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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 4.7.41
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
Guida verificata passo dopo passo1
Recognize that the integral is \( \int \frac{\csc \theta \cot \theta}{2} \, d\theta \). Since the constant \( \frac{1}{2} \) can be factored out, rewrite the integral as \( \frac{1}{2} \int \csc \theta \cot \theta \, d\theta \).
Recall the derivative of \( \csc \theta \) is \( -\csc \theta \cot \theta \). This suggests that \( \csc \theta \cot \theta \) is closely related to the derivative of \( \csc \theta \).
Use this relationship to guess that the antiderivative of \( \csc \theta \cot \theta \) is \( -\csc \theta \), because differentiating \( -\csc \theta \) gives \( \csc \theta \cot \theta \).
Therefore, the integral becomes \( \frac{1}{2} \times (-\csc \theta) + C \), where \( C \) is the constant of integration.
Finally, verify your result by differentiating \( -\frac{1}{2} \csc \theta + C \) to ensure it matches the original integrand \( \frac{\csc \theta \cot \theta}{2} \).
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Indefinite Integral and Antiderivative
An indefinite integral represents the most general form of an antiderivative of a function, including a constant of integration. It reverses differentiation, finding a function whose derivative matches the integrand. Understanding this helps in solving integrals without specified limits.
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Introduction to Indefinite Integrals
Trigonometric Functions and Identities
Knowledge of trigonometric functions like cosecant (csc) and cotangent (cot), and their relationships, is essential. Recognizing identities such as the derivative of cscθ being -cscθ cotθ aids in simplifying and integrating expressions involving these functions.
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6:04Introduction to Trigonometric Functions
Verification by Differentiation
After finding an antiderivative, differentiating it confirms the correctness of the integral. This step ensures the solution is accurate and helps identify any errors in the integration process, reinforcing understanding of the fundamental theorem of calculus.
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05:53Finding Differentials
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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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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.
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Domanda del libro di testo
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.
∫(−3csc²x)dx
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Initial Value Problems
Solve the initial value problems in Exercises 71–90.
d³y/dx³ = 6; y″(0) = −8, y′(0) = 0, y(0) = 5
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