Question

Find the function \(\mathbf{r}\) that satisfies the given conditions. $$\mathbf{r}^{\prime}(t)=\frac{t}{t^{2}+1} \mathbf{i}+t e^{-t^{2}} \mathbf{j}-\frac{2 t}{\sqrt{t^{2}+4}} \mathbf{k} ; \mathbf{r}(0)=\mathbf{i}+\frac{3}{2} \mathbf{j}-3 \mathbf{k}$$

Short answer

Question: Find the function \(\mathbf{r}(t)\) given that its derivative is \(\mathbf{r}'(t) = \frac{t}{t^2 + 1} \mathbf{i} + te^{-t^2} \mathbf{j} - \frac{2t}{\sqrt{t^2 + 4}} \mathbf{k}\) and its initial condition is \(\mathbf{r}(0) = \mathbf{i} + \frac{3}{2}\mathbf{j} - 3\mathbf{k}\). Answer: \(\mathbf{r}(t) = \left(\frac{1}{2} \ln|t^2 + 1| + 1 \right)\mathbf{i} + \left( -\frac{1}{2}e^{-t^2} + 2\right)\mathbf{j} + \left( -2\sqrt{t^2 + 4} + 1\right)\mathbf{k}\)

Step-by-step solution

Integrate the derivative components

We have the derivative of \(\mathbf{r}(t)\) given as: $$\mathbf{r}^{\prime}(t)=\frac{t}{t^{2}+1} \mathbf{i}+t e^{-t^{2}} \mathbf{j}-\frac{2 t}{\sqrt{t^{2}+4}} \mathbf{k}$$ We need to integrate each component of this vector function with respect to \(t\): $$\mathbf{r}(t)=\int \left(\frac{t}{t^{2}+1}\mathbf{i}+t e^{-t^{2}}\mathbf{j}-\frac{2 t}{\sqrt{t^{2}+4}}\mathbf{k}\right) dt$$

Integrate the i-component

Let's start by integrating the i-component: $$r_i(t) = \int \frac{t}{t^{2}+1} dt$$ To solve this integral, we can use the substitution method with \(u = t^2 + 1\) and \(du = 2t dt\). Then, it becomes: $$r_i(t) = \int \frac{1}{2} \frac{du}{u} = \frac{1}{2} \ln|u| + C_i = \frac{1}{2} \ln|t^2 + 1| + C_i$$

Integrate the j-component

Now, let's integrate the j-component: $$r_j(t) = \int t e^{-t^{2}} dt$$ To solve this integral, we can use the substitution method with \(u = -t^2\) and \(du = -2t dt\). Then, it becomes: $$r_j(t) = -\int \frac{1}{2}e^u du = -\frac{1}{2}e^u + C_j = -\frac{1}{2}e^{-t^2} + C_j$$

Integrate the k-component

Lastly, we integrate the k-component: $$r_k(t) = \int -\frac{2t}{\sqrt{t^2 + 4}} dt$$ To solve this integral, we can use the substitution method with \(u = t^2 + 4\) and \(du = 2t dt\). Then, it becomes: $$r_k(t) = -\int \frac{du}{\sqrt{u}} = -2\sqrt{u} + C_k = -2\sqrt{t^2 + 4} + C_k$$

Combine the components

Now, we combine the i, j and k components to get the function \(\mathbf{r}(t)\): $$\mathbf{r}(t) = \left(\frac{1}{2} \ln|t^2 + 1| + C_i \right)\mathbf{i} + \left( -\frac{1}{2}e^{-t^2} + C_j\right)\mathbf{j} + \left( -2\sqrt{t^2 + 4} + C_k\right)\mathbf{k}$$

Use the initial condition to find the constants

We are given the initial condition \(\mathbf{r}(0) = \mathbf{i} + \frac{3}{2}\mathbf{j} - 3\mathbf{k}\), so we plug \(t = 0\) into the function and set it equal to the given value: $$\left(\frac{1}{2} \ln|0^2 + 1| + C_i \right)\mathbf{i} + \left( -\frac{1}{2}e^{-(0)^2} + C_j\right)\mathbf{j} + \left( -2\sqrt{0^2 + 4} + C_k\right)\mathbf{k} = \mathbf{i} + \frac{3}{2}\mathbf{j} - 3\mathbf{k}$$ By comparing the components, we find the values of the constants \(C_i\), \(C_j\) and \(C_k\): $$C_i = 1, \qquad C_j = \frac{3}{2} + \frac{1}{2}e^{0} = 2, \qquad C_k = -3 + 2\sqrt{4} = 1$$

Write the final expression for the function

Now we substitute the values of the constants into the function \(\mathbf{r}(t)\): $$\mathbf{r}(t) = \left(\frac{1}{2} \ln|t^2 + 1| + 1 \right)\mathbf{i} + \left( -\frac{1}{2}e^{-t^2} + 2\right)\mathbf{j} + \left( -2\sqrt{t^2 + 4} + 1\right)\mathbf{k} $$ This is the function \(\mathbf{r}(t)\) that satisfies the given conditions.

Key concepts

Vector Functions

Vector functions, often denoted by \( \mathbf{r}(t) \), are pivotal in vector calculus. They link a variable, typically time \( t \), to vector quantities, giving us an understanding of how a vector behaves over time or along a path. Each vector function consists of multiple components, typically represented by \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) which correspond to the x, y, and z axes, respectively. For example, a vector function might look something like \( \mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j} + z(t)\mathbf{k} \).

• The vector \( \mathbf{r}^{\prime}(t) \) is the derivative of \( \mathbf{r}(t) \), depicting the vector's rate of change.
• Each component function \( x(t), y(t), \) and \( z(t) \) can be separately analyzed and manipulated using calculus techniques.

Understanding vector functions allows us to work with vector fields, describe curves' motion in space, and solve real-world physics problems related to velocity and acceleration.

Integration Techniques

Integrating vector functions involves tackling each component independently, which may require various integration techniques. In the example provided, the derivative \( \mathbf{r}^{\prime}(t) \) gave us three separate integrals where different techniques were applied. Let's break them down:- **Substitution Method:** Often used when an integral consists of a composite function, this method simplifies a complex expression into an easily integrable form.

• For the \( \mathbf{i} \) component: \( \int \frac{t}{t^2 + 1} dt \) was made simpler by letting \( u = t^2 + 1 \).
• This technique also came in handy for the \( \mathbf{k} \) component: \( \int -\frac{2t}{\sqrt{t^2 + 4}} dt \) with \( u = t^2 + 4 \).

- **Recognition of Standard Forms:** For the \( \mathbf{j} \) component, recognizing the form \( t e^{-t^2} \) as derivable directly via substitution helped in simplifying the integral.Accessibility to these techniques equips students with the tools necessary for tackling more complex integrals they may encounter in advanced mathematics and physics.

Initial Conditions

Initial conditions provide essential additional information required to determine the constants that arise from the integration of a vector function. In our example, integration resulted in the expression \( \mathbf{r}(t) = C_i\mathbf{i} + C_j\mathbf{j} + C_k\mathbf{k} \), each with an unknown constant.The provided initial condition tells us that when \( t = 0 \), the vector function equals a specific vector: \( \mathbf{r}(0) = \mathbf{i} + \frac{3}{2}\mathbf{j} - 3\mathbf{k} \). This condition:

• Guides us in substituting \( t = 0 \) into the integrated function to solve for unknown constants \( C_i, C_j, \text{and} \ C_k \).
• Using the values obtained, we achieve a unique solution that adheres to the specified initial situation.

Comprehending and applying initial conditions is crucial not only for solving specific problems but also for forming predictions in physical systems where initial states project future behavior.