Domanda del libro di testo
Identifying Extrema
In Exercises 41–52:
a. Identify the function’s local extreme values in the given domain, and say where they occur.
g(x) = −x² − 6x − 9,−4 ≤ x < ∞
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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 4.7.101a
Finding displacement from an antiderivative of velocity
a. Suppose that the velocity of a body moving along the s-axis is
ds/dt = v = 9.8t − 3.
i. Find the body’s displacement over the time interval from t = 1 to t = 3 given that s = 5 when t = 0.
Guida verificata passo dopo passo1
Identify the given velocity function: \(v(t) = \frac{ds}{dt} = 9.8t - 3\).
Recall that displacement over the interval \([t_1, t_2]\) is given by the change in position \(s(t_2) - s(t_1)\), which can be found by integrating the velocity function over that interval.
Set up the definite integral for displacement from \(t=1\) to \(t=3\): \(\Delta s = \int_{1}^{3} (9.8t - 3) \, dt\).
Find the antiderivative (indefinite integral) of the velocity function: \(S(t) = \int (9.8t - 3) \, dt = 4.9t^2 - 3t + C\), where \(C\) is the constant of integration.
Use the initial condition \(s(0) = 5\) to solve for \(C\) by substituting \(t=0\) into \(S(t)\), then evaluate \(S(3) - S(1)\) to find the displacement over the interval \([1,3]\).
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Relationship Between Velocity and Displacement
Velocity is the rate of change of displacement with respect to time, expressed as v = ds/dt. To find displacement over a time interval, we integrate the velocity function over that interval, which gives the net change in position.
Video consigliato:
Percorso guidato
06:29
Derivatives Applied To Velocity
Definite Integration to Find Displacement
Definite integration of the velocity function from the initial time to the final time calculates the total displacement. This integral sums the instantaneous velocities over time, accounting for direction and magnitude of movement.
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Percorso guidato
05:43Definition of the Definite Integral
Initial Conditions and Antiderivatives
An antiderivative of velocity gives the displacement function s(t) up to a constant. Using the initial condition s(0) = 5 allows us to solve for this constant, ensuring the displacement function accurately reflects the body's position at all times.
Video consigliato:
Percorso guidato
05:50Antiderivatives
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Domanda del libro di testo
53. Distance between two ships At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (nautical miles per hour; a nautical mile is 2000 yd) and continued to do so all day. Ship B was sailing east at 8 knots and continued to do so all day.
a. Start counting time with t=0 at noon and express the distance s between the ships as a function of t.
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Domanda del libro di testo
Theory and Examples
Sketch the graph of a differentiable function y = f(x) that has a local minimum at (1, 1) and a local maximum at (3, 3).
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Domanda del libro di testo
Checking Antiderivative Formulas
Right, or wrong? Say which for each formula and give a brief reason for each answer.
∫tanθ sec²θ dθ = sec³θ / 3 + C
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Domanda del libro di testo
Finding Antiderivatives
In Exercises 1–16, find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation.
πcos πx
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Domanda del libro di testo
Analyzing Functions from Derivatives
Answer the following questions about the functions whose derivatives are given in Exercises 1–14:
a. What are the critical points of f?
f′(x) = 1− 4/x², x ≠ 0
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