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Ch. 4 - Applications of the Derivative
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 4 - Applications of the DerivativeProblem 54b
Chapter 4, Problem 54b

Growth rate of bamboo Bamboo belongs to the grass family and is one of the fastest growing plants in the world.


b. Based on the Mean Value Theorem, what can you conclude about the instantaneous growth rate of bamboo measured in millimeters per second between 10:00 A.M. and 3:00 P.M.?

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1
First, understand the Mean Value Theorem (MVT). It states that for a continuous function f on a closed interval [a, b], there exists at least one point c in the open interval (a, b) such that the derivative f'(c) is equal to the average rate of change of the function over [a, b].
Identify the function f(t) that represents the growth of bamboo over time, where t is the time in seconds. The interval [a, b] corresponds to the time from 10:00 A.M. to 3:00 P.M. Convert these times into seconds to find the interval.
Calculate the average growth rate of bamboo over the interval [a, b]. This is done by finding the difference in bamboo height at the end points of the interval and dividing by the total time elapsed. Use the formula: \( \frac{f(b) - f(a)}{b - a} \).
Apply the Mean Value Theorem. According to MVT, there exists a time c within the interval (a, b) where the instantaneous growth rate, represented by the derivative f'(c), is equal to the average growth rate calculated in the previous step.
Conclude that at some point between 10:00 A.M. and 3:00 P.M., the instantaneous growth rate of bamboo is equal to the average growth rate over that time period. This means that the bamboo grows at a rate that matches the average rate at least once during this interval.

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Key Concepts

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

Mean Value Theorem

The Mean Value Theorem (MVT) states that if a function is continuous on a closed interval and differentiable on the open interval, there exists at least one point where the instantaneous rate of change (derivative) equals the average rate of change over that interval. This theorem is crucial for understanding how a function behaves between two points and can be applied to determine specific growth rates.
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Instantaneous Growth Rate

The instantaneous growth rate refers to the rate at which a quantity changes at a specific moment in time, represented mathematically as the derivative of a function. In the context of bamboo growth, it indicates how fast the bamboo is growing at a particular time, which can be derived from the function modeling its growth over the specified time interval.
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Average Growth Rate

The average growth rate is calculated as the change in the quantity over a specified time period divided by the length of that time period. For bamboo growth between 10:00 A.M. and 3:00 P.M., this would involve measuring the total growth in millimeters during that time and dividing it by the number of seconds in that interval, providing a baseline for comparison with the instantaneous growth rate.
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Related Practice
Textbook Question

Mean Value Theorem The population of a culture of cells grows according to the function P(t) = 100t / t+1, where t ≥ 0 is measured in weeks.


b. At what point of the interval [0, 8] is the instantaneous rate of change equal to the average rate of change?

Textbook Question

Growth rate of bamboo Bamboo belongs to the grass family and is one of the fastest growing plants in the world.


a. A bamboo shoot was 500 cm tall at 10:00 A.M. and 515 cm tall at 3:00 P.M. Compute the average growth rate of the bamboo shoot in cm/hr over the period of time from 10:00 A.M. to 3:00 P.M.

Textbook Question

{Use of Tech} Approximating reciprocals To approximate the reciprocal of a number a without using division, we can apply Newton’s method to the function f(x) = 1/x - a. 

b. Apply Newton’s method with a = 7 using a starting value of your choice. Compute an approximation with eight digits of accuracy. What number does Newton’s method approximate in this case?  

Textbook Question

First Derivative Test


a. Locate the critical points of f.

b. Use the First Derivative Test to locate the local maximum and minimum values.

c. Identify the absolute maximum and minimum values of the function on the given interval (when they exist).


f(x) = x - 2 tan⁻¹ x on [-√3,√3)

Textbook Question

{Use of Tech} Modified Newton’s method The function f has a root of multiplicity 2 at r if f(r) = f'(r) = 0 and f"(r) ≠ 0. In this case, a slight modification of Newton’s method, known as the modified (or accelerated) Newton’s method, is given by the formula xₙ + 1 = xₙ - (2f(xₙ)/(f'(xₙ), for n = 0, 1, 2, . . . . This modified form generally increases the rate of convergence.

b. Apply Newton’s method and the modified Newton’s method using x₀ = 0.1 to find the value of x₃ in each case. Compare the accuracy of these values of x₃.

Textbook Question

Verify that the following functions satisfy the conditions of Theorem 4.9 on their domains. Then find the location and value of the absolute extrema guaranteed by the theorem.


f(x) = 4x + 1/√x