Question

Find the function \(\mathrm{r}\) that satisfies the given conditions. $$\mathbf{r}^{\prime}(t)=\left\langle\sqrt{t}, \cos \pi t, \frac{4}{t}\right\rangle ; \quad \mathbf{r}(1)=\langle 2,3,4\rangle$$

Short answer

Question: Find the original function 𝑟(𝑡) given its derivative 𝑟′(𝑡)=⟨√𝑡,cos𝜋𝑡,4/𝑡⟩ and that 𝑟(1)=⟨2,3,4⟩. Answer: The original function 𝑟(𝑡) is given by 𝑟(𝑡)=⟨(2/3)𝑡^(3/2)+(4/3), (1/𝜋)sin(𝜋𝑡)+3, 4ln|𝑡|+4⟩.

Step-by-step solution

Integrate the derivative component-wise

Given the derivative of the function \(\mathbf{r}^{\prime}(t)=\langle\sqrt{t}, \cos \pi t, \frac{4}{t}\rangle\), we need to integrate each component with respect to \(t\). The integral of \( \sqrt{t}\) with respect to t is: $$\int \sqrt{t} \, dt = \int t^{\frac{1}{2}}\, dt = \frac{2}{3}t^{\frac{3}{2}} + C_1$$ The integral of \( \cos \pi t\) with respect to t is: $$\int \cos \pi t \, dt = \frac{1}{\pi} \sin \pi t + C_2$$ The integral of \( \frac{4}{t}\) with respect to t is: $$\int \frac{4}{t} \, dt = 4 \ln |t| + C_3$$ So, based on the integration of each component, our function is now: $$\mathbf{r}(t) = \left\langle \frac{2}{3}t^{\frac{3}{2}}+C_1, \frac{1}{\pi} \sin\pi t + C_2, 4\ln|t| + C_3\right\rangle$$

Find the integration constants

We are given that \(\mathbf{r}(1) = \langle 2, 3, 4 \rangle\). By plugging \(t = 1\) into the function \(\mathbf{r}(t)\), we can find the values of the integration constants \(C_1\), \(C_2\), and \(C_3\). $$\left\langle \frac{2}{3}(1)^{\frac{3}{2}}+C_1, \frac{1}{\pi} \sin\pi(1) + C_2, 4\ln|1| + C_3\right\rangle = \langle 2, 3, 4 \rangle$$ This gives us the following equations for \(C_1\), \(C_2\), and \(C_3\): 1. \( \frac{2}{3} + C_1 = 2 \) 2. \( \frac{1}{\pi} \sin \pi + C_2 = 3 \) 3. \( 4 \ln 1 + C_3 = 4\) Solving each equation for the constants, we get: 1. \( C_1 = 2 - \frac{2}{3} = \frac{4}{3} \) 2. \( C_2 = 3 - \frac{1}{\pi}\sin\pi = 3 \) (since \(\sin \pi = 0\)) 3. \( C_3 = 4 - 4\ln 1 = 4 \) (since \(\ln 1 = 0\))

Find the original function

Now that we have the values for \(C_1\), \(C_2\), and \(C_3\), we can plug them back into our function \(\mathbf{r}(t)\) to find the original function: $$\mathbf{r}(t) = \left\langle \frac{2}{3}t^{\frac{3}{2}}+\frac{4}{3}, \frac{1}{\pi} \sin\pi t + 3, 4\ln|t| + 4\right\rangle$$ This is the original function \(\mathbf{r}(t)\) that satisfies the given conditions.

Key concepts

Vector Calculus

Vector calculus is a branch of mathematics that deals with the differentiation and integration of vector fields, primarily in 3-dimensional Euclidean space. A vector field is a construction in which a vector is associated with every point in a region of space. For example, in physics, vector fields can represent the velocity of a moving fluid throughout space, or the strength and direction of some force, such as the magnetic or gravitational force, as it changes from one point to another.

In our exercise, we are given a vector-valued function's derivative, \( \mathbf{r}'(t) \), and are asked to find the original function, which is a common problem in vector calculus. The function's derivative gives us the rate of change of the function with respect to time, and by integrating this derivative, we are essentially finding the total change, which leads us back to the original function. This process is analogous to finding the antiderivative in scalar calculus, but it's applied to each component of a vector function individually.

Integration of Vector Functions

The integration of vector functions, similar to the integration of scalar functions, involves finding a function that represents the accumulated sum of the rates of change indicated by the vector derivative. Integration is often performed component-wise, meaning that each component of the vector function is integrated separately with respect to the variable of integration, usually time or space.

In the exercise, the derivative \( \mathbf{r}'(t) \) has three components: \( \sqrt{t} \) for the x-component, \( \cos \pi t \) for the y-component, and \( \frac{4}{t} \) for the z-component. Each of these is integrated separately to find the corresponding components of the original vector function \( \mathbf{r}(t) \). After integration, each component will also have an arbitrary constant, which can be determined from initial conditions or other given information.

Initial Conditions

Initial conditions specify the state of a system at the start of a process. In the context of integrating vector-valued functions, initial conditions help us determine the particular solution to a differential equation by providing specific values that the function must satisfy at a given point. Generally, these are values of the function at specific times or positions.

In this exercise, the initial condition given is \( \mathbf{r}(1) = \langle 2, 3, 4 \rangle \). This tells us exactly what the vector function should output when \( t = 1 \). We use this piece of information to solve for the constants resulting from the integration of each component. By substituting \( t = 1 \) into the integrated function and equating it to the initial condition, we solve for \( C_1 \) in the x-component, \( C_2 \) in the y-component, and \( C_3 \) in the z-component. These constants ensure that the function fulfills the given initial condition, rendering a complete, specific solution to the original problem.