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Ch. 6 - Applications of Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 6 - Applications of IntegrationProblem 6.1.48c
Chapter 6, Problem 6.1.48c

Filling a tank A 2000-liter cistern is empty when water begins flowing into it (at t=0 at a rate (in L/min) given by Q′(t) = 3√t, where t is measured in minutes.


c. When will the tank be full?

Verified step by step guidance
1
Identify the rate of change of the volume of water in the tank, which is given by the function \(Q'(t) = 3\sqrt{t}\). This represents the inflow rate in liters per minute at time \(t\) minutes.
To find the total volume of water \(Q(t)\) that has flowed into the tank by time \(t\), integrate the rate function \(Q'(t)\) with respect to \(t\): \[Q(t) = \int 3\sqrt{t} \, dt = \int 3t^{1/2} \, dt.\]
Perform the integration by applying the power rule for integrals: \[Q(t) = 3 \times \frac{2}{3} t^{3/2} + C = 2 t^{3/2} + C,\] where \(C\) is the constant of integration.
Use the initial condition that the tank is empty at \(t=0\), so \(Q(0) = 0\). Substitute \(t=0\) into the integrated function to solve for \(C\): \[0 = 2 \times 0^{3/2} + C \implies C = 0.\] Thus, the volume function simplifies to \[Q(t) = 2 t^{3/2}.\]
Set the volume function equal to the tank's capacity to find when the tank is full: \[2000 = 2 t^{3/2}.\] Solve this equation for \(t\) to determine the time at which the tank reaches 2000 liters.

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Key Concepts

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

Definite Integral as Accumulated Quantity

The definite integral of a rate function over time gives the total accumulated quantity. Here, integrating the flow rate Q′(t) from 0 to t will yield the total volume of water that has entered the tank by time t.
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Definition of the Definite Integral

Solving for Time Using Integral Equations

To find when the tank is full, set the integral of the flow rate equal to the tank's capacity (2000 liters) and solve for t. This involves evaluating the integral and then isolating t in the resulting equation.
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Solving Exponential Equations Using Logs

Integration of Power Functions

The flow rate Q′(t) = 3√t can be rewritten as 3t^(1/2). Integrating power functions involves increasing the exponent by one and dividing by the new exponent, which is essential to find the volume function Q(t).
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Related Practice
Textbook Question

A nonlinear spring Hooke’s law is applicable to idealized (linear) springs that are not stretched or compressed too far from their equilibrium positions. Consider a nonlinear spring whose restoring force is given by F(x) = 16x−0.1x³, for |x|≤7. 

c. How much work is done in compressing the spring from its equilibrium position (x=0) to x=−2?

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Textbook Question

Use the region R that is bounded by the graphs of y=1+√x,x=4, and y=1 complete the exercises.


Region R is revolved about the y-axis to form a solid of revolution whose cross sections are washers.


d. Write an integral for the volume of the solid.

Textbook Question

Let R be the region in the first quadrant bounded above by the curve y=2−x² and bounded below by the line y=x. Suppose the shell method is used to determine the volume of the solid generated by revolving R about the y-axis.

c. Write an integral for the volume of the solid using the shell method.

Textbook Question

9–10. Velocity graphs The figures show velocity functions for motion along a line. Assume the motion begins with an initial position of s(0)=0. Determine the following.

d. A piecewise function for s(t)

Textbook Question

Let R be the region bounded by the curve y=√cos x and the x-axis on [0, π/2]. A solid of revolution is obtained by revolving R about the x-axis (see figures). 


c. Write an integral for the volume of the solid.

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Textbook Question

Flow rates in the Spokane River The daily discharge of the Spokane River as it flows through Spokane, Washington, in April and June is modeled by the functions

r1(t) = 0.25t²+37.46t+722.47 (April) and

r2(t) = 0.90t²−69.06t+2053.12 (June), where the discharge is measured in millions of cubic feet per day, and t=0 corresponds to the beginning of the first day of the month (see figure).

c. The Spokane River flows out of Lake Coeur d’Alene, which contains approximately 0.67mi³ of water. Determine the percentage of Lake Coeur d’Alene’s volume that flows through Spokane in April and June.