Problem 295 - Calculus Volume 1

Question

The position of a moving hockey puck after \(t\) seconds is \(s(t)=\tan ^{-1} t\) where \(s\) is in meters. a. Find the velocity of the hockey puck at any time \(t\) b. Find the acceleration of the puck at any time \(t\) c. Evaluate a and b. for \(t=2,4,\) and 6 seconds. d. What conclusion can be drawn from the results in c.?

Step-by-Step Solution

Answer

a) Velocity is \( \frac{1}{1+t^2} \), b) Acceleration is \( \frac{-2t}{(1+t^2)^2} \). At \( t=2,4,6 \), velocity decreases and acceleration is negative. The puck slows down over time.

Step 1: Understand and Differentiate Position Function

The function for the position of the hockey puck is given by \( s(t) = \tan^{-1}(t) \). To find the velocity, we need to differentiate the position function with respect to time \( t \). The derivative of \( \tan^{-1}(t) \) is \( \frac{1}{1+t^2} \). Therefore, the velocity function is \( v(t) = \frac{1}{1+t^2} \).

Step 2: Differentiate Velocity Function

The velocity function is \( v(t) = \frac{1}{1+t^2} \). To find the acceleration, differentiate \( v(t) \) with respect to \( t \). Using the chain rule, we get \( a(t) = \frac{-2t}{(1+t^2)^2} \). This is the acceleration function of the hockey puck.

Step 3: Evaluate Velocity and Acceleration at Specific Times

Now, evaluate the velocity and acceleration at \( t = 2, 4, \text{ and } 6 \) seconds:- For \( t = 2 \): - Velocity: \( v(2) = \frac{1}{1+2^2} = \frac{1}{5} \) - Acceleration: \( a(2) = \frac{-2 imes 2}{(1+2^2)^2} = \frac{-4}{25} \)- For \( t = 4 \): - Velocity: \( v(4) = \frac{1}{1+4^2} = \frac{1}{17} \) - Acceleration: \( a(4) = \frac{-2 imes 4}{(1+4^2)^2} = \frac{-8}{289} \)- For \( t = 6 \): - Velocity: \( v(6) = \frac{1}{1+6^2} = \frac{1}{37} \) - Acceleration: \( a(6) = \frac{-2 imes 6}{(1+6^2)^2} = \frac{-12}{1369} \)

Step 4: Draw Conclusion from Calculated Values

From the values calculated, we observe that as \( t \) increases, the velocity \( v(t) \) decreases and the acceleration \( a(t) \) also decreases, becoming more negative. This indicates that the puck slows down as time progresses due to decreasing velocity and negative acceleration.

Key Concepts

Differentiation

Differentiation is a core concept in calculus, allowing us to find the rate at which a quantity changes. In simple terms, it's a mathematical way to discover how much a function is changing at any given point.
For instance, if we have a position function, which tells us where an object is at any time, differentiation helps us find the object's velocity.

• This is achieved by finding the derivative of the position function.
• The derivative represents the rate of change of position with respect to time, which is the velocity.

In our hockey puck example, the position is given by the function, \( s(t) = \tan^{-1}(t) \). When we differentiate \( s(t) \) with respect to \( t \), we find the velocity \( v(t) \) to be \( \frac{1}{1+t^2} \). This velocity function shows how fast the puck's position is changing over time.

Velocity

Velocity is an essential concept when studying motion. It tells us how quickly and in what direction an object moves.
Velocity is the result of differentiating the position function concerning time. It is a vector quantity, meaning it has both a magnitude (speed) and direction.
According to our position function \( s(t) = \tan^{-1}(t) \), the velocity function derived by differentiation is \( v(t) = \frac{1}{1+t^2} \).

• This shows the instantaneous speed of the puck at any time \( t \).
• The formula \( v(t) = \frac{1}{1+t^2} \) indicates that velocity changes as time progresses, specifically depending on \( t^2 \).

Evaluating this at specific times, you see that the velocity decreases over time, such as \( v(2) = \frac{1}{5} \), \( v(4) = \frac{1}{17} \), and \( v(6) = \frac{1}{37} \). This tells us that as the hockey puck progresses, its speed decreases.

Acceleration

Acceleration measures how quickly an object's velocity changes over time. It is found by differentiating the velocity function with respect to time.
Simply put, it's the derivative of the velocity function. Acceleration can be positive (increasing speed) or negative (decreasing speed).
For the hockey puck, we use the velocity function \( v(t) = \frac{1}{1+t^2} \) to find acceleration. Differentiating this yields \( a(t) = \frac{-2t}{(1+t^2)^2} \).

• The negative sign indicates the puck's acceleration is in the opposite direction of its velocity, meaning the puck is slowing down.
• The function \( a(t) = \frac{-2t}{(1+t^2)^2} \) shows that as \( t \) increases, acceleration decreases, becoming more negative.

When calculating acceleration at specific times, we find values like \( a(2) = \frac{-4}{25} \), \( a(4) = \frac{-8}{289} \), and \( a(6) = \frac{-12}{1369} \), confirming a steady decrease in acceleration.

Position Function

The position function is fundamental in describing the movement of an object. It gives the exact location of an object at a particular time.
This function often serves as the starting point for finding velocity and acceleration.
In our example, the position function is \( s(t) = \tan^{-1}(t) \). It describes the position of the hockey puck over time, in terms of meters.

• The inverse tangent function, \( \tan^{-1}(t) \), is used to model the puck's position.
• From this position function, one can derive how fast the position changes (velocity) and how the velocity changes (acceleration).

Understanding the position function helps in predicting the future motion of the object. By manipulating this function, you can find how an object travels and behaves over time, providing insights into its overall motion dynamics.