Problem 12 - 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

(a) Graph is decreasing curve from (1,100) to (5,20). (b) Average velocity is -20 km/h. (c) Instantaneous velocity at t=2 is -25 km/h, speed is 25 km/h.

Step 1: Graphing the Function

The function provided is \( s(t) = \frac{100}{t} \) for \( 1 \leq t \leq 5 \). To graph this, plot points for key values of \( t \) within the interval, such as \( t = 1, 2, 3, 4, \) and \( 5 \). Calculate each corresponding \( s(t) \) value: \( s(1) = 100, s(2) = 50, s(3) \approx 33.33, s(4) = 25, \) and \( s(5) = 20 \). Connect these points to form the curve.

Step 2: Finding Average Velocity

The average velocity of the train between \( t = 1 \) and \( t = 5 \) is calculated using the formula for average rate of change: \( v_{avg} = \frac{s(t_2) - s(t_1)}{t_2 - t_1} \). Here, \( t_1 = 1 \) and \( t_2 = 5 \). So the average velocity is \( v_{avg} = \frac{s(5) - s(1)}{5 - 1} = \frac{20 - 100}{4} = \frac{-80}{4} = -20 \) km/h. The average velocity is visually represented by the slope of the secant line connecting the points \((1, 100)\) and \((5, 20)\) on the graph.

Step 3: Find Instantaneous Velocity Using Calculus

To find the instantaneous velocity at \( t = 2 \), we need the derivative of \( s(t) \), \( s'(t) \). Start by differentiating \( s(t) = \frac{100}{t} \). Using the power rule, recognize \( s(t) = 100t^{-1} \), so \( s'(t) = -100t^{-2} = -\frac{100}{t^2} \). Evaluate this derivative at \( t = 2 \), which gives \( v_{inst} = s'(2) = -\frac{100}{4} = -25 \) km/h. The instantaneous velocity is visually represented by the slope of the tangent line to \( s(t) \) at \( t = 2 \).

Step 4: Understanding Speed

The speed of the train is the absolute value of its velocity. Thus, at \( t = 2 \), the speed is \(|-25| = 25 \) km/h.

Key Concepts

average velocity

The concept of average velocity is a key idea in understanding motion over a span of time. It is defined as the total displacement divided by the total time taken. In this case, we are looking at a train moving along a straight path between two time points, specifically from \( t = 1 \) to \( t = 5 \).
Average velocity is calculated using the formula:

• \( v_{\text{avg}} = \frac{s(t_2) - s(t_1)}{t_2 - t_1} \)

Here, \( s(t_2) \) and \( s(t_1) \) are the positions at the corresponding times. For the train problem, we substitute:

• \( s(5) = 20 \) km and \( s(1) = 100 \) km
• Then, calculate \( v_{\text{avg}} = \frac{20 - 100}{5 - 1} = -20 \) km/h

The negative sign indicates the train moved in the opposite direction over this period. On a graph, the average velocity is represented by the slope of the straight line, or the secant line, connecting the points at \( t = 1 \) and \( t = 5 \). This illustrates the general direction and rate of the train's movement over time.

derivative

The derivative is a powerful mathematical tool used to determine how a quantity changes instantaneously. For the train problem, we calculate the derivative of the position function \( s(t) = \frac{100}{t} \) to find the instantaneous velocity.
To differentiate, we apply the power rule and rewrite the function in a more suitable form:

• \( s(t) = 100t^{-1} \)

Using the rule, the derivative is:

• \( s'(t) = -100t^{-2} \), which can be expressed as \( -\frac{100}{t^2} \)

The derivative, \( s'(t) \), provides the instantaneous rate of change of \( s(t) \). This means at any given time, we can compute how fast the position is changing. Derivatives are essential in physics and calculus as they help relate positions to velocities. Importantly, they give us a snapshot of motion at a specific point in time.

secant line

A secant line is crucial in understanding average velocity across an interval. In our context, it connects two points on the path of the train, specifically at \( t = 1 \) and \( t = 5 \). The secant line on the graph represents the average velocity over this time interval.
Calculating the slope of this line provides the average velocity, \(-20\) km/h in our case. This slope is a visual representation of how fast the train is moving over the entire interval, not just at a single moment. The concept of the secant line allows us to transition to more specific measurements of velocity, like instantaneous velocity, highlighting how average comparisons can lead to more nuanced understandings of motion.

tangent line

The tangent line is an essential concept for finding instantaneous velocity. It is the line that just 'touches' the curve at one point without crossing it. For the train at \( t = 2 \), the tangent line to the curve represents the instantaneous velocity.Finding the slope of this tangent line involves calculus and specifically, the derivative. From the derivative \( s'(t) = -\frac{100}{t^2} \), we calculate the instantaneous velocity at \( t = 2 \):

• \( v_{\text{inst}} = s'(2) = -\frac{100}{4} = -25 \) km/h

This slope on the graph shows how fast the train is moving exactly at \( t = 2 \), unlike the secant line which represents a more general speed. The tangent line is a cornerstone of differential calculus that helps to understand the immediate direction and speed of moving objects.