Question

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

Short answer

The function $\mathbf{r}(t)$ is given by: $$\mathbf{r}(t)=⟨\frac{2}{3}{t}^{\frac{3}{2}}+\frac{4}{3},\frac{1}{\pi }\sin (\pi t)+3,4\ln (|t|)+4⟩$$

Step-by-step solution

Integrate each component of the derivative vector

To find the original function, we will integrate each component of the derivative vector: $$\int \sqrt{t}dt$$ $$\int \cos (\pi t)dt$$ $$\int \frac{4}{t}dt$$

Add the integration constants

After integrating, we need to add integration constants $C_1,C_2,$ and $C_3$ to each of the respective components: $$\int \sqrt{t}dt=\frac{2}{3}{t}^{\frac{3}{2}}+C_1$$ $$\int \cos (\pi t)dt=\frac{1}{\pi }\sin (\pi t)+C_2$$ $$\int \frac{4}{t}dt=4\ln (|t|)+C_3$$ Thus, the function looks like: $$\mathbf{r}(t)=⟨\frac{2}{3}{t}^{\frac{3}{2}}+C_1,\frac{1}{\pi }\sin (\pi t)+C_2,4\ln (|t|)+C_3⟩$$

Use the initial condition to solve for the integration constants

We're given the initial condition $\mathbf{r}(1)=⟨2,3,4⟩$. We will plug in $t=1$ to find the values of the constants: $$2=\frac{2}{3}(1{)}^{\frac{3}{2}}+C_1$$ $$3=\frac{1}{\pi }\sin (\pi (1))+C_2$$ $$4=4\ln (|1|)+C_3$$ Solving for the constants, we get: $$C_1=\frac{4}{3}$$ $$C_2=3$$ $$C_3=4$$

Combine everything to give the final form of the function

Finally, we substitute the constants' values back into the function: $$\mathbf{r}(t)=⟨\frac{2}{3}{t}^{\frac{3}{2}}+\frac{4}{3},\frac{1}{\pi }\sin (\pi t)+3,4\ln (|t|)+4⟩$$ So, the desired function $\mathbf{r}(t)$, that satisfies the given conditions is: $$\mathbf{r}(t)=⟨\frac{2}{3}{t}^{\frac{3}{2}}+\frac{4}{3},\frac{1}{\pi }\sin (\pi t)+3,4\ln (|t|)+4⟩$$

Key concepts

Integration

Integration is a fundamental concept in vector calculus that helps us find a function given its derivative. When integrating vector functions, we treat each component of the vector function independently. For our exercise, function components involve performing integration on expressions like $\sqrt{t}$, $\cos (\pi t)$, and $\frac{4}{t}$.

• For $\sqrt{t}$, we find the integral as $\frac{2}{3}{t}^{\frac{3}{2}}$.
• For $\cos (\pi t)$, the integral is computed as $\frac{1}{\pi }\sin (\pi t)$ because the sine function is the antiderivative of cosine.
• The integral of $\frac{4}{t}$ is $4\ln (|t|)$, a classic logarithmic integral.

After integrating, we attach constants $C_1,C_2,$ and $C_3$ to account for any shifts or translations of the function along each axis.

Integration not only "reverses" differentiation but also allows shaping functions in accordance with initial conditions, a crucial next step in solving differential equations.

Initial Conditions

Initial conditions are values that specify the state of a function at a particular point, giving us essential information to determine the unique solution of an integral.

In our exercise, the initial condition $\mathbf{r}(1)=⟨2,3,4⟩$ means that when $t=1$, the vector $\mathbf{r}$ should equal $⟨2,3,4⟩$.

By plugging the value $t=1$ into our integrated function, we solve for the constants:

• $2=\frac{2}{3}(1{)}^{3/2}+C_1$ gives $C_1=\frac{4}{3}$.
• $3=\frac{1}{\pi }\sin (\pi )+C_2$ results in $C_2=3$ as $\sin (\pi )$ is zero.
• $4=4\ln (|1|)+C_3$ leads to $C_3=4$ since $\ln (1)=0$.

Having initial conditions means our final solution isn't just some function, but the precise function that behaves as expected at a specific point. This concept ensures that integration leads us not to many possible solutions, but one correct and plausible answer.

Vector Functions

Vector functions are essentially functions where each output is a vector. In this exercise, the vector function $\mathbf{r}(t)$ was derived from its derivative ${\mathbf{r}}^{\prime }(t)$, using integration and specific initial conditions.

Given the derivative vector ${\mathbf{r}}^{\prime }(t)=⟨\sqrt{t},\cos \pi t,\frac{4}{t}⟩$, we focused on finding the original vector function $\mathbf{r}(t)$. Each component of ${\mathbf{r}}^{\prime }(t)$ corresponds to a direction in space:

• $\sqrt{t}$ influences movement in the $x$-direction.
• $\cos \pi t$ characterizes the behavior along the $y$-axis.
• $\frac{4}{t}$ dictates changes in the $z$-axis.

These components reflect how a point moves through space over time, with each axis having its own rate of change. Vector functions are vital in physics and engineering as they succinctly describe multi-dimensional dynamics and processes.