)

Substitute these constants back into the vector function: $$\mathbf{r}(t) = \langle 4, 2t+1, t^2-6 \rangle$$ The function \(\mathbf{r}(t)\) 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 used to find original functions from their derivatives. When we integrate, we essentially reverse the process of differentiation. In the given exercise, we started with the derivative of a vector function \( \mathbf{r}^{\prime}(t) = \langle 0, 2, 2t \rangle \).
To find \( \mathbf{r}(t) \), we must integrate each component of \( \mathbf{r}^{\prime}(t) \):

• The constant function \(0\) integrates to \(c_1\), a constant of integration, since a constant function means no change—or movement in this case—exists for that dimension of the vector.
• Integrating \(2\) gives \(2t + c_2\), as we simply multiply \(2\) by \(t\) and add the constant \(c_2\).
• The integration of \(2t\) results in \(t^2 + c_3\), applying the power rule for integration which changes the exponent from \(1\) to \(2\) and divides by the new power.

It's important to integrate each component of the vector function independently, allowing us to write \( \mathbf{r}(t) \) as a vector of integrated components with constants of integration.

Initial Conditions

Initial conditions are crucial when finding the specific solution of an equation that includes constants of integration. They give us the value of the function at a specific point, allowing us to solve for these constants.
In the exercise, we were given the initial condition \(\mathbf{r}(1) = \langle 4, 3, -5 \rangle\). This means that when \(t = 1\), the vector function \(\mathbf{r}(t)\) equals \(\langle 4, 3, -5 \rangle\).

• To use the initial condition, we substituted \( t = 1 \) into our integrated vector function \( \mathbf{r}(t) \).
• We compared each rewritten component of the vector with the given initial condition vector.

For each component equation, we solved for the constants \(c_1, c_2, c_3\). This step ensures that the function fits the specific initial condition provided, pinpointing the exact function among many possible solutions.

Vector Functions

Vector functions extend calculus into higher dimensions by dealing with vectors rather than scalar quantities. A vector function assigns a vector to each value in the domain. This is different from regular functions which assign single number values.
For example, a vector function like \( \mathbf{r}(t) = \langle 4, 2t+1, t^2-6 \rangle \) describes a curve in three-dimensional space, rather than along a single line. The number of dimensions (or components) relies on the number of directions present in the problem.

• The i-component (first) could represent the x-axis, determining forward or backward movement.
• The j-component (second) might relate to up or down motion on the y-axis.
• Meanwhile, the k-component (third) can indicate movement along the z-axis.

In this exercise, our function \( \mathbf{r}(t) \) models the motion path in space, with each component determining position along one of the axes in a three-dimensional coordinate system. Understanding vector functions is essential in physics and engineering, where paths, forces, or fields often depend on multiple dimensions.