Problem 37 - Calculus Volume 1

Question

For the following position functions \(y=s(t),\) 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 \((\) i) \(h=0.1\) (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

a. The average velocity expression is \(24 + 12h + 2h^2\). b. Specific values: (i) 25.22, (ii) 24.1202, (iii) 24.012002, (iv) 24.00120002. c. Instantaneous velocity: 24 m/s at \(t = 2\).

Step 1: Find the difference in position

To find the simplified expression for the average velocity, we must first calculate the difference in position for the time interval from \( t = 2 \) to \( t = 2 + h \). The position function is given as \( s(t) = 2t^3 + 3 \). Calculate \( s(2+h) \) and \( s(2) \):\[s(2+h) = 2(2+h)^3 + 3\]\[s(2) = 2(2)^3 + 3 = 16 + 3 = 19\]

Step 2: Simplify \( s(2+h) \)

Expand the expression \( s(2+h) = 2(2+h)^3 + 3 \):First calculate \( (2+h)^3 \):\[(2+h)^3 = (2+h)(2+h)(2+h) = (4 + 4h + h^2)(2+h) = 8 + 12h + 6h^2 + h^3\]Thus,\[2(2+h)^3 = 2(8 + 12h + 6h^2 + h^3) = 16 + 24h + 12h^2 + 2h^3\]Add 3 to this result:\[s(2+h) = 16 + 24h + 12h^2 + 2h^3 + 3 = 19 + 24h + 12h^2 + 2h^3\]

Step 3: Calculate the difference quotient

Now, calculate the difference in position: \( \Delta s = s(2+h) - s(2) \):\[\Delta s = (19 + 24h + 12h^2 + 2h^3) - 19 = 24h + 12h^2 + 2h^3\]The average velocity is given by:\[\text{Average Velocity} = \frac{\Delta s}{\Delta t} = \frac{24h + 12h^2 + 2h^3}{h}\]

Step 4: Simplify the average velocity expression

Simplify the expression for average velocity:\[\frac{24h + 12h^2 + 2h^3}{h} = \frac{h(24 + 12h + 2h^2)}{h} = 24 + 12h + 2h^2\]Therefore, the simplified expression for the average velocity is:\[24 + 12h + 2h^2\]

Step 5: Calculate the average velocity for specific \( h \) values

Using the expression from Step 4, substitute the different values for \( h \).(i) For \( h = 0.1 \):\[24 + 12(0.1) + 2(0.1)^2 = 24 + 1.2 + 0.02 = 25.22\](ii) For \( h = 0.01 \):\[24 + 12(0.01) + 2(0.01)^2 = 24 + 0.12 + 0.0002 = 24.1202\](iii) For \( h = 0.001 \):\[24 + 12(0.001) + 2(0.001)^2 = 24 + 0.012 + 0.000002 = 24.012002\](iv) For \( h = 0.0001 \):\[24 + 12(0.0001) + 2(0.0001)^2 = 24 + 0.0012 + 0.00000002 = 24.00120002\]

Step 6: Estimate instantaneous velocity at \( t=2 \)

To find the instantaneous velocity at \( t=2 \), we take the limit of the average velocity expression \( 24 + 12h + 2h^2 \) as \( h \to 0 \).\[\text{Instantaneous Velocity} = \, \lim_{{h \to 0}} (24 + 12h + 2h^2) = 24\]Thus, the instantaneous velocity at \( t = 2 \) is 24 meters per second.

Key Concepts

Average Velocity

Average velocity helps us understand the overall rate of change of position over a specific time period. It shows us how fast an object is moving on average between two time points. If you think about driving a car from one city to another, average velocity would be similar to the average speed shown on your car's trip computer for the whole journey.
For an object with a position function given by \(s(t) = 2t^3 + 3\), we can find the average velocity over an interval \([t_1, t_2]\). The key steps involve computing the change in position, \(\Delta s = s(t_2) - s(t_1)\), and the change in time, \(\Delta t = t_2 - t_1\).
Therefore, the average velocity \(v_{\text{avg}}\) is calculated by:

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

For our specific case, we examined the interval from \(t = 2\) to \(t = 2+h\), and derived a general expression for average velocity, simplified as \(24 + 12h + 2h^2\). This formula allows us to plug in values for \(h\) to examine average velocity over very short time spans.

Instantaneous Velocity

Instantaneous velocity refers to how fast an object is moving at exactly one specific point in time. It's like reading the speedometer on your car at a precise instant. To find the instantaneous velocity, we use calculus to explore what happens as the time interval becomes infinitesimally small.
In the context of our problem, we applied the position function \(s(t) = 2t^3 + 3\) to derive a general expression for average velocity. Then, we took the limit of this expression as \(h\) approaches zero. Essentially, this means that we're looking at what happens to our average velocity formula, \(24 + 12h + 2h^2\), as \(h\) gets smaller and smaller, until it approaches zero.
As we calculated, the limit was \(24\). This means the instantaneous velocity at \(t = 2\) seconds is \(24\) meters per second. It's important because it tells us the exact speed of the object at that particular moment in time.

Position Function

A position function describes the location of an object along a path over time. The function provides a mathematical way to understand how an object moves. In our exercise, the position function is given as \(s(t) = 2t^3 + 3\). Here, \(s(t)\) represents the position in meters at any time \(t\) in seconds.
As you analyze this function, you'll notice that it illustrates non-linear motion. The cubic term \(2t^3\) implies that the object's position changes at varying rates as time progresses. The constant \(3\) indicates the starting position at time \(t=0\).
Understanding position functions is essential for calculating velocities as they are the foundational element from which we derive velocity functions. By differentiating a position function, we can uncover the velocity function which provides insights into an object's speed and direction at any given moment.

Velocity Calculation

The calculation of velocity, whether average or instantaneous, provides us with insights into how fast an object moves and in what direction. To compute velocity, we begin with a position function, such as \(s(t) = 2t^3 + 3\) from our exercise.
For average velocity, the procedure involves finding changes in position and dividing this by the change in time. We can visualize it as taking snapshots of where the object is over time and determining how far it has traveled over a certain period.

• For example, the average velocity between \(t = 2\) and \(t = 2+h\) is: \(\frac{\Delta s}{\Delta t}\), leading to \(24 + 12h + 2h^2\).

Instantaneous velocity, on the other hand, requires us to look at derivatives. By differentiating the position function \(s(t)\), we calculate the rate of position change at any specific moment. This gives us our instantaneous velocity.
In our problem set, by evaluating the limit of the average velocity expression as \(h\) approaches zero, we were able to determine that the instantaneous velocity at \(t = 2\) seconds is \(24\) meters per second, providing insights into the object's motion at that precise instant.