Question

Find the function \(\mathbf{r}\) that satisfies the given condition. $$\mathbf{r}^{\prime}(t)=\langle\sqrt{t}, \cos \pi t, 4 / t\rangle ; \mathbf{r}(1)=\langle 2,3,4\rangle$$

Short answer

Question: Given the derivative of a function as \(\mathbf{r}'(t) = \langle\sqrt{t}, \cos\pi t, 4/t\rangle\) and an initial condition \(\mathbf{r}(1) = \langle 2, 3, 4\rangle\), find the function \(\mathbf{r}(t)\). Answer: The function \(\mathbf{r}(t)\) is given by \(\mathbf{r}(t) = \left\langle\frac{2}{3} t^{\frac{3}{2}} + \frac{4}{3}, \frac{1}{\pi}\sin\pi t + 3, 4\ln|t| + 4\right\rangle\).

Step-by-step solution

Integrate the derivative component-wise

We're given the derivative \(\mathbf{r}'(t) = \langle\sqrt{t}, \cos\pi t, 4/t\rangle\). We can find \(\mathbf{r}(t)\) by integrating each component with respect to \(t\): $$\int \sqrt{t} \,dt, \int \cos\pi t\,dt, \int\frac{4}{t}\,dt$$

Evaluate integrals

Calculate each integral separately: 1. \(\int \sqrt{t} \,dt = \frac{2}{3} t^{\frac{3}{2}} + C_1\) 2. \(\int \cos\pi t \,dt = \frac{1}{\pi}\sin\pi t + C_2\) 3. \(\int\frac{4}{t} \,dt = 4\ln|t| + C_3\)

Combine components

Combine the integrated components to form the function \(\mathbf{r}(t)\): $$\mathbf{r}(t) = \left\langle\frac{2}{3} t^{\frac{3}{2}} + C_1, \frac{1}{\pi}\sin\pi t + C_2, 4\ln|t| + C_3\right\rangle$$

Use the initial condition

We're given the initial condition \(\mathbf{r}(1) = \langle 2, 3, 4\rangle\). Plug in \(t = 1\) into \(\mathbf{r}(t)\) and set it equal to the initial condition: $$\left\langle\frac{2}{3}(1)^{\frac{3}{2}} + C_1, \frac{1}{\pi}\sin\pi(1) + C_2, 4\ln|1| + C_3\right\rangle = \langle 2, 3, 4\rangle$$

Solve for constants

From the initial condition, we can find the values of the constants \(C_1, C_2\), and \(C_3\): 1. \(\frac{2}{3}(1)^{\frac{3}{2}} + C_1 = 2 \Rightarrow C_1 = \frac{4}{3}\) 2. \(\frac{1}{\pi}\sin\pi(1) + C_2 = 3 \Rightarrow C_2 = 3\) 3. \(4\ln|1| + C_3 = 4 \Rightarrow C_3 = 4\)

Write the final function

Plug the values of the constants we found back into \(\mathbf{r}(t)\) to obtain the final function: $$\mathbf{r}(t) = \left\langle\frac{2}{3} t^{\frac{3}{2}} + \frac{4}{3}, \frac{1}{\pi}\sin\pi t + 3, 4\ln|t| + 4\right\rangle$$

Key concepts

Integration

Integration is a fundamental concept in calculus used to find the quantity where rates of change are known. In our vector calculus scenario, we have a given derivative of a vector function, \(\mathbf{r}^{\prime}(t)= \langle\sqrt{t}, \cos \pi t, 4 / t\rangle\), which requires us to integrate to find the original function, \(\mathbf{r}(t)\).
Here, we employ integration component-wise, meaning we integrate each part of the vector separately:

• \(\int \sqrt{t} \,dt = \frac{2}{3} t^{\frac{3}{2}} + C_1\).
• \(\int \cos\pi t \,dt = \frac{1}{\pi}\sin\pi t + C_2\).
• \(\int\frac{4}{t} \,dt = 4\ln|t| + C_3\).

Integration not only provides us with the function form but also involves constants \(C_1, C_2,\) and \(C_3\) that appear during indefinite integration. This highlights the importance of combining solutions carefully to capture all terms based on their derivatives.

Initial Conditions

Initial conditions play a crucial role in determining the specific solution to a differential equation. An initial condition allows us to find the constants that come from the integration process. In this context, the initial condition is given as:
\(\mathbf{r}(1)=\langle 2,3,4\rangle\).
Using this, we substitute \( t = 1 \) into \(\mathbf{r}(t)\) to find the constants \(C_1, C_2,\) and \(C_3\).

• For the \(x\)-component: \(\frac{2}{3}(1)^{\frac{3}{2}} + C_1 = 2\) giving \(C_1 = \frac{4}{3}\).
• For the \(y\)-component: \(\frac{1}{\pi}\sin\pi(1) + C_2 = 3\) providing \(C_2 = 3\).
• For the \(z\)-component: \(4\ln|1| + C_3 = 4\), solving \(C_3 = 4\).

Each constant is dictated by the condition \(\mathbf{r}(1)=\langle 2,3,4\rangle\), allowing the solution of \(\mathbf{r}(t)\) to be uniquely determined.

Vector Functions

A vector function is a function with a vector as its range. These functions can describe many objects, such as lines, curves, or more complex shapes in space. In this exercise, \(\mathbf{r}(t)\) is a vector function with three components:

• First component: \(\frac{2}{3} t^{\frac{3}{2}} + \frac{4}{3}\)
• Second component: \(\frac{1}{\pi}\sin\pi t + 3\)
• Third component: \(4\ln|t| + 4\)

Each component represents a separate dimension in space, and the combination of these dimensions provides a path or a shape in multi-dimensional space. Vector functions are powerful tools in calculus allowing us to model and understand geometric and physical realities in two, three, or more dimensions.