Problem 3 - The Calculus with Analytic Geometry

Question

\(a=800 ; v=20\) when \(s=1\). Find an equation involving \(v\) and s. (HINT: \(\left.a=\frac{d v}{d t}=\frac{d v}{d s} \frac{d s}{d t}=v \frac{d v}{d s}\right)\)

Step-by-Step Solution

Answer

The equation involving \(v\) and \(s\) is \(v^2 = 1600 s - 1200\).

Step 1: Understand the given information

The problem gives the acceleration, initial velocity, and initial position. The given values are: - Acceleration, \(a = 800\)- Velocity, \(v = 20\)- Position, \(s = 1\).

Step 2: Use the hint to derive the relationship

The hint provided gives the relationship for acceleration: \(\frac{d v}{d t} = \frac{d v}{d s} \frac{d s}{d t} = v \frac{d v}{d s}\). Given that acceleration \(a = 800\), we substitute into the equation: \[800 = v \frac{d v}{d s} \]

Step 3: Solve the differential equation

Rearrange the equation to separate variables: \[ \frac{d v}{v} = \frac{800}{v^2} \frac{d s}{1} \] This simplifies to: \[ v \frac{d v}{d s} = 800 \] Integrate both sides with respect to their respective variables: \[ \frac{1}{2} v^2 = 800 s + C \].

Step 4: Determine the constant of integration

Use the given values \(v = 20\) and \(s = 1\) to find the constant \(C\): \[ \frac{1}{2} (20)^2 = 800(1) + C \] Simplifying, we get: \[ 200 = 800 + C \] Thus, \( C = -600 \).

Step 5: Formulate the final equation

Substitute the constant \(C\) back into the equation: \[ \frac{1}{2} v^2 = 800 s - 600 \] Multiply both sides by 2 to simplify the equation: \[ v^2 = 1600 s - 1200 \]

Key Concepts

Differential Equations in Calculus

Differential equations play a crucial role in calculus, especially when dealing with rates of change. They help link functions to their derivatives, making them indispensable in modeling how quantities evolve over time. In this exercise, we use the given hint to transform a relation into an understandable differential equation. We start with the hint that gives the relationship for acceleration: \[\[\begin{equation}a = \frac{d v}{d t} = v \frac{d v}{d s}\end{equation}\]\]Later, by separating the variables and integrating, we find an equation that involves velocity (v) and position (s). The final form of this differential equation is:\[v^2 = 1600 s - 1200:\]This equation showcases how differential equations can transform the relationship between acceleration, velocity, and position into a clear mathematical form. Understanding this concept is vital because it lays the foundation for further studies in advanced calculus and mathematical modeling.

Understanding Kinematics

Kinematics is the branch of physics that studies motion without considering the forces that cause it. It focuses on parameters like position, velocity, and acceleration. In this problem, we investigate the motion of an object by examining how its velocity changes with position.

• Position (s): The location of the object at any point in time.
• Velocity (v): The rate of change of position with respect to time.
• Acceleration (a): The rate of change of velocity with respect to time.

Given values are:- Acceleration, \(a = 800\)- Velocity, \(v = 20\)- Position, \(s = 1\).By using these concepts, we derive the relationship between velocity and position. The object's motion is characterized by the equation, highlighting how kinematics helps us understand the principles of motion and the interconnectedness of these fundamental parameters. These principles are core to fields like mechanics, robotics, and any domain involving movement.

The Role of Integration

Integration is a vital technique in calculus that helps us find quantities like areas, volumes, and in this case, accumulated changes. When we solve the differential equation \[ v \frac{d v}{d s} = 800 \], we integrate both sides to find a relationship between velocity and position.

• Separating Variables: \[ \frac{dv}{v} = 800 \frac{ds}{v^2}\]
• Integrating: Both sides are integrated to find: \[ \frac{1}{2}v^2 = 800s + C\]

In the final solution, the constant of integration is determined using initial conditions:\ \[ \frac{1}{2}(20)^2 = 800(1) + C \Rightarrow 200 = 800 + C \Rightarrow C = -600 \]. Substituting C back, we get \( v^2 = 1600s - 1200 \), showing how integration helps us form meaningful relationships in mathematical models. This approach is pivotal when working with any differential equations in calculus-based scenarios, providing immense value in fields like physics, engineering, and economics.