Ch. 3 - Derivatives

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 3 - Derivatives

Problem 3.6.59a

Chapter 3, Problem 3.6.59a

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.
Determine her velocity at t = 1 and t = 3.

Verified step by step guidance

To find the velocity of the woman at a given time, we need to determine the derivative of her height function y(t) with respect to time t. The derivative, y'(t), represents the velocity.

The height function is given as y(t) = 15(1 + e^(-t) * cos(t)). We will apply the product rule and chain rule to differentiate this function.

First, identify the components of the function: u(t) = e^(-t) and v(t) = cos(t). The product rule states that the derivative of u(t) * v(t) is u'(t) * v(t) + u(t) * v'(t).

Calculate the derivatives: u'(t) = -e^(-t) (using the chain rule for e^(-t)) and v'(t) = -sin(t) (derivative of cos(t)).

Substitute these derivatives into the product rule formula: y'(t) = 15 * [(-e^(-t) * cos(t)) + (e^(-t) * (-sin(t)))]. Evaluate y'(t) at t = 1 and t = 3 to find the velocity at these times.

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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. To determine the velocity of the woman at specific times, we need to differentiate her height function y(t) with respect to t, yielding the velocity function v(t).

Exponential Functions

Exponential functions are mathematical functions of the form f(t) = a * e^(bt), where e is Euler's number. In the given height function y(t), the term e^(-t) indicates that the height changes exponentially over time. Understanding how exponential decay affects the height is crucial for accurately calculating the velocity at different time points.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, are periodic functions that relate angles to ratios of sides in right triangles. In the height function y(t), the cosine term introduces oscillatory behavior to the height over time. Recognizing how the cosine function influences the overall height is important for evaluating the velocity at specific moments during the jump.

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Related Practice

Textbook Question

Surface area of a cone The lateral surface area of a cone of radius r and height h (the surface area excluding the base) is A = πr√r²+h².

a. Find dr/dh for a cone with a lateral surface area of A=1500π.

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Textbook Question

13-26 Implicit differentiation Carry out the following steps.

a. Use implicit differentiation to find dy/dx.

³√x+³√y⁴ = 2;(1,1)

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Textbook Question

Volume of a torus The volume of a torus (doughnut or bagel) with an inner radius of a and an outer radius of b is V=π²(b+a)(b−a)²/4.

a. Find db/da for a torus with a volume of 64π².

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Textbook Question

Comparing velocities Two stones are thrown vertically upward, each with an initial velocity of 48 ft/s at time t=0. One stone is thrown from the edge of a bridge that is 32 feet above the ground, and the other stone is thrown from ground level. The height above the ground of the stone thrown from the bridge after t seconds is f(t) = − 16t²+48t+32. and the height of the stone thrown from the ground after t seconds is g(t) = −16t²+48t.

a. Show that the stones reach their high points at the same time.

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Textbook Question

Derivatives and tangent lines

a. For the following functions and values of a, find f′(a).

f(x) = √3x; a= 12

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Textbook Question

13-26 Implicit differentiation Carry out the following steps.

a. Use implicit differentiation to find dy/dx.