Problem 37 - Calculus Volume 1

Question

IT] For the following position functions \(y=s(t), \quad\) an object is moving along a straight line, where \(t\) is in seconds and \(s\) is in meters. Find a. the simplified expression for the average velocity from \(t=2\) to \(t=2+h\) b. the average velocity between \(t=2\) and \(t=2+h, \quad\) where \((\mathrm{i}) h=0.1, \quad\) (ii) \(h=0.01\) (iii) \(h=0.001,\) and (iv) \(h=0.0001 ;\) and c. use the answer from a. to estimate the instantaneous velocity at \(t=2\) second. $$ s(t)=2 t^{3}+3 $$

Step-by-Step Solution

Answer

The instantaneous velocity at \(t=2\) seconds is 24 m/s.

Step 1: Compute the Difference Quotient

To find the average velocity from \(t=2\) to \(t=2+h\), we start by calculating the difference quotient:\[\text{Average velocity} = \frac{s(2+h) - s(2)}{h}\]First, let's find \(s(2)\) and \(s(2+h)\).For \(s(2)\):\[s(2) = 2(2)^3 + 3 = 2 \times 8 + 3 = 16 + 3 = 19\]Next, for \(s(2+h)\):\[s(2+h) = 2(2+h)^3 + 3\]Expanding \((2+h)^3\) using the binomial theorem:\[(2+h)^3 = 2^3 + 3 \cdot 2^2 \cdot h + 3 \cdot 2 \cdot h^2 + h^3 = 8 + 12h + 6h^2 + h^3\]Thus, we have:\[s(2+h) = 2(8 + 12h + 6h^2 + h^3) + 3 = 16 + 24h + 12h^2 + 2h^3 + 3 = 19 + 24h + 12h^2 + 2h^3\]

Step 2: Simplify the Difference Quotient

Substituting the expressions for \(s(2+h)\) and \(s(2)\) into the difference quotient:\[\frac{s(2+h) - s(2)}{h} = \frac{(19 + 24h + 12h^2 + 2h^3) - 19}{h} = \frac{24h + 12h^2 + 2h^3}{h}\]Simplifying by dividing each term by \(h\):\[\frac{24h}{h} + \frac{12h^2}{h} + \frac{2h^3}{h} = 24 + 12h + 2h^2\]So, the simplified expression for the average velocity is:\[24 + 12h + 2h^2\]

Step 3: Calculate Specific Average Velocities

Substitute specific values of \(h\) into the simplified expression for the average velocity:(i) When \(h = 0.1\):\[24 + 12(0.1) + 2(0.1)^2 = 24 + 1.2 + 0.02 = 25.22\](ii) When \(h = 0.01\):\[24 + 12(0.01) + 2(0.01)^2 = 24 + 0.12 + 0.0002 = 24.1202\](iii) When \(h = 0.001\):\[24 + 12(0.001) + 2(0.001)^2 = 24 + 0.012 + 0.000002 = 24.012002\](iv) When \(h = 0.0001\):\[24 + 12(0.0001) + 2(0.0001)^2 = 24 + 0.0012 + 0.00000002 = 24.00120002\]

Step 4: Estimate Instantaneous Velocity

The instantaneous velocity at \(t=2\) is the limit of the average velocity as \(h\) approaches zero. From the simplified expression for average velocity \(24 + 12h + 2h^2\), as \(h\to 0\), we have:\[\lim_{h\to 0} (24 + 12h + 2h^2) = 24\]Therefore, the instantaneous velocity at \(t=2\) seconds is 24 meters per second.

Key Concepts

Instantaneous Velocity

When we talk about instantaneous velocity, we refer to the velocity of an object at a particular point in time. It's like finding the speed of a car at a specific instant captured by a snapshot. Unlike average velocity, which considers a time interval, instantaneous velocity is more precise. Mathematically, it is the limit of the average velocity as the time interval approaches zero. For example, in our exercise, we estimated the instantaneous velocity by taking the limit of average velocity as the interval size, represented by \(h\), gets smaller and approaches zero. At \(t=2\) seconds, that instantaneous velocity was calculated to be 24 meters per second.

Difference Quotient

The difference quotient is a fundamental concept in calculus and is used to find average rates of change. You can think of it as a tool that helps determine how a quantity changes over an interval. In our problem, the difference quotient was used to calculate the average velocity, which is

• \(\frac{s(2+h) - s(2)}{h}\)

Here, it measures how the position \(s(t)\) changes from time \(t=2\) to time \(t=2+h\). This expression is important because it sets the stage for finding instantaneous velocity by evaluating its limit as \(h\) approaches zero. In simpler terms, we first find how much the position has changed, then divide by the time interval to get average velocity. As this interval becomes smaller, this quotient becomes a prediction of instantaneous velocity.

Binomial Theorem

The binomial theorem is a powerful tool used to expand expressions raised to powers, like \((a+b)^n\). In this exercise, we used it to expand \((2+h)^3\) to help find the expression for average velocity. The expansion process follows a specific formula:

• \( (2+h)^3 = 2^3 + 3 \cdot 2^2 \cdot h + 3 \cdot 2 \cdot h^2 + h^3 = 8 + 12h + 6h^2 + h^3 \)

This expansion turns a complex polynomial into something manageable. It simplifies the computation needed to determine changes in the position function \( s(t) = 2t^3 + 3 \), which is crucial for finding the difference quotient. Knowing how to apply the binomial theorem can make problems involving polynomial expressions much easier, whether it's for calculus or general algebra.

Limit of a Function

Limits are at the heart of calculus and allow us to analyze behavior at specific points. In this exercise, the concept of a limit helped us determine the instantaneous velocity of the position function at \(t=2\) seconds. The limit takes the expression for average velocity

• \( 24 + 12h + 2h^2 \)

and analyzes its behavior as \(h\) tends to zero. The expression converges to a single value, the instantaneous velocity of 24 meters per second. Limits help formalize the idea of approaching a value without necessarily reaching it, making them extremely useful in many fields of mathematics. They allow us to handle continuously changing quantities, which makes understanding concepts like velocity both meaningful and mathematically robust.