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Ch. 9 - Differential Equations
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 9, Problema 9.3.49a

{Use of Tech} Torricelli’s law An open cylindrical tank initially filled with water drains through a hole in the bottom of the tank according to Torricelli’s law (see figure). If h(t) is the depth of water in the tank for t≥0 s, then Torricelli’s law implies h′(t)=−k√h, where k is a constant that includes g=9.8m/s², the radius of the tank, and the radius of the drain. Assume the initial depth of the water is h(0)=Hm. 
a. Find the solution of the initial value problem.
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Guida verificata passo dopo passo
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Recognize that the differential equation given is \(h'(t) = -k \sqrt{h(t)}\), where \(h(t)\) represents the water depth at time \(t\), and \(k\) is a positive constant.
Rewrite the differential equation in separable form: \(\frac{dh}{dt} = -k \sqrt{h}\) can be rearranged as \(\frac{dh}{\sqrt{h}} = -k \, dt\).
Integrate both sides: integrate \(\int \frac{1}{\sqrt{h}} \, dh\) on the left and \(\int -k \, dt\) on the right. Recall that \(\int h^{-1/2} \, dh = 2 \sqrt{h}\).
After integration, you will have an equation involving \(\sqrt{h}\) and \(t\) plus an integration constant \(C\). Use the initial condition \(h(0) = H_m\) to solve for \(C\).
Finally, solve the resulting equation for \(h(t)\) explicitly to express the water depth as a function of time.

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Torricelli’s Law

Torricelli’s law describes the speed of fluid flowing out of an orifice under gravity, stating that the outflow velocity is proportional to the square root of the fluid’s height above the hole. In this problem, it leads to the differential equation h'(t) = -k√h, relating the rate of change of water depth to the current depth.
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Newton's Law of Cooling

Separable Differential Equations

A separable differential equation can be written as a product of a function of t and a function of h, allowing variables to be separated on opposite sides of the equation. Here, h'(t) = -k√h can be rearranged to integrate both sides and find an explicit solution for h(t).
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Solving Separable Differential Equations

Initial Value Problem (IVP)

An initial value problem involves solving a differential equation with a given initial condition, such as h(0) = Hm. This condition allows determination of the constant of integration after solving the differential equation, yielding a unique solution that models the water depth over time.
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Percorso guidato
05:03
Initial Value Problems
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Growth rate functions

a. Show that the logistic growth rate function f(P)=rP(1−P/K) has a maximum value of rK/4 at the point P=K/2.

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43–44. Motion in a gravitational field: An object is fired vertically upward with initial velocity v(0)=v₀ from initial position s(0)=s₀.

a. For the following values of v₀ and s₀, find the position and velocity functions for all times at which the object is above the ground (s = 0).

v₀ = 49 m/s, s₀ = 60 m

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Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.

a. The general solution of the differential equation y'(t) = 1 is y(t) = t

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17–20. Increasing and decreasing solutions Consider the following differential equations. A detailed direction field is not needed.


a. Find the solutions that are constant, for all t ≥ 0 (the equilibrium solutions).


y'(t) = (y−2)(y+1)

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42–43. Implicit solutions for separable equations For the following separable equations, carry out the indicated analysis.

a. Find the general solution of the equation.


e⁻ʸᐟ²y'(x) = 4x sin x² − x; y(0) = 0, y(0) = ln(1/4), y(√(π/2)) = 0


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33–36. {Use of Tech} Computing Euler approximations Use a calculator or computer program to carry out the following steps.

a. Approximate the value of y(T) using Euler’s method with the given time step on the interval [0,T].


y′(t) = -2y, y(0) = 1; Δt = 0.2, T = 2; y(t) = e⁻²ᵗ

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