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Chapter 3: Problem 9

Technicians working for the Ministry of Natural 91影视 found that the amount of a pollutant in a certain river can be represented by \(P(t)=2 t+\frac{1}{(162 t+1)}, 0 \leq t \leq 1,\) where \(t\) is the time, in years, since a cleanup campaign started. At what time was the pollution at its lowest level?

Short Answer

Expert verified
Pollution is lowest at \(t = \frac{4}{81}\).

Step by step solution

01

Understand the Problem

We need to find the time \(t\) within the interval \(0 \leq t \leq 1\) when the pollution function \(P(t)=2t+\frac{1}{162t+1}\) is at its lowest level.
02

Determine the Function's Derivative

To find the time \(t\) when the pollution is at its lowest, we first calculate the derivative of \(P(t)\) with respect to \(t\):\[ P'(t) = \frac{d}{dt}(2t + \frac{1}{162t+1}). \] This requires using the sum and quotient rules for differentiation.
03

Differentiate Using Product and Quotient Rules

Using the quotient rule \( \frac{d}{dt}\left( \frac{1}{162t+1} \right) = -\frac{162}{(162t+1)^2}\), the derivative \(P'(t)\) becomes:\[ P'(t) = 2 - \frac{162}{(162t+1)^2}. \]
04

Set the Derivative to Zero

To find critical points, we set the derivative equal to zero:\[ 2 - \frac{162}{(162t+1)^2} = 0. \] This equation helps identify points where the pollution might be at a minimum.
05

Solve for t

From the equation \( 2 - \frac{162}{(162t+1)^2} = 0 \), solve for \(t\):\[ \frac{162}{(162t+1)^2} = 2 \] \[ (162t+1)^2 = \frac{162}{2} \] \[ (162t+1)^2 = 81 \] \[ 162t+1 = 9 \; \text{or} \; 162t+1 = -9. \] Since \(t\) must be non-negative, solve \( 162t+1 = 9 \): \[ 162t = 8 \] \[ t = \frac{8}{162} \] \[ t \approx \frac{4}{81}. \]
06

Verify the Interval

Check that the found \(t\) value, \( t \approx \frac{4}{81} \), lies within the open interval \([0,1]\). It does, so it is valid for our problem.
07

Confirm a Minimum

To confirm, evaluate \(P''(t)\) at \(t = \frac{4}{81}\). If \(P''(t) > 0\), then \(t = \frac{4}{81}\) is a minimum. We find \(P''(t)\) at this point to be positive, confirming a local minimum.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Critical Points
When analyzing a function to determine where it reaches minimum or maximum values, identifying its critical points is a crucial step. Critical points are values of \( t \) where the derivative of the function equals zero or is undefined. These points are potential candidates for locations where the function could reach its lowest or highest values. In the context of the pollution problem, finding the critical points helps us locate where the pollution level might be at its minimum within the given time interval.
Derivative
In calculus, the derivative of a function provides vital information about how the function's value changes as its input changes. It gives the function's rate of change. In our example, the derivative \( P'(t) \) of the pollution function \( P(t) = 2t + \frac{1}{162t+1} \) is calculated to examine how pollution changes over time. By finding the derivative, we were able to transform the problem into one of finding when this derivative equals zero. Finding where the derivative equals zero is the key to determining the critical points, as these might indicate a minimum value of pollution within the specified interval.
Minimum Value
After identifying critical points, the next step is to verify if these points correspond to a minimum value of the function. In examining the pollution levels, we found that the critical point \( t = \frac{4}{81} \) was inside our search interval. By checking the second derivative \( P''(t) \) at this point, we confirm that it is indeed a minimum since \( P''(t) > 0 \).
This step assures us that the pollution levels were at their lowest at \( t = \frac{4}{81} \) years, which demonstrates the practical purpose of finding minimum values in real-world problems, like minimizing pollution.
Quotient Rule
The quotient rule is a method used in calculus for differentiating functions that are ratios of two differentiable functions. It is crucial in finding derivatives of functions of the form \( \frac{f(t)}{g(t)} \). In the equation for pollution \( P(t) = 2t + \frac{1}{162t+1} \), the quotient rule is used to differentiate the term \( \frac{1}{162t+1} \). The rule states that for \( \frac{f(t)}{g(t)} \), the derivative is \( \frac{f'(t)g(t) - f(t)g'(t)}{(g(t))^2} \). Applying this technique allows us to compute the derivative accurately, which is essential when finding critical points and ultimately the minimum value of the function.

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Most popular questions from this chapter

During a cough, the diameter of the trachea decreases. The velocity, \(v\) of air in the trachea during a cough may be modelled by the formula \(v(r)=A r^{2}\left(r_{0}-r\right),\) where \(A\) is a constant, \(r\) is the radius of the trachea during the cough, and \(r_{0}\) is the radius of the trachea in a relaxed state. Find the radius of the trachea when the velocity is the greatest, and find the associated maximum velocity of air. Note that the domain for the problem is \(0 \leq r \leq r_{0}.\)

The position function of a moving object is \(s(t)=t^{\frac{3}{2}}(7-t), t \geq 0,\) in metres, at time \(t,\) in seconds. a. Calculate the object's velocity and acceleration at any time \(t\) b. After how many seconds does the object stop? c. When does the motion of the object change direction? d. When is its acceleration positive? e. When does the object return to its original position?

Determine the minimal distance from point (-3,3) to the curve given by \(y=(x-3)^{2}\)

A piece of wire, \(100 \mathrm{cm}\) long, needs to be bent to form a rectangle. Determine the dimensions of a rectangle with the maximum area.

A north-south highway intersects an east-west highway at point \(P .\) A vehicle crosses \(P\) at \(1: 00 \mathrm{p} . \mathrm{m} .,\) travelling east at a constant speed of \(60 \mathrm{km} / \mathrm{h} .\) At the same instant, another vehicle is \(5 \mathrm{km}\) north of \(P,\) travelling south at \(80 \mathrm{km} / \mathrm{h}\). Find the time when the two vehicles are closest to each other and the distance between them at this time.
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