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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 3 - DerivativesProblem 3.6.11c
Chapter 3, Problem 3.6.11c

Highway travel A state patrol station is located on a straight north-south freeway. A patrol car leaves the station at 9:00 A.M. heading north with position function s = f(t) that gives its location in miles t hours after 9:00 A.M. (see figure). Assume s is positive when the car is north of the patrol station. <IMAGE>
c. Find the average velocity of the car over the interval [1.75, 2.25]. Estimate the velocity of the car at 11:00 A.M. and determine the direction in which the patrol car is moving.

Verified step by step guidance
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To find the average velocity of the car over the interval [1.75, 2.25], use the formula for average velocity: \( v_{avg} = \frac{f(b) - f(a)}{b - a} \), where \( a = 1.75 \) and \( b = 2.25 \).
Substitute the values of \( a \) and \( b \) into the formula: \( v_{avg} = \frac{f(2.25) - f(1.75)}{2.25 - 1.75} \).
To estimate the velocity of the car at 11:00 A.M., which corresponds to \( t = 2 \) hours after 9:00 A.M., find the derivative of the position function \( s = f(t) \) to get the velocity function \( v(t) = f'(t) \).
Evaluate the derivative at \( t = 2 \) to estimate the velocity: \( v(2) = f'(2) \).
Determine the direction of the patrol car by checking the sign of \( v(2) \). If \( v(2) > 0 \), the car is moving north; if \( v(2) < 0 \), the car is moving south.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Average Velocity

Average velocity is defined as the change in position divided by the time interval over which that change occurs. Mathematically, it is calculated as (s(t2) - s(t1)) / (t2 - t1), where s(t) is the position function. In this context, it helps determine how fast the patrol car is moving on average between two specific times.
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Average Value of a Function

Position Function

The position function s = f(t) describes the location of an object at any given time t. In this scenario, it represents the distance of the patrol car from the station in miles, where t is measured in hours after 9:00 A.M. Understanding this function is crucial for calculating both average and instantaneous velocities.
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Instantaneous Velocity

Instantaneous velocity refers to the velocity of an object at a specific moment in time, which can be found by taking the derivative of the position function with respect to time. This concept is essential for estimating the patrol car's speed at 11:00 A.M. and determining its direction of movement, as it indicates whether the car is moving north or south.
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Related Practice
Textbook Question

{Use of Tech} Spring runoff The flow of a small stream is monitored for 90 days between May 1 and August 1. The total water that flows past a gauging station is given by v(t) = <matrix 2x2> where V is measured in cubic feet and t is measured in days, with t=0 corresponding to May 1.

c. Describe the flow of the stream over the 3-month period. Specifically, when is the flow rate a maximum?

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Textbook Question

Another way to approximate derivatives is to use the centered difference quotient: f' (a) ≈ f(a+h) - f(a- h) / 2h. Again, consider f(x) = √x.

c. Explain why it is not necessary to use negative values of h in the table of part (b).

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62–65. {Use of Tech} Graphing f and f'

c. Verify that the zeros of f' correspond to points at which f has a horizontal tangent line.

f(x)=e^−x tan^−1 x on [0,∞)

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City urbanization City planners model the size of their city using the function A(t) = - 1/50t² + 2t +20, for 0 ≤ t ≤ 50, where A is measured in square miles and t is the number of years after 2010.

c. Suppose the population density of the city remains constant from year to year at 1000 people mi². Determine the growth rate of the population in 2030.

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Computing the derivative of f(x) = e^-x

c. Use parts (a) and (b) to find the derivative of f(x) = e^-x.

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Suppose a stone is thrown vertically upward from the edge of a cliff on Earth with an initial velocity of 19.6 m/s from a height of 24.5 m above the ground. The height (in meters) of the stone above the ground t seconds after it is thrown is s(t) = -4.9t²+19.6t+24.5.

c. What is the height of the stone at the highest point?

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