Problem 40 - Calculus for Biology and Medicine

Question

A train moves along a straight line. Its location at time \(t\) is given by $$ s(t)=\frac{100}{t}, \quad 1 \leq t \leq 5 $$ where \(t\) is measured in hours and \(s(t)\) is measured in kilometers. (a) Graph \(s(t)\) for \(1 \leq t \leq 5\). (b) Find the average velocity of the train between \(t=1\) and \(t=5\). Where on the graph of \(s(t)\) can you find the average velocity? (c) Use calculus to find the instantaneous velocity of the train at \(t=2\). Where on the graph of \(s(t)\) can you find the instantaneous velocity? What is the speed of the train at \(t=2\) ?

Step-by-Step Solution

Answer

The average velocity is -20 km/h and the instantaneous velocity at \( t=2 \) is -25 km/h. The velocity values represent slopes on the graph.

Step 1: Graphing the Function

To graph the function \( s(t) = \frac{100}{t} \) for \( 1 \leq t \leq 5 \), plot several key points. At \( t=1 \), \( s(1) = 100 \) km. At \( t=2 \), \( s(2) = 50 \) km. At \( t=3 \), \( s(3) \approx 33.3 \) km. At \( t=4 \), \( s(4) = 25 \) km. Finally, at \( t=5 \), \( s(5) = 20 \) km. By plotting these points, you'll see that the graph is a hyperbola that decreases as \( t \) increases.

Step 2: Finding Average Velocity

The average velocity \( v_{avg} \) over an interval \([t_1, t_2]\) is given by \( v_{avg} = \frac{s(t_2) - s(t_1)}{t_2 - t_1} \). For \( t_1 = 1 \) and \( t_2 = 5 \), it becomes \( v_{avg} = \frac{s(5) - s(1)}{5 - 1} = \frac{20 - 100}{4} = -20 \) km/h. This average velocity is the slope of the secant line connecting points \( (1, 100) \) and \( (5, 20) \) on the graph.

Step 3: Calculating Instantaneous Velocity

To find the instantaneous velocity \( v(t) \) at \( t=2 \), first derive \( s(t) \). The derivative \( s'(t) \) represents the instantaneous velocity and is calculated using \( s(t) = 100t^{-1} \). Therefore, \( s'(t) = -100t^{-2} = -\frac{100}{t^2} \). At \( t=2 \), \( s'(2) = -\frac{100}{4} = -25 \) km/h. This instantaneous velocity is the slope of the tangent line at \( t=2 \) on the graph.

Key Concepts

Instantaneous Velocity

Instantaneous velocity is a measure of how fast an object is moving at a specific moment in time. Unlike average velocity, which considers the total displacement over a time interval, instantaneous velocity zeroes in on the exact rate of change at a single point.
This is akin to what a speedometer in a car shows at any particular moment rather than the average speed over a trip.

• It gives a snapshot of motion in an instant.
• It's determined by the limit of average velocities as the time interval shrinks to zero.

This precise calculation is crucial for understanding motion as it provides more detailed information than the average velocity.
In this exercise, the instantaneous velocity of the train at time, \( t = 2 \), is calculated by taking the derivative of the position function \( s(t) = \frac{100}{t} \) and evaluating it at that point. A key result was the instantaneous velocity of \(-25\) km/h, which is the slope of the tangent line to the curve at that specific time.

Derivatives

Derivatives are foundational in calculus and are used to find instantaneous rates of change, such as velocity. The derivative of a function at a particular point gives the slope of the tangent to the function at that point.
In simpler terms, it tells you how fast something is changing at any given point.

• For a position function like \( s(t) \), its derivative \( s'(t) \) tells you the velocity.
• The operation of differentiation can be thought of as a mathematical form of zooming in until the curve straightens out sufficiently to estimate an accurate slope.

For the train problem, we start with the position function \( s(t) = \frac{100}{t} \). By applying the rules of differentiation, one finds that the derivative is \( s'(t) = -\frac{100}{t^2} \).
This expression precisely gives the instantaneous velocity at any time \( t \). Specifically, at \( t = 2 \), substituting in yields \(-25\) km/h as the velocity, indicating the train is slowing down as it moves along the track.

Graphing Functions

Graphing functions is a visual way to understand mathematical relationships and changes over time. It combines data points with visual interpretation, providing an immediate way to grasp how a function behaves.
For a function like \( s(t) \), the graph shows how displacement varies with time.

• A well-plotted graph allows for identification of trends, behaviors, and critical points.
• Understanding the graph can help predict future behavior and analyze past performance.

In this exercise, graphing the function \( s(t) = \frac{100}{t} \) presents a hyperbolic curve descending steeply as time goes from \( 1 \) to \( 5 \).
The graph is key in visualizing both average and instantaneous velocities: - The average velocity can be found by the slope of the secant line through two points on the graph.- The instantaneous velocity is represented by the slope of the tangent line at any given point.The graph overall illustrates the diminishing speed of the train as time progresses.