Problem 4 - Applied Calculus

Question

In a time of \(t\) seconds, a particle moves a distance of \(s\) meters from its starting point, where \(s=4 t^{2}+3\). (a) Find the average velocity between \(t=1\) and \(t=\) \(1+h\) if: (i) \(\quad h=0.1\) (ii) \(\quad h=0.01,\) (iii) \(\quad h=0.001\) (b) Use your answers to part (a) to estimate the instantaneous velocity of the particle at time \(t=1\).

Step-by-Step Solution

Answer

(i) 8.4 m/s, (ii) 8.04 m/s, (iii) 8.004 m/s; instantaneous velocity is 8 m/s.

Step 1: Understanding Average Velocity

The average velocity of a particle over a time interval \([t, t+h]\) is given by the formula \( v_{avg} = \frac{s(t+h) - s(t)}{h} \). First, we'll need to calculate the position of the particle at \(t\) and \(t+h\), using the given function \(s(t) = 4t^2 + 3\).

Step 2: Calculating Position at \(t=1\) and \(t=1+h\)

First, calculate \(s(1)\): \[s(1) = 4(1)^2 + 3 = 4 + 3 = 7\] meters.Next, calculate \(s(1+h)\): \[s(1+h) = 4(1+h)^2 + 3\] Expanding the square term, we have:\[= 4(1 + 2h + h^2) + 3 = 4+8h+4h^2+3 = 7 + 8h + 4h^2\] meters.

Step 3: Calculating Average Velocity for Part (i) \(h = 0.1\)

Substitute \(s(1+h)\) and \(s(1)\) in the average velocity formula:\[v_{avg} = \frac{(7 + 8(0.1) + 4(0.1)^2) - 7}{0.1}\]\[= \frac{(7 + 0.8 + 0.04) - 7}{0.1}\]\[= \frac{0.84}{0.1} = 8.4\] m/s.

Step 4: Calculating Average Velocity for Part (ii) \(h = 0.01\)

For \(h=0.01\), using the average velocity formula:\[v_{avg} = \frac{(7 + 8(0.01) + 4(0.01)^2) - 7}{0.01}\]\[= \frac{(7 + 0.08 + 0.0004) - 7}{0.01}\]\[= \frac{0.0804}{0.01} = 8.04\] m/s.

Step 5: Calculating Average Velocity for Part (iii) \(h = 0.001\)

For \(h=0.001\), use the formula:\[v_{avg} = \frac{(7 + 8(0.001) + 4(0.001)^2) - 7}{0.001}\]\[= \frac{(7 + 0.008 + 0.000004) - 7}{0.001}\]\[= \frac{0.008004}{0.001} = 8.004\] m/s.

Step 6: Estimating Instantaneous Velocity at \(t=1\)

The instantaneous velocity is approximated by observing the trend as \(h\) approaches zero. The average velocities calculated are 8.4, 8.04, and 8.004 m/s for decreasing values of \(h\). Thus, the instantaneous velocity at \(t=1\) is estimated as 8 m/s.

Key Concepts

Instantaneous Velocity

In understanding motion, especially when a particle is moving along a straight path, it's important to distinguish between average and instantaneous velocity. While average velocity gives us the rate of change over a specific time interval, instantaneous velocity tells us how fast a particle is moving at an exact moment.

Think of it like the speedometer in a car, which shows you the car's speed at a specific point in time. To calculate instantaneous velocity, we need to find the limit of the average velocity as the time interval approaches zero. Mathematically, this is represented as \(v_{inst} = \lim_{h \to 0} \frac{s(t+h) - s(t)}{h}\).

Applying this to the problem at hand, we calculate the average velocity for smaller and smaller values of \(h\) and observe the trend as \(h\) moves closer to zero. Here, it leads us to conclude that the instantaneous velocity at \(t=1\) second is about 8 m/s.

Particle Motion

Understanding particle motion involves tracking how a particle changes its position over time. In this calculus problem, we model the particle's motion using a position function \(s(t) = 4t^2 + 3\). This function tells us where the particle is at any given time \(t\).

Let's break down what this function means:

• \(4t^2\): This quadratic term suggests that as time goes on, the particle moves faster, indicating acceleration.
• \(+ 3\): This constant term is the starting point or the initial position of the particle.

By following the changes in this function over time, we can determine how fast or slow a particle moves, known as its velocity. The focus on velocity helps us analyze whether the particle is speeding up or slowing down as it moves along its path.

Calculus Problem Solving

Calculus plays a central role in solving real-world problems related to motion and change. The problem demonstrates calculus in action by using derivatives to understand motion dynamics. Specifically, we use the concept of average velocity and then transition towards finding the instantaneous velocity.

To solve calculus problems involving motion:

• First, identify the position function that describes the motion.
• Calculate average velocities over increasingly smaller intervals to estimate the instantaneous velocity.
• Finally, apply the idea of limits to transition from average to instantaneous velocity as the interval approaches zero.

This approach allows us to transition from understanding average changes over time to gaining insight into exact changes at specific moments. By mastering these concepts, you can solve more complex problems involving rates of change, not just in motion, but in various fields like economics, biology, and physics.