Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

Suppose that an object moves along a coordinate line so that its directed distance from the origin after \(t\) seconds is \(\sqrt{2 t+1}\) feet. (a) Find its instantaneous velocity at \(t=\alpha, \alpha>0\). (b) When will it reach a velocity of \(\frac{1}{2}\) foot per second? (see Example 5.)

Step-by-Step Solution

Answer

(a) \( \frac{1}{\sqrt{2\alpha + 1}} \); (b) \( t = \frac{3}{2} \) seconds.

Step 1: Understand the Problem

We have a function for the position of an object: \( s(t) = \sqrt{2t + 1} \), representing distance in feet from the origin with respect to time \( t \) in seconds. We need to find the instantaneous velocity at time \( t = \alpha \) and when this velocity will be \( \frac{1}{2} \) feet per second.

Step 2: Calculate the Velocity Function

The instantaneous velocity of an object at any time \( t \) is given by the derivative of the position function with respect to time. We need to differentiate \( s(t) = \sqrt{2t + 1} \). Using the chain rule: \[ \frac{d}{dt}[\sqrt{2t + 1}] = \frac{1}{2\sqrt{2t + 1}} \cdot (2) = \frac{1}{\sqrt{2t + 1}}. \]Thus, the velocity function \( v(t) = \frac{1}{\sqrt{2t + 1}} \).

Step 3: Find Instantaneous Velocity at \( t = \alpha \)

Substitute \( t = \alpha \) into the velocity function: \[ v(\alpha) = \frac{1}{\sqrt{2\alpha + 1}}. \]

Step 4: Determine When Velocity is \( \frac{1}{2} \)

Set the velocity function equal to \( \frac{1}{2} \) and solve for \( t \): \[ \frac{1}{\sqrt{2t + 1}} = \frac{1}{2}. \]Cross-multiplying gives \( \sqrt{2t + 1} = 2 \). Square both sides to get: \[ 2t + 1 = 4. \] Solving for \( t \), we get: \[ 2t = 3 \Rightarrow t = \frac{3}{2}. \]

Step 5: Validate all Conditions

Confirm that \( t = \frac{3}{2} \) satisfies the condition \( t > 0 \) given in the problem and is appropriate according to the initial domain of the function.

Key Concepts

Calculus Problem Solving

Solving calculus problems usually involves a step-by-step approach, ensuring that each stage of the solution is addressed clearly. Here is how you can navigate through a calculus exercise like the one provided:

• **Understanding the Problem:** Break down the problem statement carefully. Ask yourself what you are trying to find, which formulas might apply, and any conditions that have been given, like time constraints or values.
• **Establish the Right Approach:** Use your calculus knowledge to decide which methods or rules are applicable. In this problem, determining instantaneous velocity involves using differentiation.
• **Sequential Steps:** Follow each required mathematical step as accurately as possible. For instance, differentiating the position function to find the velocity function is key in this exercise.
• **Check Your Work:** After finding a solution, verify that your answer makes sense, adheres to any conditions, and works within the context of the problem.

By sticking to these principles, you can tackle calculus problems more effectively and confidently.

Differentiation

Differentiation is a fundamental tool in calculus that helps in finding the rate at which a quantity changes. It's crucial for determining instantaneous velocity as shown in this problem.
The key idea in differentiation is to compute the derivative of a function. A derivative represents the slope of the tangent to the curve at any given point.

• **Basic Rules:** Familiarize yourself with rules like the power rule, product rule, quotient rule, and chain rule, as they form the backbone of solving differentiation problems.
• **Chain Rule Application:** The chain rule is used when you have a function within another function, just like in our position function, \( s(t) = \sqrt{2t + 1} \). When differentiating, find the derivative of the outside function and multiply it by the derivative of the inside function.

The derivative of our function, using the chain rule, leads us to the velocity function \( v(t) = \frac{1}{\sqrt{2t + 1}} \). This derivative tells us how quickly the position of the object changes with time.

Velocity Function

The velocity function gives us the speed and direction at which an object is moving at any point in time. Determining this involves taking the derivative of a position function.
In this exercise, the position of the object is denoted by \( s(t) = \sqrt{2t + 1} \). To find the instantaneous velocity, we differentiate this function to obtain the velocity function \( v(t) = \frac{1}{\sqrt{2t + 1}} \).

• **Interpreting Instantaneous Velocity:** Instantaneous velocity is the exact speed and direction of an object at an exact moment in time. For instance, \( v(\alpha) = \frac{1}{\sqrt{2\alpha + 1}} \) gives the velocity at the moment when time equals \( \alpha \).
• **Solving for a Specific Velocity:** To find the time when the velocity will be \( \frac{1}{2} \) feet per second, equate the velocity function to \( \frac{1}{2} \), and solve for \( t \).

This approach provides a clear picture of the behavior and movement of the object along the coordinate line, illustrating how quickly it is moving and in which direction at any specific time.