Problem 70 - Calculus Early Transcendentals

Question

Consider the logistic equation $$P^{\prime}(t)=0.1 P\left(1-\frac{P}{300}\right), \text { for } t \geq 0$$, with \(P(0)>0 .\) Show that the solution curve is concave down for \(150300\).

Step-by-Step Solution

Answer

Answer: The solution curve is concave up for \(0300\), and concave down for \(150

Step 1: Identify the First Derivative

The given logistic equation is already in the form of the first derivative: \(P^{\prime}(t)=0.1 P\left(1-\frac{P}{300}\right)\).

Step 2: Find the Second Derivative

To find the concavity of the solution curve, we need to compute the second derivative \(P''(t)\). Applying the product rule and chain rule for differentiation, we get: $$P''(t) = \frac{d}{dt} \left[ 0.1 P\left(1-\frac{P}{300}\right) \right] = 0.1 \left[\left(1-\frac{P}{300}\right)\frac{dP}{dt} - \frac{1}{300}\cdot P\cdot\frac{dP}{dt}\right] = 0.1 \left[\left(1-\frac{P}{300}\right)P^{\prime}(t) - \frac{1}{300}\cdot P\cdot P^{\prime}(t)\right]$$ Now, substitute the original given first derivative equation back into the second derivative equation: $$P''(t) = 0.1 \left[\left(1-\frac{P}{300}\right)(0.1 P\left(1-\frac{P}{300}\right)) - \frac{1}{300}\cdot P\cdot(0.1 P\left(1-\frac{P}{300}\right))\right]$$

Step 3: Simplify the Second Derivative Expression

Simplify the second derivative expression to find: $$P''(t) = 0.01 P\left(1-\frac{P}{300}\right)^2 - \frac{0.01}{300} P^2\left(1-\frac{P}{300}\right) $$

Step 4: Determine the Concavity in the given Intervals

Now, we analyze the sign of \(P''(t)\) in the intervals \(0300\). 1. In the interval \(0 0\) in this interval, and the solution curve is concave up. 2. In the interval \(150300\), the term \(\left(1-\frac{P}{300}\right)^2\) is still positive, but the term \(1-\frac{P}{300}\) is now more negative, which leads to \(P''(t) > 0\). So, in this interval, the solution curve is again concave up. In conclusion, the solution curve is concave down for \(150300\).

Key Concepts

Concavity of a Curve

When studying the shape of a graph, we often want to know where the graph curves up or down. This is known as the concavity of a curve. A curve is said to be concave up if it bends upwards like a cup and concave down if it bends downwards like a frown.

To determine the concavity of a function's graph, we look at the function's second derivative. If the second derivative is positive over an interval, the graph is concave up on that interval. Conversely, if the second derivative is negative, the graph is concave down. It's crucial for understanding the behavior of graphs, as it gives us a sense of how the function behaves between its high and low points.

In the exercise, you are examining the solution curve of a logistic differential equation, which models growth processes. By finding the second derivative and assessing its sign, you're able to deduce the concavity of the solution curve in different intervals.

Second Derivative Test

The second derivative test is a useful tool in calculus for determining whether a given point on a function's graph is a maximum, a minimum, or a point of inflection. This test involves taking the second derivative of the function and substiting values of the first derivative's critical points into this second derivative.

If the second derivative at a critical point is positive, the function has a relative minimum at that point. If it's negative, the function has a relative maximum. If the second derivative is zero, the test is inconclusive; the point could be an inflection point, or we may need more information to decide.

In the context of our logistic equation, the second derivative test isn't directly used for finding maximums or minimums, but it helps us understand where the growth rate of the population increases or decreases, based on the concavity of the curve.

Product Rule

When taking the derivative of a function that is the product of two or more functions, the product rule becomes essential. The rule states that if you have two functions, say, \(u(t)\) and \(v(t)\), then the derivative of their product \(u(t)v(t)\) is given by:

• \(u'(t)v(t) + u(t)v'(t)\)

where \(u'(t)\) and \(v'(t)\) are the derivatives of \(u(t)\) and \(v(t)\), respectively.

The product rule is particularly useful in finding the rate of change of quantities that can be expressed as the product of functions. For example, when finding the second derivative of the population function \(P(t)\) in our logistic equation, we apply the product rule since \(P(t)\) itself is part of the rate of change expression.

Chain Rule

The chain rule is a fundamental derivative rule used for finding the derivative of a composition of functions. If we have a function \(u\) which is a function of \(v\), and \(v\) is a function of \(t\), that is, \(u(v(t))\), then the derivative of \(u\) with respect to \(t\) is given by the product of the derivative of \(u\) with respect to \(v\) and the derivative of \(v\) with respect to \(t\). The formula appears as:

• \(\frac{du}{dt} = \frac{du}{dv} \times \frac{dv}{dt}\)

It allows us to differentiate complex functions by breaking them down into simpler parts. In the exercise's logistic differential equation, the chain rule enables us to differentiate the compound function inside the derivative and is critical for finding the second derivative accurately.

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Key Concepts

Concavity of a Curve

When studying the shape of a graph, we often want to know where the graph curves up or down. This is known as the concavity of a curve. A curve is said to be concave up if it bends upwards like a cup and concave down if it bends downwards like a frown.

To determine the concavity of a function's graph, we look at the function's second derivative. If the second derivative is positive over an interval, the graph is concave up on that interval. Conversely, if the second derivative is negative, the graph is concave down. It's crucial for understanding the behavior of graphs, as it gives us a sense of how the function behaves between its high and low points.

In the exercise, you are examining the solution curve of a logistic differential equation, which models growth processes. By finding the second derivative and assessing its sign, you're able to deduce the concavity of the solution curve in different intervals.

Second Derivative Test

The second derivative test is a useful tool in calculus for determining whether a given point on a function's graph is a maximum, a minimum, or a point of inflection. This test involves taking the second derivative of the function and substiting values of the first derivative's critical points into this second derivative.

If the second derivative at a critical point is positive, the function has a relative minimum at that point. If it's negative, the function has a relative maximum. If the second derivative is zero, the test is inconclusive; the point could be an inflection point, or we may need more information to decide.

In the context of our logistic equation, the second derivative test isn't directly used for finding maximums or minimums, but it helps us understand where the growth rate of the population increases or decreases, based on the concavity of the curve.

Product Rule

When taking the derivative of a function that is the product of two or more functions, the product rule becomes essential. The rule states that if you have two functions, say, \(u(t)\) and \(v(t)\), then the derivative of their product \(u(t)v(t)\) is given by:

• \(u'(t)v(t) + u(t)v'(t)\)

where \(u'(t)\) and \(v'(t)\) are the derivatives of \(u(t)\) and \(v(t)\), respectively.

The product rule is particularly useful in finding the rate of change of quantities that can be expressed as the product of functions. For example, when finding the second derivative of the population function \(P(t)\) in our logistic equation, we apply the product rule since \(P(t)\) itself is part of the rate of change expression.

Chain Rule

The chain rule is a fundamental derivative rule used for finding the derivative of a composition of functions. If we have a function \(u\) which is a function of \(v\), and \(v\) is a function of \(t\), that is, \(u(v(t))\), then the derivative of \(u\) with respect to \(t\) is given by the product of the derivative of \(u\) with respect to \(v\) and the derivative of \(v\) with respect to \(t\). The formula appears as:

• \(\frac{du}{dt} = \frac{du}{dv} \times \frac{dv}{dt}\)

It allows us to differentiate complex functions by breaking them down into simpler parts. In the exercise's logistic differential equation, the chain rule enables us to differentiate the compound function inside the derivative and is critical for finding the second derivative accurately.