Chapter 10: Problem 65
Find \(\mathbf{r}(t)\) if \(\mathbf{r}^{\prime}(t)=2 t \mathbf{i}+3 t^{2} \mathbf{j}+\sqrt{t} \mathbf{k}\) and \(\mathbf{r}(1)=\mathbf{i}+\mathbf{j}\)
Short Answer
Expert verified
\( \mathbf{r}(t) = t^2 \mathbf{i} + t^3 \mathbf{j} + \left( \frac{2}{3} t^{3/2} - \frac{2}{3} \right) \mathbf{k} \)
Step by step solution
01
Understanding the Problem
We need to find the vector function \( \mathbf{r}(t) \) given its derivative \( \mathbf{r}^{\prime}(t) = 2t \mathbf{i} + 3t^2 \mathbf{j} + \sqrt{t} \mathbf{k} \), and an initial condition \( \mathbf{r}(1) = \mathbf{i} + \mathbf{j} \). This requires integrating each component of the derivative vector to find the original vector function.
02
Integrate Each Component
To find \( \mathbf{r}(t) \), we integrate each component of the derivative vector separately:- \( \mathbf{i} \)-component: \( \int 2t \, dt = t^2 + C_1 \)- \( \mathbf{j} \)-component: \( \int 3t^2 \, dt = t^3 + C_2 \)- \( \mathbf{k} \)-component: \( \int \sqrt{t} \, dt = \frac{2}{3} t^{3/2} + C_3 \)Thus, initially, \( \mathbf{r}(t) = (t^2 + C_1) \mathbf{i} + (t^3 + C_2) \mathbf{j} + \left( \frac{2}{3} t^{3/2} + C_3 \right) \mathbf{k} \).
03
Apply the Initial Condition
We use the initial condition \( \mathbf{r}(1) = \mathbf{i} + \mathbf{j} \) to solve for the constants:- From \( \mathbf{i} \)-component: \( 1^2 + C_1 = 1 \) implies \( C_1 = 0 \).- From \( \mathbf{j} \)-component: \( 1^3 + C_2 = 1 \) implies \( C_2 = 0 \).- From \( \mathbf{k} \)-component: \( \frac{2}{3}(1)^{3/2} + C_3 = 0 \) implies \( C_3 = -\frac{2}{3} \).
04
Construct the Final Function
Using the values of \( C_1 \), \( C_2 \), and \( C_3 \), the vector function \( \mathbf{r}(t) \) becomes:\[ \mathbf{r}(t) = t^2 \mathbf{i} + t^3 \mathbf{j} + \left( \frac{2}{3} t^{3/2} - \frac{2}{3} \right) \mathbf{k} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration
Integration is a fundamental concept in calculus involving the process of finding antiderivatives or the original function given its derivative. In this problem, we use integration to determine the function \( \mathbf{r}(t) \) from its derivative \( \mathbf{r}^{\prime}(t) \).
We achieve this by integrating each vector component separately:
We achieve this by integrating each vector component separately:
- For the \( \mathbf{i} \)-component: Start with the integrand \( 2t \), integrating it yields \( \int 2t \, dt = t^2 + C_1 \).
- For the \( \mathbf{j} \)-component: The integrand is \( 3t^2 \), so \( \int 3t^2 \, dt = t^3 + C_2 \).
- For the \( \mathbf{k} \)-component: Begin with \( \sqrt{t} \), integrating gives \( \int \sqrt{t} \, dt = \frac{2}{3} t^{3/2} + C_3 \).
Vector Functions
Vector functions are mathematical expressions that describe a vector's magnitude and direction as functions of a parameter, usually time \( t \). In this exercise, \( \mathbf{r}(t) \) is our vector function. It is a position vector that includes three components corresponding to the \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) directions.
Each component is a scalar function of \( t \) and is determined by the integration of the corresponding component of the derivative \( \mathbf{r}^{\prime}(t) \). Thus, the combined equation of the vector function is initially expressed as:
Each component is a scalar function of \( t \) and is determined by the integration of the corresponding component of the derivative \( \mathbf{r}^{\prime}(t) \). Thus, the combined equation of the vector function is initially expressed as:
- \( \mathbf{r}(t) = (t^2 + C_1) \mathbf{i} + (t^3 + C_2) \mathbf{j} + \left( \frac{2}{3} t^{3/2} + C_3 \right) \mathbf{k} \)
Initial Conditions
Initial conditions are particular constraints that are used to determine the specific solution of a differential equation among a family of possible solutions. In this context, the initial condition helps us find the values of the constants of integration, ensuring our solution satisfies certain real-world constraints or reference points.
We have the initial condition \( \mathbf{r}(1) = \mathbf{i} + \mathbf{j} \), which denotes the vector's specific value at \( t=1 \). This leads to the following individual equations by plugging in \( t = 1 \):
We have the initial condition \( \mathbf{r}(1) = \mathbf{i} + \mathbf{j} \), which denotes the vector's specific value at \( t=1 \). This leads to the following individual equations by plugging in \( t = 1 \):
- For the \( \mathbf{i} \)-component: \( 1 + C_1 = 1 \) gives \( C_1 = 0 \).
- For the \( \mathbf{j} \)-component: \( 1 + C_2 = 1 \) yields \( C_2 = 0 \).
- For the \( \mathbf{k} \)-component: \( \frac{2}{3}(1)^{3/2} + C_3 = 0 \), resulting in \( C_3 = -\frac{2}{3} \).