Free Math Worksheets

- Write the net rate as inflow minus outflow. - The initial amount is zero because the tank begins with pure water. - Find the equilibrium amount by setting the derivative equal to zero. - Examine the exponential term as time becomes large.

1. The rate of change equals the constant inflow minus the proportional outflow: \(m'(t)=15-0.025m(t)\), with \(m(0)=0\). 2. At equilibrium, \(15-0.025S=0\), so \(S=600\). The solution satisfying \(m(0)=0\) is \(m(t)=600(1-e^{-0.025t})\). 3. After \(60\) minutes, \(m(60)=600(1-e^{-1.5})\approx 466.12\,\text{g}\). 4. As \(t\to\infty\), \(e^{-0.025t}\to 0\), so \(m(t)\to 600\,\text{g}\).

a) \(m'(t)=15-0.025m(t)\), with \(m(0)=0\) b) \(m(t)=600(1-e^{-0.025t})\) c) \(m(60)\approx 466.12\,\text{g}\) d) \(600\,\text{g}\)

- What happens to units when a quantity is differentiated with respect to time? - What would it mean if twice as much bacterial mass produced twice as much new mass? - What happens to the slope of a curve at a logistic inflection point? - Can a culture in a closed dish grow exponentially forever?

1. Since mass is measured in grams and time in hours, \(m'(t)\) has units \(\frac{\text{g}}{\text{h}}\) and represents the instantaneous growth rate. The second derivative \(m''(t)\) has units \(\frac{\text{g}}{\text{h}^2}\) and represents how the growth rate changes over time. 2. The equation \(m'(t)=km(t)\) states that the growth rate is proportional to the current mass. Biologically, it assumes that each bacterium contributes to reproduction at a constant average rate, producing unrestricted exponential growth. 3. A sign change in \(m''(t)\) identifies an inflection point. At this time, \(m'(t)\) reaches its maximum, so the culture is growing as rapidly as possible before its growth rate begins to decrease. 4. Exponential growth assumes unlimited resources. In a petri dish, limited space and nutrients and the accumulation of waste eventually slow and stop growth.

a) \(m'(t)\) has units \(\frac{\text{g}}{\text{h}}\) and is the growth rate. \(m''(t)\) has units \(\frac{\text{g}}{\text{h}^2}\) and is the rate of change of the growth rate. b) The growth rate is proportional to the current bacterial mass. c) It is the time of maximum growth rate, where growth changes from accelerating to slowing. d) Real cultures have limited space and nutrients, so unrestricted exponential growth cannot continue indefinitely.

- Write the net change as inflow minus removal. - Find the equilibrium amount from the differential equation. - The long-term value is the equilibrium amount. - For part d), express the equilibrium in terms of the unknown inflow.

1. The net rate is inflow minus proportional removal: \(P'(t)=20-0.08P(t)\), with \(P(0)=50\). 2. The equilibrium amount is \(S=\frac{20}{0.08}=250\). Therefore, \(P(t)=250+(50-250)e^{-0.08t}=250-200e^{-0.08t}\). 3. \(P(10)=250-200e^{-0.8}\approx 160.13\,\text{kg}\). 4. The long-term amount is \(250\,\text{kg}\), which is below \(300\,\text{kg}\), so the requirement is satisfied. 5. If the inflow is \(Z\,\text{kg/day}\), the equilibrium amount is \(\frac{Z}{0.08}\). Requiring \(\frac{Z}{0.08}\le 200\) gives \(Z\le 16\).

a) \(P'(t)=20-0.08P(t)\), with \(P(0)=50\) b) \(P(10)\approx 160.13\,\text{kg}\) c) Yes; the long-term amount is \(250\,\text{kg}\). d) At most \(16\,\text{kg/day}\)