Ch. 3 - Derivatives

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 3 - Derivatives

Problem 3.6.59b

Chapter 3, Problem 3.6.59b

{Use of Tech} Bungee jumper A woman attached to a bungee cord jumps from a bridge that is 30 m above a river. Her height in meters above the river t seconds after the jump is y(t) = 15(1+e^−t cos t), for t ≥ 0.
b. Use a graphing utility to determine when she is moving downward and when she is moving upward during the first 10 s.

Verified step by step guidance

To determine when the woman is moving downward or upward, we need to analyze the velocity of her motion. The velocity is the derivative of her height function y(t) with respect to time t.

Calculate the derivative y'(t) of the function y(t) = 15(1 + e^(-t) cos(t)). Use the product rule and chain rule for differentiation. The product rule states that if you have a function u(t)v(t), its derivative is u'(t)v(t) + u(t)v'(t).

Apply the product rule to the term e^(-t) cos(t). Let u(t) = e^(-t) and v(t) = cos(t). Then, u'(t) = -e^(-t) and v'(t) = -sin(t). The derivative of e^(-t) cos(t) is u'(t)v(t) + u(t)v'(t) = -e^(-t) cos(t) - e^(-t) sin(t).

Substitute the derivative back into the expression for y'(t): y'(t) = 15(0 + [-e^(-t) cos(t) - e^(-t) sin(t)]). Simplify this to get y'(t) = -15e^(-t)(cos(t) + sin(t)).

Use a graphing utility to plot y'(t) over the interval from t = 0 to t = 10 seconds. The woman is moving downward when y'(t) < 0 and moving upward when y'(t) > 0. Identify these intervals on the graph.

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Key Concepts

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

Differentiation

Differentiation is a fundamental concept in calculus that involves finding the derivative of a function. The derivative represents the rate of change of the function with respect to its variable, which in this case is time. By differentiating the height function y(t), we can determine the velocity of the bungee jumper, indicating whether she is moving upward or downward based on the sign of the derivative.

Critical Points

Critical points occur where the derivative of a function is zero or undefined. These points are essential for analyzing the behavior of the function, as they can indicate local maxima, minima, or points of inflection. In the context of the bungee jumper's height function, identifying critical points will help determine when her velocity changes from positive to negative, indicating a transition from upward to downward movement.

Graphing Utilities

Graphing utilities are software tools that allow users to visualize mathematical functions and their behaviors. They can plot functions, calculate derivatives, and analyze critical points graphically. In this scenario, using a graphing utility will help visualize the height function y(t) over time, making it easier to identify intervals where the bungee jumper is moving upward or downward during the first 10 seconds of her jump.

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Related Practice

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b. Evaluate f'(a) for the given values of a.

f(x) = 4x²+1; a= 2,4

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13-26 Implicit differentiation Carry out the following steps.

b. Find the slope of the curve at the given point.

(x+y)^2/3=y; (4, 4)

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13-26 Implicit differentiation Carry out the following steps.

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sin y = 5x⁴−5; (1, π)

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Textbook Question

21–30. Derivatives

b. Evaluate f'(a) for the given values of a.

f(s) = 4s³+3s; a= -3, -1

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Textbook Question

Throwing a stone Suppose a stone is thrown vertically upward from the edge of a cliff on Earth with an initial velocity of 32 ft/s from a height of 48 ft above the ground. The height (in feet) of the stone above the ground t seconds after it is thrown is s(t) = -16t²+32t+48.

b. When does the stone reach its highest point?

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Textbook Question

21–30. Derivatives

b. Evaluate f'(a) for the given values of a.

f(t) = 3t⁴; a= -2, 2

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