Determine the area of the shaded region in the following figures.
Ch. 6 - Applications of Integration
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 6, Problema 6.1.32
29–36. Position and velocity from acceleration Find the position and velocity of an object moving along a straight line with the given acceleration, initial velocity, and initial position. Use the Fundamental Theorem of Calculus (Theorems 6.1 and 6.2).
a(t) = e^−t; v(0) = 60; s(0) = 40
Guida verificata passo dopo passo1
Identify the given acceleration function: \(a(t) = e^{-t}\), the initial velocity \(v(0) = 60\), and the initial position \(s(0) = 40\).
Recall that velocity is the integral of acceleration with respect to time: \(v(t) = \int a(t) \, dt + C_1\). Here, \(C_1\) is the constant of integration that we will find using the initial velocity.
Integrate the acceleration function: \(v(t) = \int e^{-t} \, dt + C_1\). The integral of \(e^{-t}\) is \(-e^{-t}\), so \(v(t) = -e^{-t} + C_1\).
Use the initial velocity condition \(v(0) = 60\) to solve for \(C_1\): substitute \(t=0\) into \(v(t)\) to get \(60 = -e^{0} + C_1\), then solve for \(C_1\).
Next, find the position function by integrating the velocity function: \(s(t) = \int v(t) \, dt + C_2\). Use the initial position \(s(0) = 40\) to solve for the constant of integration \(C_2\).

Risposta video verificata per un problema simile:
Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.
Durata del video:
4mConcetti chiave
Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.
Acceleration, Velocity, and Position Relationship
Acceleration is the rate of change of velocity with respect to time, and velocity is the rate of change of position. Given acceleration, velocity can be found by integrating acceleration, and position can be found by integrating velocity. Initial conditions help determine the constants of integration.
Video consigliato:
Percorso guidato
Using The Acceleration Function
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus links differentiation and integration, stating that integration can be reversed by differentiation. It allows us to find a function from its derivative by integrating, and use initial values to solve for constants, which is essential when finding velocity and position from acceleration.
Video consigliato:
Percorso guidato
Fundamental Theorem of Calculus Part 1
Initial Conditions in Differential Equations
Initial conditions specify the value of a function at a particular point, enabling the determination of integration constants after integrating. For motion problems, initial velocity and position are used to find the exact velocity and position functions from their derivatives.
Video consigliato:
Solutions to Basic Differential Equations
Pratica correlata
Domanda del libro di testo
145
views
Domanda del libro di testo
64–68. Shell method Use the shell method to find the volume of the following solids.
A hole of radius r≤R is drilled symmetrically along the axis of a bullet. The bullet is formed by revolving the parabola y = 6(1−x²/R²) about the y-axis, where 0≤x≤R.
108
views
Domanda del libro di testo
Determine the area of the shaded region in the following figures.
52
views
Domanda del libro di testo
A solid has a circular base; cross sections perpendicular to the base are squares. What method should be used to find the volume of the solid?
145
views
Domanda del libro di testo
Why is the disk method a special case of the general slicing method?
100
views
Domanda del libro di testo
Work from force How much work is required to move an object from x=1 to x=3 (measured in meters) in the presence of a force (in N) given by F(x) = 2x² acting along the x-axis?
84
views
