Ch. 3 - Derivatives

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 3 - Derivatives

Problem 27

Chapter 3, Problem 27

Vehicular stopping distance Based on data from the U.S. Bureau of Public Roads, a model for the total stopping distance of a moving car in terms of its speed is s = 1.1v + 0.054v², where s is measured in ft and v in mph. The linear term 1.1v models the distance the car travels during the time the driver perceives a need to stop until the brakes are applied, and the quadratic term 0.054v² models the additional braking distance once they are applied. Find ds/dv at v = 35 and v = 70 mph, and interpret the meaning of the derivative.

Verified step by step guidance

First, identify the function given for the stopping distance: s = 1.1v + 0.054v². This function represents the total stopping distance in terms of the car's speed v.

To find the derivative ds/dv, apply the rules of differentiation to each term in the function. The derivative of a constant times a variable, such as 1.1v, is simply the constant. The derivative of 0.054v² is found using the power rule, which states that d/dx of x^n is n*x^(n-1).

Calculate the derivative of the linear term: d/dv of 1.1v is 1.1.

Calculate the derivative of the quadratic term: d/dv of 0.054v² is 2 * 0.054 * v^(2-1) = 0.108v.

Combine the derivatives to find ds/dv: ds/dv = 1.1 + 0.108v. Evaluate this derivative at v = 35 mph and v = 70 mph to understand how the stopping distance changes with speed. The derivative represents the rate of change of stopping distance with respect to speed, indicating how much the stopping distance increases for each additional mph of speed.

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Key Concepts

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

Derivative

The derivative of a function represents the rate at which the function's value changes with respect to changes in its input. In this context, ds/dv is the derivative of the stopping distance s with respect to speed v, indicating how the stopping distance changes as the speed of the car changes. Calculating this derivative helps understand the sensitivity of stopping distance to changes in speed.

Power Rule for Differentiation

The power rule is a basic rule in calculus used to find the derivative of a function of the form f(x) = ax^n. The derivative is given by f'(x) = n*ax^(n-1). For the stopping distance function s = 1.1v + 0.054v², applying the power rule allows us to differentiate each term separately, resulting in ds/dv = 1.1 + 0.108v, which can then be evaluated at specific speeds.

Interpretation of Derivatives in Context

Interpreting the derivative in the context of the problem involves understanding what the calculated rate of change signifies in real-world terms. For the stopping distance model, ds/dv at a specific speed tells us how much the stopping distance is expected to increase for each additional mph of speed. This interpretation is crucial for assessing safety and understanding the impact of speed on stopping distance.

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Derivatives

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[Technology Exercise]

Draining a tank It takes 12 hours to drain a storage tank by opening the valve at the bottom. The depth y of fluid in the tank t hours after the valve is opened is given by the formula

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a. Find the rate dy/dt (m/h) at which the tank is draining at time t.

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[Technology Exercise]

Draining a tank It takes 12 hours to drain a storage tank by opening the valve at the bottom. The depth y of fluid in the tank t hours after the valve is opened is given by the formula

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