Problem 65
Question
NET ASSET VALUE It is estimated that \(t\) days from now a farmer's crop will be increasing at the rate of \(0.5 t^{2}+4(t+1)^{-1}\) bushels per day. By how much will the value of the crop increase during the next 6 days if the market price remains fixed at \(\$ 2\) per bushel?
Step-by-Step Solution
Verified Answer
The crop's value will increase by approximately \$ 87.57.
1Step 1: Understand the Rate of Increase
The rate at which the farmer's crop is increasing is given by the function: \(0.5t^2 + \frac{4}{t+1}\). This represents the bushels per day increase at time \(t\).
2Step 2: Set the Integration Boundaries
We need to find out the increase in bushels from today \(t = 0\) to 6 days from now \(t = 6\). Therefore, we will integrate the rate function from \(t = 0\) to \(t = 6\).
3Step 3: Integrate the Function
Calculate the definite integral of the rate function from \(0\) to \(6\): \[\int_{0}^{6} \left( 0.5t^2 + \frac{4}{t+1} \right)\, dt\].
4Step 4: Solve the Integral
First, find the antiderivative of the rate function. The antiderivative of \(0.5t^2\) is \(\frac{0.5t^3}{3} = \frac{t^3}{6}\), and the antiderivative of \(\frac{4}{t+1}\) is \(4\ln(|t+1|)\). Thus, the antiderivative of the entire function is: \frac{t^3}{6} + 4\ln(|t+1|)\. Next, evaluate this function from \(t=0\) to \(t=6\).
5Step 5: Evaluate the Antiderivative at the Bounds
Substitute the limits into the antiderivative: \[\left( \frac{6^3}{6} + 4\ln(7) \right) - \left( \frac{0^3}{6} + 4\ln(1) \right)\]. This simplifies to: \[36 + 4\ln(7) - 0\]. Since \(\ln(1) = 0\), the result is \[36 + 4\ln(7)\].
6Step 6: Simplify and Evaluate Numerically
Calculate the numerical value: \( \approx 36 + 4(1.9459) = 36 + 7.7836 = 43.7836 \) bushels.
7Step 7: Calculate the Increase in Value
Since each bushel is valued at \$ 2 per bushel, the total increase in value is \[43.7836 \times 2 = 87.5672 \text{ dollars}\].
Key Concepts
Definite IntegrationRate of ChangeAntiderivativesEconomic Mathematics
Definite Integration
In this exercise, we use definite integration to find out how much a farmer's crop will increase over a given period. Definite integration allows us to calculate the exact total of quantities that vary over time.
We have the rate of increase of the crop given by the function: \(0.5t^2 + \frac{4}{t+1}\).
To find the total increase in bushels from day 0 to day 6, we integrate this function between the limits of 0 and 6.
That means we compute the integral:\[ \int_{0}^{6} \left( 0.5t^2 + \frac{4}{t+1} \right) dt \].
This process accumulates the small increases day by day over the 6 days, to give us the total increase.
We have the rate of increase of the crop given by the function: \(0.5t^2 + \frac{4}{t+1}\).
To find the total increase in bushels from day 0 to day 6, we integrate this function between the limits of 0 and 6.
That means we compute the integral:\[ \int_{0}^{6} \left( 0.5t^2 + \frac{4}{t+1} \right) dt \].
This process accumulates the small increases day by day over the 6 days, to give us the total increase.
Rate of Change
The rate of change is an essential concept that defines how a quantity varies with time.
In this scenario, the rate of change of the crop's growth, given by the function \(0.5t^2 + \frac{4}{t+1}\), tells us how many bushels the farmer's crop increases each day.
At day 0, the rate might be different than at day 3 or day 6.
Understanding the rate of change helps in knowing not just how much the crop increases, but how rapidly or slowly this process happens over different periods. The function itself can change at each point in time, and by integrating it, we get a comprehensive view of the total change over the desired timeframe.
In this scenario, the rate of change of the crop's growth, given by the function \(0.5t^2 + \frac{4}{t+1}\), tells us how many bushels the farmer's crop increases each day.
At day 0, the rate might be different than at day 3 or day 6.
Understanding the rate of change helps in knowing not just how much the crop increases, but how rapidly or slowly this process happens over different periods. The function itself can change at each point in time, and by integrating it, we get a comprehensive view of the total change over the desired timeframe.
Antiderivatives
Antiderivatives are the opposite of derivatives.
While a derivative gives us the rate of change of a function, an antiderivative helps us find the original function given its rate of change.
For example, to integrate \(0.5t^2\), whose derivative is \(t^2\), we find its antiderivative to be \(\frac{t^3}{6}\).
Similarly, the antiderivative of \(\frac{4}{t+1}\) is \(4\ln(|t+1|)\).
Putting these together, we get the antiderivative of the function \(0.5t^2 + \frac{4}{t+1}\), which is:\[ \frac{t^3}{6} + 4\ln(|t+1|) \],
Evaluating this antiderivative from 0 to 6 gives us the accumulated increase in the crop over the 6-day period.
While a derivative gives us the rate of change of a function, an antiderivative helps us find the original function given its rate of change.
For example, to integrate \(0.5t^2\), whose derivative is \(t^2\), we find its antiderivative to be \(\frac{t^3}{6}\).
Similarly, the antiderivative of \(\frac{4}{t+1}\) is \(4\ln(|t+1|)\).
Putting these together, we get the antiderivative of the function \(0.5t^2 + \frac{4}{t+1}\), which is:\[ \frac{t^3}{6} + 4\ln(|t+1|) \],
Evaluating this antiderivative from 0 to 6 gives us the accumulated increase in the crop over the 6-day period.
Economic Mathematics
Economic Mathematics plays a crucial role in this problem.
After determining the total increase in bushels over 6 days through integration, we still need to find out the economic value of this increase.
We use the given fixed market price of \(\$2\) per bushel.
By multiplying the total increase in bushels, which we found to be about 43.7836 bushels, by the price, we get the total increase in value:
\[ 43.7836 \text{ bushels} \times 2 \text{ dollars per bushel} = 87.5672 \text{ dollars} \].
This approach allows us to convert the physical increase of the crop into a measurable economic benefit, reflecting real-world worth.
After determining the total increase in bushels over 6 days through integration, we still need to find out the economic value of this increase.
We use the given fixed market price of \(\$2\) per bushel.
By multiplying the total increase in bushels, which we found to be about 43.7836 bushels, by the price, we get the total increase in value:
\[ 43.7836 \text{ bushels} \times 2 \text{ dollars per bushel} = 87.5672 \text{ dollars} \].
This approach allows us to convert the physical increase of the crop into a measurable economic benefit, reflecting real-world worth.
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