Problem 14 - Calculus for Biology and Medicine

Question

Suppose a particle moves along a straight line. The position at time \(t\) is given by $$ s(t)=3 t-t^{2}, \quad t \geq 0 $$ where \(t\) is measured in seconds and \(s(t)\) is measured in meters. (a) Graph \(s(t)\) for \(t \geq 0\). (b) Use the graph in (a) to answer the following questions: (i) Where is the particle at time \(0 ?\) (ii) Is there another time at which the particle visits the location where it was at time 0 ? (iii) How far to the right on the straight line does the particle travel? (iv) How far to the left on the straight line does the particle travel? (v) Where is the velocity positive? negative? equal to 0 ? (c) Find the velocity of the particle. (d) When is the velocity of the particle equal to \(1 \mathrm{~m} / \mathrm{s}\) ?

Step-by-Step Solution

Answer

(a) The graph is a downward-opening parabola with a vertex at \( t = \frac{3}{2} \). (b) (i) The particle is at 0 meters initially. (ii) It returns to 0 meters at \( t = 3 \). (iii) It moves 2.25 meters to the right. (iv) It doesn't move left beyond 0 meters. (v) Velocity is positive for \( 0 \leq t < \frac{3}{2} \), negative for \( t > \frac{3}{2} \), and zero at \( t = \frac{3}{2} \). (c) Velocity is \( v(t) = 3 - 2t \). (d) Velocity is 1 m/s at \( t = 1 \).

Step 1: Graph the Function

To graph the function \( s(t) = 3t - t^2 \) for \( t \geq 0 \), sketch a parabola opening downwards because the coefficient of \( t^2 \) is negative. The vertex is at \( t = \frac{-b}{2a} = \frac{-3}{-2} = \frac{3}{2} \). The point \( t = 0 \) gives \( s(0) = 0 \). Evaluate \( s(\frac{3}{2}) = \frac{9}{4} \) to find the maximum height. Make sure to plot these key points and use a smooth curve to connect them.

Step 2: Analyze the Position

(i) Compute the initial position: \( s(0) = 0 \). Thus, the particle is at the origin at time 0. (ii) Solve \( 3t - t^2 = 0 \) to determine when the particle is at the origin again: \( t(t-3) = 0 \), so \( t = 0 \) or \( t = 3 \) are solutions, meaning the particle returns to the origin at \( t = 3 \) seconds.

Step 3: Determine Right and Left Movement

(iii) For motion to the right, find the maximum point: \( s(\frac{3}{2}) \). The maximum forward position is \( \frac{9}{4} \) meters.(iv) Since \( s(t) \) only decreases after \( \frac{3}{2} \) without going below the origin, the particle does not move left beyond the origin, so it moves left back to 0, which is 0 meters to the left.

Step 4: Analyze Velocity

(v) Derive the velocity: \( v(t) = \frac{d}{dt}(3t - t^2) = 3 - 2t \). Solve for when velocity is zero: \( 3 - 2t = 0 \implies t = \frac{3}{2} \). Velocity is positive when \( 0 \leq t < \frac{3}{2} \) and negative when \( \frac{3}{2} < t \). The velocity is zero at \( t = \frac{3}{2} \).

Step 5: Solve for Specific Velocity

(c) As found, \( v(t) = 3 - 2t \).(d) To find when the velocity is \( 1 \) m/s, solve \( 3 - 2t = 1 \). This gives \( 2t = 2 \), or \( t = 1 \).

Key Concepts

Velocity in Particle Motion

In particle motion, velocity is essential as it defines how quickly the position changes with respect to time.

• Velocity is a function of time, often represented as \( v(t) \).
• It is obtained by differentiating the position function \( s(t) \) with respect to time \( t \).
• For our example, where \( s(t) = 3t - t^2 \), we use calculus to find velocity: \[ v(t) = \frac{d}{dt}(3t - t^2) = 3 - 2t \]

This resulting function tells us:

• The velocity is positive when \( 3 - 2t > 0 \), indicating the particle is moving forward.
• When \( 3 - 2t < 0 \), the velocity is negative, showing backward movement.
• The velocity becomes zero at \( t = \frac{3}{2} \), meaning the particle stops momentarily.

Position Function and Its Role

A position function describes where a particle is located on a line as a function of time. It’s crucial for determining the path and behavior of a particle.

• In mathematics, a position function is typically expressed as \( s(t) \).
• For the function \( s(t) = 3t - t^2 \), this tells us that the particle follows a parabolic path.

At time \( t = 0 \), the particle’s position \( s(0) = 0 \), placing it at the origin.
This function also helps identify other key positions, such as:

• Maximum displacement occurs at the vertex of the parabola, \( t = \frac{3}{2} \), yielding a positioned at \( \frac{9}{4} \) meters.
• Returns to the origin at \( t = 3 \) since \( s(3) = 0 \).

Graphing Functions to Visualize Motion

Graphing a function provides a visual representation of the motion of a particle over time. It’s a powerful tool in calculus and physics to interpret behavior.
For the position function \( s(t) = 3t - t^2 \):

• The graph is a parabola opening downwards due to the negative coefficient of \( t^2 \).
• The vertex \( (\frac{3}{2}, \frac{9}{4}) \) marks the highest point, representing the furthest point the particle travels to the right.
• The points where it crosses the x-axis, \( t = 0 \) and \( t = 3 \), indicate the instances when the particle is at the origin.

These graphs help students see intersecting points and changing directions, visualizing forward and backward motion.

Applications of Calculus in Understanding Motion

Calculus is fundamental in analyzing particle motion through derivatives and graph interpretation.

• It allows us to derive the velocity function from the position function, offering insights into speed and direction.

Using calculus:

• We find points where velocity is zero, indicating a single change in direction \( (t = \frac{3}{2}) \).
• Determine the velocity when it equals specific values, like \( 1 \) m/s at \( t = 1 \).

In particle motion, calculus helps us deeply understand changes in position and speed, which are crucial for exploring more complex physics concepts.