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Chapter 13: Problem 12

Solve the initial value problems in Exercises \(11-20\) for \(\mathbf{r}\) as a vector function of \(t .\) $$ \begin{array}{ll}{\text { Differential equation: }} & {\frac{d \mathbf{r}}{d t}=(180 t) \mathbf{i}+\left(180 t-16 t^{2}\right) \mathbf{j}} \\ {\text { Initial condition: }} & {\mathbf{r}(0)=100 \mathbf{j}}\end{array} $$

Short Answer

Expert verified
The vector function is \( \mathbf{r}(t) = (90t^2)\mathbf{i} + (90t^2 - \frac{16t^3}{3} + 100)\mathbf{j} \).

Step by step solution

01

Understand the Problem

We need to find the vector function \( \mathbf{r}(t) \) given its derivative with respect to \( t \) and an initial condition. The derivative is \( \frac{d \mathbf{r}}{dt}=(180t)\mathbf{i}+(180t-16t^2)\mathbf{j} \). The initial condition is \( \mathbf{r}(0)=100\mathbf{j} \).
02

Integrate the i-Component

The i-component of the derivative is \( 180t \). To find \( r_i(t) \), integrate \( 180t \) with respect to \( t \). \[ r_i(t) = \int 180t \, dt = 180 \cdot \frac{t^2}{2} + C_i = 90t^2 + C_i \] where \( C_i \) is a constant.
03

Integrate the j-Component

The j-component of the derivative is \( 180t - 16t^2 \). To find \( r_j(t) \), integrate \( 180t - 16t^2 \) with respect to \( t \). \[ r_j(t) = \int (180t - 16t^2) \, dt = 180\cdot\frac{t^2}{2} - 16\cdot\frac{t^3}{3} + C_j = 90t^2 - \frac{16t^3}{3} + C_j \] where \( C_j \) is a constant.
04

Apply the Initial Condition

Given \( \mathbf{r}(0) = 100\mathbf{j} \), we know that \( r_i(0) = 0 \) and \( r_j(0) = 100 \). Substitute \( t = 0 \) into the expressions from Steps 2 and 3 to solve for \( C_i \) and \( C_j \). \[ 90(0)^2 + C_i = 0 \Rightarrow C_i = 0 \] \[ 90(0)^2 - \frac{16(0)^3}{3} + C_j = 100 \Rightarrow C_j = 100 \]
05

Write the Vector Function

Substitute the constants obtained from Step 4 back into the integrated functions to write the full vector function \( \mathbf{r}(t) \). \[ \mathbf{r}(t) = (90t^2)\mathbf{i} + (90t^2 - \frac{16t^3}{3} + 100)\mathbf{j} \] This is the vector function that satisfies both the differential equation and the initial condition.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Initial Value Problem
An Initial Value Problem (IVP) in the context of differential equations is a problem where we need to find a function that not only satisfies a given differential equation but also meets certain initial conditions. This is crucial in many real-world applications, where initial conditions represent the starting state of a system.

In our specific example, we are tasked with solving for the vector function \( \mathbf{r}(t) \) given its derivative and the initial condition \( \mathbf{r}(0) = 100\mathbf{j} \).

  • The differential equation describes how \( \mathbf{r}(t) \) changes over time.
  • The initial condition provides a constraint or starting point at time \( t=0 \).
This combination allows us to uniquely determine the function that fits both the rate of change and the starting state.
Differential Equation
A Differential Equation involves derivatives, which express rates of change, and these equations are vital in modeling dynamic systems. In mathematics, especially in vector calculus, these equations might involve vector functions, as in our example.

The given differential equation is \( \frac{d \mathbf{r}}{dt}=(180t)\mathbf{i}+(180t-16t^2)\mathbf{j} \) and tells us how the vector \( \mathbf{r}(t) \) changes with respect to time \( t \):

  • \( (180t)\mathbf{i} \) gives the rate of change in the \( i \)-direction, or horizontal direction.
  • \( (180t-16t^2)\mathbf{j} \) represents the rate of change in the \( j \)-direction, often referred to as the vertical direction.
The goal is to find the original vector function \( \mathbf{r}(t) \) from this equation.
Vector Function
Vector Functions are functions with vectors as outputs rather than simple scalar values. These functions are crucial in representing entities that have both magnitude and direction, such as velocity or force.

In our exercise, \( \mathbf{r}(t) \) is a vector function that combines two components:
  • The \( i \)-component, which relates to the x-axis or horizontal motion.
  • The \( j \)-component, which relates to the y-axis or vertical motion.
By solving the initial value problem, we find \( \mathbf{r}(t) \) which describes the motion of an object or point with respect to time, incorporating both direction and magnitude.
Integration
Integration is the mathematical process used to find the original function from its derivative, effectively 'reversing' the differentiation process. It is a key step in solving differential equations.

While integrating each component of the vector differential equation, we:
  • Integrate the i-component \( 180t \) to get \( 90t^2 + C_i \).
  • Integrate the j-component \( 180t - 16t^2 \) to get \( 90t^2 - \frac{16t^3}{3} + C_j \).

Each integration provides an additional term, the constant of integration, reflecting the family of functions that could fit the differential equation, before applying any initial conditions.
Constants of Integration
When integrating, we introduce constants of integration, \( C_i \) and \( C_j \), which are unknowns that need to be determined. These constants reflect any shifts or changes in the initial conditions of the function being integrated.

To determine these constants:
  • Apply the initial condition \( \mathbf{r}(0) = 100\mathbf{j} \) to solve for \( C_i \) and \( C_j \).
  • Set \( t = 0 \) and substitute into the integrated equations to find \( C_i = 0 \) and \( C_j = 100 \).
Determining these constants is crucial because it ensures that the resultant function satisfies both the differential equation and its initial condition, providing a unique solution to the problem.

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Most popular questions from this chapter

Can anything be said about the acceleration of a particle that is moving at a constant speed? Give reasons for your answer.

Use a CAS to perform the following steps in Exercises \(49-52 .\) \(\begin{array}{l}{\text { a. Plot the space curve traced out by the position vector } \mathbf{r} \text { . }} \\ {\text { b. Find the components of the velocity vector } d \mathbf{r} / d t \text { . }} \\ {\text { c. Evaluate } d \mathbf{r} / d t \text { at the given point } t_{0} \text { and determine the equa- }} \\ {\text { tion of the tangent line to the curve at } \mathbf{r}\left(t_{0}\right) .} \\ {\text { d. Plot the tangent line together with the curve over the given }} \\ {\text { interval. }}\end{array}\) $$ \begin{array}{l}{\mathbf{r}(t)=(\sin 2 t) \mathbf{i}+(\ln (1+t)) \mathbf{j}+t \mathbf{k}, \quad 0 \leq t \leq 4 \pi} \\ {t_{0}=\pi / 4}\end{array} $$

In Exercises \(27-30\) , find the value(s) of \(t\) so that the tangent line to the given curve contains the given point. $$ \mathbf{r}(t)=-t \mathbf{i}+t^{2} \mathbf{j}+(\ln t) \mathbf{k} ; \quad(2,-5,-3) $$

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