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Chapter 12: Problem 43

The demand for personal computers in the home goes up with household income. For a given community, we can approximate the average number of computers in a home as $$q=0.3454 \ln x-3.047 \quad 10,000 \leq x \leq 125,000$$ where \(x\) is mean household income. \({ }^{60}\) Your community has a mean income of \(\$ 30,000\), increasing at a rate of \(\$ 2,000\) per year. How many computers per household are there, and how fast is the number of computers in a home increasing? (Round your answer to four decimal places.)

Short Answer

Expert verified
There are currently approximately \(1.8628\) computers per household in the community, and this number is increasing at a rate of approximately \(0.0230\) computers per year.

Step by step solution

01

Calculate q for the given income

We have the formula: $$q = 0.3454 \ln x - 3.047$$ With x = 30,000, we need to plug in this value to find the average number of computers in a home: $$q = 0.3454 \ln(30000) - 3.047$$ Evaluating this expression, we get: $$q \approx 1.8628$$ So, currently, there are approximately \(1.8628\) computers per household in the community.
02

Calculating the rate of change in the number of computers (dq/dt)

First, we need to find the derivative of q with respect to x: $$\frac{dq}{dx} = \frac{d}{dx}(0.3454 \ln x - 3.047)$$ Using the rules of differentiation, we get: $$\frac{dq}{dx} = 0.3454 \frac{1}{x} $$ Now, we need to find the rate of change of the number of computers with respect to time, i.e. \(\frac{dq}{dt}\). To do this, we can use the chain rule, as follows: $$\frac{dq}{dt} = \frac{dq}{dx} \cdot \frac{dx}{dt} $$ We are given that the mean income of the community is increasing at a rate of $2000 per year, which means \(\frac{dx}{dt} = 2000\). So, we can calculate \(\frac{dq}{dt}\) as follows: $$\frac{dq}{dt} = 0.3454 \frac{1}{x} \cdot 2000$$ Plug in the given mean income, x = 30000: $$\frac{dq}{dt} = 0.3454 \frac{1}{30000} \cdot 2000$$ Evaluating this expression, we get: $$\frac{dq}{dt} \approx 0.0230$$ Hence, the number of computers in a home is increasing at a rate of approximately \(0.0230\) computers per year. To summarize, there are currently approximately \(1.8628\) computers per household in the community, and this number is increasing at a rate of approximately \(0.0230\) computers per year.

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Key Concepts

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

Differentiation
Differentiation is a fundamental concept in calculus. It is a technique used to find the rate at which a quantity changes based on changes in another variable.

In simple terms, it helps us understand how a function behaves as the input changes. In our example, differentiation allows us to determine how the number of computers in a household changes as the income varies.

The formula for this is known as the derivative. The derivative represents the slope or rate of change of a function at any given point. Calculating a derivative can be done using different rules, such as the power rule, product rule, or chain rule.

In our exercise, we found the derivative of the function \( q = 0.3454 \ln x - 3.047 \) with respect to \( x \).
* When differentiating \( \ln(x) \), because it is a logarithmic function, the derivative is \( \frac{1}{x} \). As a result, the derivative of our function is \( \frac{dq}{dx} = 0.3454 \frac{1}{x} \).

This result allows us to see how sensitive the average number of computers per household is to changes in income.
Logarithmic Functions
Logarithmic functions are essential in various real-world applications. They are particularly helpful when modeling phenomena that change rapidly and then level off, such as population growth or demand for products.

The logarithmic function appears in our problem as \( \ln x \), which is the natural logarithm of \( x \). The natural logarithm has a base of \( e \), approximately equal to 2.718.

In our exercise, \( q = 0.3454 \ln x - 3.047 \), the logarithmic term \( \ln(x) \) indicates that the number of computers per household increases as income (\( x \)) increases. However, it does so at a decreasing rate.
* When income doubles, the number of computers does not necessarily double. This property is due to the nature of logarithms, making them suitable for modeling such relationships.

The logarithmic function's behavior is crucial for understanding real-world scenarios where growth constraints exist.
Rate of Change
The rate of change is a vital concept when dealing with dynamic situations, especially when we want to predict future outcomes.

In our example, we are interested in how fast the number of computers per household is increasing as the mean income rises over time. This involves using the chain rule of calculus to relate different rates.

To comprehend this better, we calculate \( \frac{dq}{dt} \), which represents the rate of change of the number of computers with respect to time.
* We achieve this by multiplying the rate of change of computers with respect to income \( \frac{dq}{dx} \) by the rate of change of income with respect to time \( \frac{dx}{dt} \).
* Given in the exercise, \( \frac{dx}{dt} = 2000 \), indicating income rises by $2000 per year.

By using the chain rule, \( \frac{dq}{dt} = 0.3454 \frac{1}{x} \cdot 2000 \), we determine the computers' increase rate when the income is \( 30000 \). Evaluating this derivative gave a result of approximately \( 0.0230 \), meaning households see an increase of about 0.0230 computers per year.

This calculation tells us how changes in income affect the growth of technology in homes.

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Most popular questions from this chapter

Let $$f(x)=\frac{N}{1+A e^{-k x}}$$ for constants \(N, A\), and \(k(A\) and \(k\) positive). Show that \(f\) has a single point of inflection at \(x=\ln A / k\).

Daily oil imports to the United States from Mexico can be approximated by \(q(t)=-0.015 t^{2}+0.1 t+1.4\) million barrels \((0 \leq t \leq 8)\) where \(t\) is time in years since the start of \(2000 .^{56}\) At the start of 2004 the price of oil was \(\$ 30\) per barrel and increasing at a rate of \(\$ 40\) per year. \({ }^{57}\) How fast was (daily) oil expenditure for imports from Mexico changing at that time?

Given that the demand \(q\) is a differentiable function of income \(x\), show that the quantity \(R=q / x\) has a stationary point when $$q-x \frac{d q}{d x}=0$$ Deduce that stationary points of \(R\) are the same as the points of unit income elasticity of demand. HINT [Differentiate \(R\) with respect to \(x\).]

The distance of the Mars orbiter from your location in Utarek, Mars is given by \(s=2(t-1)^{3}\) \(-3(t-1)^{2}+100 \mathrm{~km}\) after \(t\) seconds \((t \geq 0) .\) Obtain the extrema, points of inflection, and behavior at infinity. Sketch the curve and interpret these features in terms of the movement of the Mars orbiter.

The radius of a circular puddle is growing at a rate of \(5 \mathrm{~cm} / \mathrm{s}\) a. How fast is its area growing at the instant when the radius is \(10 \mathrm{~cm}\) ? HINT [See Example 1.] b. How fast is the area growing at the instant when it equals \(36 \mathrm{~cm}^{2}\) ? HINT [Use the area formula to determine the radius at that instant.]
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