Chapter 11: Problem 5
A quantity \(A\) satisfies the differential equation $$\frac{d A}{d t}=k A(1-0.0002 A), \quad \text { with } k>0$$ Sketch a graph of \(d A / d t\) as a function of \(A\)
Short Answer
Expert verified
The graph of \(\frac{dA}{dt}\) is a downward-opening parabola in terms of \(A\).
Step by step solution
01
Understand the Differential Equation
The differential equation \(\frac{dA}{dt} = kA(1-0.0002 A)\) is a first-order, separable equation. Here \(k\) is a constant greater than zero, and \(A\) is a function of time \(t\). The equation suggests that \(\frac{dA}{dt}\) depends on both \(A\) and itself.
02
Analyze the Expression for \(\frac{dA}{dt}\)
The expression \(kA(1-0.0002A)\) can be analyzed as a function \(f(A) = kA(1-0.0002A)\). This is a quadratic function in \(A\), more specifically, a downward opening parabola since the term \(-0.0002A^2\) is negative.
03
Find Critical Points of \(f(A)\)
To find the critical points, solve \(f(A) = 0\). Since \(f(A) = kA(1-0.0002A)\), it equals zero when either \(A = 0\) or \(1 - 0.0002A = 0\), which gives \(A = 5000\). Thus, \(A = 0\) and \(A = 5000\) are critical points where \(\frac{dA}{dt} = 0\).
04
Determine Behavior Between Critical Points
For values of \(A\) between the critical points, analyze the sign of \(f(A) = kA(1-0.0002A)\):- For \(A \in (0, 5000)\), \(1-0.0002A > 0\), thus \(\frac{dA}{dt} > 0\).- For \(A > 5000\), \(1-0.0002A < 0\), thus \(\frac{dA}{dt} < 0\).This indicates that \(\frac{dA}{dt}\) increases from zero at \(A=0\) to a maximum point and then decreases back to zero at \(A=5000\).
05
Sketch the Graph
Plot \(\frac{dA}{dt}\) against \(A\):- Start at \( (0, 0) \).- Increase to a peak value before \(A=5000\).- Return to zero at \(A=5000\).The graph will look like an upside-down U (parabolic arc), with the vertex below the peak at \(A=5000\).
06
Conclusion
The graph of \(\frac{dA}{dt}\) as a function of \(A\) is a downward-opening parabola with intercepts at \(A=0\) and \(A=5000\), reaching a maximum value between. The behavior reflects a logistic-like growth dynamic.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Separable Equations
In the realm of differential equations, a separable equation is one where we can isolate the variables on each side of the equation. This allows us to integrate both sides separately. Our given differential equation, \( \frac{dA}{dt} = kA(1-0.0002A) \), is a classic example of a separable equation. The form indicates that the rate of change, \( \frac{dA}{dt} \), depends on both \( A \) and another factor involving \( A \).To solve such equations, we manipulate them so that all terms involving \( A \) are on one side and terms involving \( t \) are on the other. This separates the equation into two integrable sides:
- \( \frac{1}{A(1-0.0002A)} dA = k dt \)
Quadratic Function
The right-hand side of the differential equation, \( f(A) = kA(1-0.0002A) \), is a quadratic function in terms of \( A \). Quadratic functions generally take the form \( ax^2 + bx + c \). In our case, it translates to:
- \( f(A) = -0.0002kA^2 + kA \).
Critical Points
Critical points are values of \( A \) where the derivative \( \frac{dA}{dt} \) is zero. To discover these points, we set \( f(A) \) to zero:
- \( kA(1-0.0002A) = 0 \)
Graph Sketching
Graphically representing \( \frac{dA}{dt} \) as a function of \( A \) offers intuitive insights into the differential equation's solution. In this case, the sketch begins at the origin, where \( A=0 \) and \( \frac{dA}{dt} = 0 \). From there, as \( A \) increases, \( \frac{dA}{dt} \) rises to a maximum before falling back to zero at \( A=5000 \).
- Start at \( A=0 \) with \( \frac{dA}{dt}=0 \).
- Increase up to a certain point \( (A, \frac{dA}{dt}) \).
- Decreases back down to zero at \( A=5000 \).