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Ch. 9 - Differential Equations
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 9, Problem 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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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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Key Concepts

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

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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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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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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Related Practice
Textbook Question

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

Textbook Question

38–43. Equilibrium solutions A differential equation of the form y′(t)=f(y) is said to be autonomous (the function f depends only on y). The constant function y=y0 is an equilibrium solution of the equation provided f(y0)=0 (because then y'(t)=0 and the solution remains constant for all t). Note that equilibrium solutions correspond to horizontal lines in the direction field. Note also that for autonomous equations, the direction field is independent of t. Carry out the following analysis on the given equations.

a. Find the equilibrium solutions. 


y′(t) = y(y - 3)

Textbook Question

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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Textbook Question

38–43. Equilibrium solutions A differential equation of the form y′(t)=f(y) is said to be autonomous (the function f depends only on y). The constant function y=y0 is an equilibrium solution of the equation provided f(y0)=0 (because then y'(t)=0 and the solution remains constant for all t). Note that equilibrium solutions correspond to horizontal lines in the direction field. Note also that for autonomous equations, the direction field is independent of t. Carry out the following analysis on the given equations.

a. Find the equilibrium solutions. 


y′(t) = y(2 - y)

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Textbook Question

38–43. Equilibrium solutions A differential equation of the form y′(t)=f(y) is said to be autonomous (the function f depends only on y). The constant function y=y0 is an equilibrium solution of the equation provided f(y0)=0 (because then y'(t)=0 and the solution remains constant for all t). Note that equilibrium solutions correspond to horizontal lines in the direction field. Note also that for autonomous equations, the direction field is independent of t. Carry out the following analysis on the given equations.

a. Find the equilibrium solutions.


y′(t) = 2y + 4

Textbook Question

46–48. Analyzing models The following models were discussed in Section 9.1 and reappear in later sections of this chapter. In each case, carry out the indicated analysis using direction fields.

Drug infusion The delivery of a drug (such as an antibiotic) through an intravenous line may be modeled by the differential equation m′(t)+km(t)=I, where m(t) is the mass of the drug in the blood at time t≥0, K is a constant that describes the rate at which the drug is absorbed, and I is the infusion rate. Let I=10mg/hr and k=0.05 hr^−1.

a. Draw the direction field, for 0≤t≤100, 0≤y≤600.

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