Table of contents

0. Functions7h 52m
1. Limits and Continuity2h 2m
2. Intro to Derivatives1h 33m
3. Techniques of Differentiation3h 18m
4. Applications of Derivatives2h 38m
5. Graphical Applications of Derivatives6h 2m
6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
7. Antiderivatives & Indefinite Integrals1h 26m
8. Definite Integrals4h 44m
9. Graphical Applications of Integrals2h 27m
- Area Between Curves
  1h 18m
- Introduction to Volume & Disk Method
  1h 8m
10. Physics Applications of Integrals 3h 16m
- Kinematics
  1h 13m
- Work
  2h 3m
11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
12. Techniques of Integration7h 39m
13. Intro to Differential Equations2h 55m
14. Sequences & Series5h 36m
15. Power Series2h 19m
- Introduction to Power Series
  1h 9m
- Taylor Series & Taylor Polynomials
  1h 10m
16. Parametric Equations & Polar Coordinates7h 58m

4. Applications of Derivatives

Differentials

Calculus

4. Applications of Derivatives

Differentials

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2:09 minutes

Problem 35c

Textbook Question

Finding Functions from Derivatives

In Exercises 31–36, find all possible functions with the given derivative.

c. y' = sin (2t) + cos (t/2)

Verified step by step guidance

Recognize that the problem involves finding the original function y(t) from its derivative y'(t). This process is called integration, as the derivative is the rate of change of the function.

Set up the integral of the given derivative: y'(t) = sin(2t) + cos(t/2). The goal is to compute the indefinite integral of this expression with respect to t: y(t) = ∫[sin(2t) + cos(t/2)] dt.

Break the integral into two separate terms for easier computation: y(t) = ∫sin(2t) dt + ∫cos(t/2) dt.

For the first term, ∫sin(2t) dt, use the substitution u = 2t, which implies du = 2 dt or dt = du/2. This simplifies the integral to (1/2)∫sin(u) du, which evaluates to -(1/2)cos(u) + C₁. Substituting back u = 2t, this becomes -(1/2)cos(2t) + C₁.

For the second term, ∫cos(t/2) dt, use the substitution v = t/2, which implies dv = (1/2) dt or dt = 2 dv. This simplifies the integral to 2∫cos(v) dv, which evaluates to 2sin(v) + C₂. Substituting back v = t/2, this becomes 2sin(t/2) + C₂. Combine the results to get the general solution: y(t) = -(1/2)cos(2t) + 2sin(t/2) + C, where C = C₁ + C₂ is the constant of integration.

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Key Concepts

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

Antiderivative

An antiderivative of a function is another function whose derivative is the original function. In calculus, finding a function from its derivative involves determining the antiderivative, which can be done using integration. The process often includes adding a constant of integration, as the derivative of a constant is zero, meaning multiple functions can share the same derivative.

Integration Techniques

To find the antiderivative of a function, various integration techniques may be employed, such as substitution, integration by parts, or recognizing standard integral forms. For the given derivative, y' = sin(2t) + cos(t/2), one would apply these techniques to integrate each term separately, ensuring to handle any necessary adjustments for constants or variable changes.

Trigonometric Integrals

Trigonometric integrals involve integrating functions that include sine and cosine. In the context of the given derivative, integrating sin(2t) and cos(t/2) requires knowledge of their respective antiderivatives. For instance, the integral of sin(kt) is -1/k cos(kt), and the integral of cos(kt) is 1/k sin(kt), where k is a constant that affects the frequency of the sine and cosine functions.

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