Problem 3 - Single Variable Calculus

Question

In Exercises, \(s(t)\) is the position function of a body moving along a coordinate line; \(s(t)\) is measured in feet and \(t\) in seconds, where \(t \geq 0 .\) Find the position, velocity, and speed of the body at the indicated time. $$s(t)=2 t^{4}-8 t^{2}+4 ; \quad t=1$$

Step-by-Step Solution

Answer

At \(t=1\), the position, velocity, and speed of the body are as follows: 1. Position: \(s(1) = 2(1)^4 - 8(1)^2 + 4 = -2\) 2. Velocity: \(v(t) = 8t^3 - 16t\), so \(v(1) = 8(1)^3 - 16(1) = -8\) 3. Speed: \(|v(1)| = |-8| = 8\) Thus, at \(t=1\), the body is at position \(-2\) feet, has a velocity of \(-8\) ft/s, and a speed of \(8\) ft/s.

Step 1: Evaluate Position Function at \(t=1\)

To find the position of the body at \(t=1\), simply evaluate the position function at \(t=1\): $$ s(1) = 2(1)^4 - 8(1)^2 + 4 $$

Step 2: Determine the Velocity Function

To find the velocity function, we take the first derivative of the position function, \(s(t)\), with respect to \(t\): $$ v(t) = \frac{d s(t)}{d t} = \frac{d}{d t} (2 t^{4} - 8 t^{2} + 4) $$

Step 3: Evaluate Velocity Function at \(t=1\)

We found the velocity function in Step 2, and now we need to evaluate it at \(t=1\) to determine the velocity of the body at that time: $$ v(1) = \text{(evaluated expression for the velocity function at t=1)} $$

Step 4: Calculate the Speed of the Body

Finally, to find the speed of the body at \(t=1\), we take the absolute value of the velocity at that time: $$ \text{Speed} = |v(1)| $$

Key Concepts

Position Function

The position function, often denoted as \(s(t)\), describes the location of an object along a coordinate line at any given time \(t\). It is a crucial component in calculus that allows us to track movements over time. In the provided exercise, the position function is \(s(t) = 2t^4 - 8t^2 + 4\). By substituting \(t=1\) into \(s(t)\), we can determine the exact position of the object at that specific moment.

• \(s(1) = 2(1)^4 - 8(1)^2 + 4\)
• This simplifies to \(s(1) = 2(1) - 8(1) + 4\).
• Finally, \(s(1) = 2 - 8 + 4 = -2\) feet.

This result tells us that at \(t=1\), the object is located at -2 feet on the coordinate line, possibly indicating a position towards the negative direction of a predefined origin.

Velocity Function

The velocity of an object is derived from its position function by computing the first derivative. It reveals how fast the position of the object is changing with respect to time. In calculus terms, it is expressed as \(v(t) = \frac{ds(t)}{dt}\).For the given position function \(s(t) = 2t^4 - 8t^2 + 4\), we find the velocity function by differentiating:

• The power rule tells us that the derivative of \(t^n\) is \(nt^{n-1}\).
• Thus, \(\frac{d}{dt}(2t^4) = 8t^3\), \(\frac{d}{dt}(-8t^2) = -16t\), and \(\frac{d}{dt}(4) = 0\).
• Therefore, \(v(t) = 8t^3 - 16t\).

This velocity function provides insights into how fast or slow the object is moving at any time \(t\) along the coordinate line.

Speed Calculation

Speed is the magnitude of velocity, a scalar quantity, which means it has no direction. Instead of indicating movement direction, speed simply measures how fast the object travels regardless of direction.To calculate speed, take the absolute value of the velocity:

• If \(v(1)\) is the velocity at \(t=1\), then the speed is \(\text{Speed} = |v(1)|\).
• This ensures that speed is always a non-negative value.

For example, if the calculated velocity at \(t=1\) is \(v(1) = -8\), then the speed is \(|-8| = 8\) feet per second. Speed provides a straightforward answer to "how fast" without worrying about the movement direction.

Derivative

A derivative in calculus represents the rate of change. When applied to position functions, the derivative is used to find the velocity. The derivative tells us how one quantity changes as another quantity changes, crucial for understanding motion.The process of differentiation is based on limit concepts:

• Start with the function of interest, like \(s(t) = 2t^4 - 8t^2 + 4\).
• Using rules like the power rule, differentiate to find \(v(t) = 8t^3 - 16t\).
• The derivative at any point, such as \(t=1\), gives the instantaneous velocity, or rate of position change, at that specific time.

Understanding derivatives and their calculations are key to solving problems about motion in calculus. They form the foundation for many physics principles and real-world applications.