Skip to main content
Ch. 4 - Applications of the Derivative
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 4, Problem 4.R.105

104–107. Functions from derivatives Find the function f with the following properties.


ƒ'(t) = sin t + 2t; ƒ(0) = 5

Verified step by step guidance
1
Step 1: Recognize that the problem involves finding the original function ƒ(t) given its derivative ƒ'(t) = sin(t) + 2t and an initial condition ƒ(0) = 5.
Step 2: To find ƒ(t), integrate the derivative ƒ'(t). Start by breaking ƒ'(t) into two terms: sin(t) and 2t. The integral of ƒ'(t) will be the sum of the integrals of these terms.
Step 3: Compute the integral of sin(t). Recall that the integral of sin(t) with respect to t is -cos(t).
Step 4: Compute the integral of 2t. Recall that the integral of t^n (where n = 1) is t^(n+1)/(n+1). Therefore, the integral of 2t is t^2.
Step 5: Combine the results from the integration: ƒ(t) = -cos(t) + t^2 + C, where C is the constant of integration. Use the initial condition ƒ(0) = 5 to solve for C by substituting t = 0 into ƒ(t).

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
2m
Was this helpful?

Key Concepts

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

Derivatives

A derivative represents the rate of change of a function with respect to its variable. It provides information about the function's slope at any given point. In this context, ƒ'(t) = sin t + 2t indicates how the function f changes as t varies, which is essential for finding the original function f.
Recommended video:
05:44
Derivatives

Integration

Integration is the process of finding the antiderivative of a function, essentially reversing differentiation. To find the function f from its derivative ƒ'(t), we need to integrate the expression sin t + 2t. This will yield a family of functions, which we can then refine using initial conditions.
Recommended video:
06:18
Integration by Parts for Definite Integrals

Initial Conditions

Initial conditions are specific values that help determine a unique solution to a differential equation or an integral. In this case, the condition ƒ(0) = 5 allows us to find the constant of integration after performing the integration, ensuring that the function f satisfies the given requirement at t = 0.
Recommended video:
05:03
Initial Value Problems