Question

Compute the indefinite integral of the following functions. $$\mathbf{r}(t)=e^{3 t} \mathbf{i}+\frac{1}{1+t^{2}} \mathbf{j}-\frac{1}{\sqrt{2 t}} \mathbf{k}$$

Short answer

Question: Compute the indefinite integral of the vector-valued function \(\mathbf{r}(t)=e^{3t}\mathbf{i}+\frac{1}{1+t^2}\mathbf{j}-\frac{1}{\sqrt{2t}}\mathbf{k}\). Answer: The indefinite integral of the vector-valued function \(\mathbf{r}(t)\) is \(\int \mathbf{r}(t) dt = \left(\frac{1}{3} e^{3t} + C_1\right) \mathbf{i} + \left(\arctan(t) + C_2\right) \mathbf{j} + \left(-2\sqrt{2t} + C_3 \right) \mathbf{k}\), where \(C_1, C_2, C_3\) are constants of integration.

Step-by-step solution

Identify the components of the vector-valued function

The given vector-valued function is $$\mathbf{r}(t)=e^{3t}\mathbf{i}+\frac{1}{1+t^2}\mathbf{j}-\frac{1}{\sqrt{2t}}\mathbf{k}.$$ The components are: - \(x(t) = e^{3t}\), - \(y(t) = \frac{1}{1+t^2}\), and - \(z(t) = -\frac{1}{\sqrt{2t}}\).

Integrate the x-component of the vector-valued function

To compute the indefinite integral for the \(x(t)\) component, we have: $$\int x(t) dt = \int e^{3t} dt.$$ We can apply the rule of integration for exponential functions, which is \(\int e^{at} dt = \frac{1}{a} e^{at} + C\). Thus, $$\int e^{3t} dt = \frac{1}{3} e^{3t} + C_1.$$

Integrate the y-component of the vector-valued function

To compute the indefinite integral for the \(y(t)\) component, we have: $$\int y(t) dt = \int \frac{1}{1+t^2} dt.$$ This integral is the arctangent integral, which has the rule \(\int \frac{1}{1+t^2} dt = \arctan(t) + C\). Hence, $$\int \frac{1}{1+t^2} dt = \arctan(t) + C_2.$$

Integrate the z-component of the vector-valued function

To compute the indefinite integral for the \(z(t)\) component, we have: $$\int z(t) dt = \int -\frac{1}{\sqrt{2t}} dt.$$ We can write the integrand more conveniently by using a rational exponent: $$-\int \frac{1}{\sqrt{2t}} dt = -\int \frac{1}{(2t)^{1/2}} dt = -\int (2t)^{-1/2} dt.$$ Now we can apply the power rule of integration, \(\int t^n dt = \frac{1}{n+1} t^{n+1} + C\), to get $$-\int (2t)^{-1/2} dt = -\frac{1}{(-1/2+1)}(2t)^{-1/2+1} + C_3 = -2(2t)^{1/2} + C_3 = -2\sqrt{2t} + C_3.$$

Write down the indefinite integral of the vector-valued function

Now we have the indefinite integrals for all components. So the indefinite integral of \(\mathbf{r}(t)\) is $$\int \mathbf{r}(t) dt = \left(\frac{1}{3} e^{3t} + C_1\right) \mathbf{i} + \left(\arctan(t) + C_2\right) \mathbf{j} + \left(-2\sqrt{2t} + C_3 \right) \mathbf{k}.$$

Key concepts

Vector-Valued Functions

A vector-valued function takes in a parameter, like time, and outputs a vector. This function can represent an object's path through space. For example, imagine tracking a flying drone; at any given instant, you can represent its position using a vector containing its location in space.
In the given exercise, the vector-valued function \( \mathbf{r}(t) = e^{3t} \mathbf{i} + \frac{1}{1+t^{2}} \mathbf{j} - \frac{1}{\sqrt{2t}} \mathbf{k} \) combines three separate component functions:

• \( x(t) = e^{3t} \)
• \( y(t) = \frac{1}{1+t^{2}} \)
• \( z(t) = -\frac{1}{\sqrt{2t}} \)

Each of these functions describes motion along one of the coordinate axes. By integrating each component separately, and then combining the results, you obtain the integral of the vector-valued function, giving insight into changes in path or accumulated distance over time.

Exponential Function Integration

Integration of exponential functions is quite straightforward. The rule you should remember is \[ \int e^{at} \, dt = \frac{1}{a} e^{at} + C \] where \( a \) is a constant and \( C \) is the constant of integration.
In our exercise, the \( x(t) \) component of the vector-valued function is \( e^{3t} \). To integrate it, apply the rule directly:

• \( \int e^{3t} \, dt = \frac{1}{3} e^{3t} + C_1 \)

This gives the indefinite integral for the \( x(t) \) component. Integrating exponential functions helps define accumulated effects like growth or decay over time.

Arctangent Function

The arctangent, often referred to as \( \text{arctan} \) or \( \tan^{-1} \), is an inverse trigonometric function. It appears in calculus commonly when dealing with integrals of the form \[ \int \frac{1}{1+t^2} \, dt \] The result of this integral is actually the arctangent function plus a constant:

• \( \int \frac{1}{1+t^2} \, dt = \arctan(t) + C \)

In the given exercise, this rule is used to integrate the \( y(t) \) component \( \frac{1}{1+t^{2}} \). Understanding how to solve such integrals is crucial in applications ranging from physics to engineering, where you often need to calculate angles or rotations.

Power Rule of Integration

The power rule is a key technique in calculus for integrating functions of the form \( t^n \). It states: \[ \int t^n \, dt = \frac{1}{n+1} t^{n+1} + C \] where \( n eq -1 \).
In the given exercise, the \( z(t) \) component is expressed using a negative power: \( -\frac{1}{\sqrt{2t}} \), or equivalently \( -(2t)^{-1/2} \). By applying the power rule, you change the function’s power to simplify integration:

• \( \int -(2t)^{-1/2} \, dt = -\frac{1}{(-1/2+1)}(2t)^{1/2} + C_3 = -2\sqrt{2t} + C_3 \)

Using the power rule enables you to transform functions into a form that is straightforward to integrate, making it one of the most universal tools in calculus.