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Theory and Examples
Maximum height of a vertically moving body The height of a body moving vertically is given by s = −12gt² + υ₀t + s₀, g > 0, with s in meters and t in seconds. Find the body’s maximum height.
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Ch. 4 - Applications of Derivatives
Hass15th EditionThomas' CalculusISBN: 9780137616077
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TOC.tableOfContents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions490
- Ch. 8 - Techniques of Integration399
- Ch. 9 - First-Order Differential Equations66
- Ch. 10 - Infinite Sequences and Series132
- Ch. 11 - Parametric Equations and Polar Coordinates58
Capitolo 4, Problema 4.5.45
The 8-ft wall shown here stands 27 ft from the building. Find the length of the shortest straight beam that will reach to the side of the building from the ground outside the wall.
Guida verificata passo dopo passo1
Visualize the problem as a right triangle where the beam is the hypotenuse, the wall is one leg, and the distance from the wall to the building is the other leg.
Let the point where the beam touches the wall be (x, 8) and the point where it touches the building be (27, y). The beam forms a right triangle with the ground and the wall.
The length of the beam can be expressed using the Pythagorean theorem: \( L = \sqrt{x^2 + (y - 8)^2} \).
To minimize the length of the beam, we need to express y in terms of x using the similar triangles formed by the beam, wall, and building. The ratio of the sides of the triangles gives us \( \frac{y}{x} = \frac{y - 8}{27} \).
Solve the equation \( \frac{y}{x} = \frac{y - 8}{27} \) for y, substitute back into the expression for L, and find the minimum value of L using calculus by taking the derivative and setting it to zero.
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Pythagorean Theorem
The Pythagorean Theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. This theorem is essential for calculating the length of the beam, as it forms a right triangle with the wall and the distance from the wall to the building.
Video consigliato:
Percorso guidato
06:11
Fundamental Theorem of Calculus Part 1
Optimization
Optimization in calculus involves finding the maximum or minimum values of a function. In this problem, we need to minimize the length of the beam, which can be modeled as a function of the height at which it touches the building. Understanding how to set up and solve optimization problems is crucial for finding the shortest beam.
Video consigliato:
10:13Intro to Applied Optimization: Maximizing Area
Related Rates
Related rates are used in calculus to find the rate at which one quantity changes in relation to another. In this scenario, as the height of the beam changes, the length of the beam also changes. Understanding how to relate these rates will help in deriving the function that describes the beam's length in terms of its height.
Video consigliato:
Percorso guidato
04:16Intro To Related Rates
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Finding Indefinite Integrals
In Exercises 17–56, find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation.
∫(cscθ cotθ) / 2 dθ
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Business and Economics
62. Production level Suppose that c(x)=x^3-20x^2 + 20,000x is the cost of manufacturing x items. Find a production level that will minimize the average cost of making x items.
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56. Airplane landing path An airplane is flying at altitude H when it begins its descent to an airport runway that is at horizontal ground distance L from the airplane, as shown in the accompanying figure. Assume that the landing path of the airplane is the graph of a cubic polynomial function y = ax^3+bx^2+cx+d, where y(-L)= H and y(0)=0.
a. What is dy/dx at x = 0?
b. What is dy/dx at x = -L?
c. Use the values for dy/dx at x = 0 and x =- L together with y(0) = 0 and y(-L) = H to show that y(x)=H[2(x/L)^3+3(x/L)^2]
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Absolute Extrema on Finite Closed Intervals
In Exercises 21–36, find the absolute maximum and minimum values of each function on the given interval. Then graph the function. Identify the points on the graph where the absolute extrema occur, and include their coordinates.
g(x) = √(4 − x²), −2 ≤ x ≤ 1
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Checking Antiderivative Formulas
Verify the formulas in Exercises 57–62 by differentiation.
∫csc²((x − 1)/3)dx = −3cot((x − 1)/3) + C
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