Problem 6 - Applied Calculus

Question

The distance (in feet) of an object from a point is given by \(s(t)=t^{2},\) where time \(t\) is in seconds. (a) What is the average velocity of the object between \(t=2\) and \(t=5 ?\) (b) By using smaller and smaller intervals around \(2,\) estimate the instantaneous velocity at time \(t=2\)

Step-by-Step Solution

Answer

(a) 7 ft/s (b) 4 ft/s

Step 1: Understand Average Velocity

To find the average velocity, we need to use the formula for average velocity over a time interval \([t_1, t_2]\), which is given by \(\text{Average Velocity} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}\). Here, \(s(t)=t^2\).

Step 2: Calculate \(s(t)\) at Given Points

Substitute \(t=2\) and \(t=5\) into \(s(t)=t^2\) to find the positions.\[s(2) = (2)^2 = 4 \s(5) = (5)^2 = 25\]

Step 3: Compute Average Velocity between \(t=2\) and \(t=5\)

Substitute the values into the average velocity formula: \[\text{Average Velocity } = \frac{25 - 4}{5 - 2} = \frac{21}{3} = 7 \text{ feet per second}\].

Step 4: Understand Instantaneous Velocity Concept

Instantaneous velocity is the derivative of the position function \(s(t)\) at a specific time \(t\). Here, we aim to find \(\frac{ds}{dt}\) at \(t=2\).

Step 5: Compute Derivative of \(s(t)\)

Find \(\frac{ds}{dt}\) by differentiating \(s(t)=t^2\). The derivative is \(s'(t) = \frac{d}{dt}(t^2) = 2t\).

Step 6: Evaluate Derivative at \(t=2\)

Substitute \(t=2\) into \(s'(t)=2t\). \[s'(2) = 2(2) = 4 \text{ feet per second}\].

Step 7: Conclusion

The average velocity between \(t=2\) and \(t=5\) is \(7\) ft/s. The instantaneous velocity at \(t=2\) is \(4\) ft/s.

Key Concepts

Average Velocity

The concept of average velocity is crucial in understanding how an object moves over a period of time. Average velocity is calculated as the change in position divided by the change in time, over a specific time interval. This is expressed with the formula:

• \( \text{Average Velocity} = \frac{s(t_2) - s(t_1)}{t_2 - t_1} \),

where \(s(t)\) is the position function.
For an example, consider the position function \(s(t) = t^2\). If you want to find the average velocity of an object between \(t = 2\) seconds and \(t = 5\) seconds:

• Calculate the object's position at \(t = 2\) seconds, which is \(s(2) = 2^2 = 4\) feet.
• Calculate the position at \(t = 5\) seconds, where \(s(5) = 5^2 = 25\) feet.
• Subtract the initial position from the final position: \(25 - 4 = 21\) feet.
• Divide this difference by the time interval \(5 - 2 = 3\) seconds.
• The average velocity is thus \(\frac{21}{3} = 7\) feet per second.

Average velocity provides insight into the object's overall speed across the interval, rather than at any single moment.

Instantaneous Velocity

Instantaneous velocity sheds light on how fast an object is moving at a specific moment in time. It can be thought of as the speed that a speedometer would read at that exact moment.
To estimate the instantaneous velocity, you can think of it as taking the average velocity over an infinitely small time interval. Practically, this involves a process of moving from larger to smaller intervals around the time in question until the interval is minuscule.
In calculus, instantaneous velocity of an object at a specific time \(t\) is found by taking the derivative of the position function \(s(t)\) with respect to time, which tells us the rate of change of the position.

• For our position function \(s(t) = t^2\), we determined earlier that the derivative is \(s'(t) = 2t\).
• To find the instantaneous velocity at \(t = 2\), substitute \(t\) into the derivative: \(s'(2) = 2 \times 2 = 4\) feet per second.

This approach gives you the velocity of the object right at that particular instant, which is essential for understanding dynamic changes in motion.

Derivative

The concept of a derivative is foundational in calculus, especially when discussing changes in physical quantities. A derivative essentially measures how a function changes as its input changes. In physics, it often represents rates of change, such as velocity being the rate of change of position with respect to time.
To find a derivative, you perform a differentiation. Consider again our position function \(s(t) = t^2\). By finding its derivative, we obtain:

• \(s'(t) = \frac{d}{dt}(t^2) = 2t\).

This expression, \(2t\), tells us that the rate of change of position (or velocity) depends linearly on time. Calculating derivatives is fundamental not only for finding instantaneous velocities but also for understanding and predicting changes in diverse contexts, from economics to engineering.
The derivative offers a precise method for analyzing how variables evolve, making it crucial for both theoretical insights and practical applications.