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Ch. 4 - Applications of the Derivative
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 4, Problema 4.1.87b

{Use of Tech} Every second counts You must get from a point P on the straight shore of a lake to a stranded swimmer who is 50 from a point Q on the shore that is 50 m from you (see figure). Assuming that you can swim at a speed of 2 m/s and run at a speed of 4 m/s, the goal of this exercise is to determine the point along the shore, x meters from Q, where you should stop running and start swimming to reach the swimmer in the minimum time. <IMAGE>


b. Find the critical point of T on (0, 50).

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First, understand the problem setup: You need to minimize the time taken to reach the swimmer by running along the shore and then swimming. The swimmer is 50 meters from point Q, and you are initially 50 meters from Q along the shore.
Define the variables: Let x be the distance from point Q where you stop running and start swimming. The total distance you run is (50 - x) meters, and the distance you swim is 50 meters.
Express the time taken for each segment: The time to run is given by \( \frac{50 - x}{4} \) seconds, and the time to swim is \( \frac{50}{2} \) seconds. The total time T is the sum of these two times.
Formulate the function for total time T: \( T(x) = \frac{50 - x}{4} + \frac{50}{2} \). Simplify this expression to find T as a function of x.
Find the critical points: To find the critical points of T on the interval (0, 50), take the derivative of T with respect to x, set it equal to zero, and solve for x. This will give you the point where the time is minimized.

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Optimization

Optimization in calculus involves finding the maximum or minimum values of a function. In this context, we need to minimize the total time taken to reach the swimmer by determining the optimal point along the shore to switch from running to swimming. This requires setting up a function that represents the total time as a function of the distance run and then finding its critical points.
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Intro to Applied Optimization: Maximizing Area

Critical Points

Critical points are values in the domain of a function where the derivative is either zero or undefined. These points are essential in optimization problems as they help identify potential maxima or minima. In this scenario, finding the critical point of the time function T will allow us to determine the optimal distance to run before swimming.
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Critical Points

Distance and Speed Relationships

Understanding the relationship between distance, speed, and time is crucial for solving this problem. The time taken to reach the swimmer consists of the time spent running and the time spent swimming, which can be expressed using the formula time = distance/speed. By breaking down the total distance into components based on the chosen point along the shore, we can formulate the time function needed for optimization.
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Derivatives Applied To Velocity
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{Use of Tech} Basketball shot A basketball is shot with an initial velocity of v ft/s at an angle of 45° to the floor. The center of the basketball is 8 ft above the floor at a horizontal distance of 18 feet from the center of the basketball hoop when it is released. The height h (in feet) of the center of the basketball after it has traveled a horizontal distance of x feet is modeled by the function h(x) = 32x² / v² + x + 8 (see figure). <IMAGE>



b. During the flight of the basketball, show that the distance s from the center of the basketball to the front of the hoop is s = √ (x - 17.25)² + ( -(4x² / 81) + x - 2)² (Hint: The diameter of the basketball hoop is 18 inches.) 

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b. Does either g or h have a local extreme value at x = 2? Explain.

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{Use of Tech} A damped oscillator The displacement of an object as it bounces vertically up and down on a spring is given by y(t) = 2.5e⁻ᵗ cos 2t, where the initial displacement is y(0) = 2.5 and y = 0 corresponds to the rest position (see figure). <IMAGE>

b. Find the time and the displacement when the object reaches its lowest point.

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b. Consider the polynomial g(x) = f(f(x)). Write g in terms of a and powers of x. What is its degree?

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{Use of Tech} Demand functions and elasticity Economists use demand functions to describe how much of a commodity can be sold at varying prices. For example, the demand function D(p) = 500 - 10p says that at a price of p = 10, a quantity of D(10) = 400 units of the commodity can be sold. The elasticity E = dD/dp p/D of the demand gives the approximate percent change in the demand for every 1% change in the price. (See Section 3.6 or the Guided Project Elasticity in Economics for more on demand functions and elasticity.)


b. If the price is \$12 and increases by 4.5%, what is the approximate percent change in the demand? 

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