Skip to main content
Ch. 4 - Applications of Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 4, Problem 4.7.81

Initial Value Problems


Solve the initial value problems in Exercises 71–90.


dv/dt = (1/2)sec t tan t, v(0) = 1

Verified step by step guidance
1
Identify the given differential equation and initial condition: \(\frac{dv}{dt} = \frac{1}{2} \sec t \tan t\), with \(v(0) = 1\).
Recognize that this is a separable differential equation where the right side is a function of \(t\) only, so you can integrate both sides with respect to \(t\) to find \(v(t)\).
Set up the integral: \(v(t) = \int \frac{1}{2} \sec t \tan t \, dt + C\), where \(C\) is the constant of integration.
Recall the integral formula: \(\int \sec t \tan t \, dt = \sec t + C\). Use this to integrate the right side.
Apply the initial condition \(v(0) = 1\) to solve for the constant \(C\) by substituting \(t=0\) and \(v=1\) into the expression for \(v(t)\).

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.

Differential Equations

A differential equation relates a function with its derivatives. Solving it involves finding the original function that satisfies the given relationship between the function and its rate of change.
Recommended video:
07:39
Classifying Differential Equations

Initial Value Problems (IVP)

An initial value problem specifies the value of the unknown function at a particular point, allowing for a unique solution to the differential equation by applying this initial condition.
Recommended video:
05:03
Initial Value Problems

Integration of Trigonometric Functions

Solving the differential equation requires integrating trigonometric expressions like sec t and tan t. Knowing standard integrals and techniques for these functions is essential to find the explicit solution.
Recommended video:
6:04
Introduction to Trigonometric Functions