Q1: A population of bacteria doubles every 3 hours. If the initial population is 500, which function models the population P(t) after t hours? (a) \(P(t) = 500(2)^{3t}\) (b) \(P(t) = 500(2)^{t/3}\) (c) \(P(t) = 500(3)^{2t}\) (d) \(P(t) = 500(2)^t\)
Solution:
Ans: (b) Explanation: Since the population doubles every 3 hours, the exponent must be t/3. When t = 3, we get \(P(3) = 500(2)^{3/3} = 500(2)^1 = 1000\), which correctly shows the population doubling. Option (a) would cause the population to grow too rapidly, option (c) uses the wrong base, and option (d) doesn't account for the 3-hour doubling period.
Q2: The value of a car depreciates by 15% each year. If the car costs $24,000 initially, which exponential function represents its value V(t) after t years? (a) \(V(t) = 24000(0.15)^t\) (b) \(V(t) = 24000(0.85)^t\) (c) \(V(t) = 24000(1.15)^t\) (d) \(V(t) = 24000 - 0.15t\)
Solution:
Ans: (b) Explanation:Depreciation by 15% means the car retains 85% of its value each year, so the decay factor is 0.85. Option (a) uses only the depreciation rate, option (c) would represent 15% growth rather than decay, and option (d) represents linear rather than exponential decay.
Q3: A scientist models the temperature T (in °F) of a cooling liquid as a function of time m (in minutes) with \(T(m) = 68 + 132e^{-0.04m}\). What does the value 68 represent in this context? (a) The initial temperature of the liquid (b) The ambient room temperature (c) The rate of cooling (d) The temperature after 1 minute
Solution:
Ans: (b) Explanation: In an exponential decay model for cooling, the constant term represents the ambient temperature that the liquid approaches as time goes to infinity. As m increases, \(e^{-0.04m}\) approaches 0, so T(m) approaches 68°F. The initial temperature is found by evaluating \(T(0) = 68 + 132 = 200\)°F.
Q4: The height h(t) in feet of a projectile after t seconds is modeled by \(h(t) = -16t^2 + 64t + 5\). At what time does the projectile reach its maximum height? (a) 1 second (b) 2 seconds (c) 3 seconds (d) 4 seconds
Solution:
Ans: (b) Explanation: For a quadratic function in the form \(h(t) = at^2 + bt + c\), the maximum occurs at \(t = -\frac{b}{2a}\). Here, a = -16 and b = 64, so \(t = -\frac{64}{2(-16)} = -\frac{64}{-32} = 2\) seconds. The negative value of a confirms the parabola opens downward, giving a maximum.
Q5: A company's profit P(x) in thousands of dollars from producing x hundred units is given by \(P(x) = -2x^2 + 12x - 10\). How many hundred units should be produced to maximize profit? (a) 2 hundred units (b) 3 hundred units (c) 4 hundred units (d) 6 hundred units
Solution:
Ans: (b) Explanation: Using the vertex formula \(x = -\frac{b}{2a}\) where a = -2 and b = 12: \(x = -\frac{12}{2(-2)} = -\frac{12}{-4} = 3\) hundred units. This represents the x-coordinate of the vertex, which gives the maximum profit since the parabola opens downward.
Q6: The population of a town is modeled by \(P(t) = \frac{25000}{1 + 49e^{-0.3t}}\), where t is years after 2000. What type of model is this? (a) Linear growth (b) Exponential growth (c) Logistic growth (d) Quadratic growth
Solution:
Ans: (c) Explanation: This is a logistic growth model, characterized by the form \(\frac{L}{1 + ae^{-kt}}\). Logistic models describe populations that initially grow exponentially but eventually level off at a carrying capacity (here, 25,000). This is more realistic than pure exponential growth for most populations.
Q7: A ball is dropped from a height of 100 feet. Each bounce reaches 60% of the previous height. Which equation models the height h(n) after n bounces? (a) \(h(n) = 100(0.6)^n\) (b) \(h(n) = 100(0.4)^n\) (c) \(h(n) = 100 - 0.6n\) (d) \(h(n) = 60(0.6)^n\)
Solution:
Ans: (a) Explanation: The ball retains 60% of its height after each bounce, so the decay factor is 0.6. Starting at 100 feet, after n bounces the height is \(h(n) = 100(0.6)^n\). Option (b) uses the wrong decay factor (0.4 = 40%, not 60%), option (c) is linear rather than exponential, and option (d) has the wrong initial height.
Q8: A pharmacologist models the concentration C(t) of a drug in the bloodstream (in mg/L) as \(C(t) = 20te^{-0.5t}\), where t is in hours. What happens to the concentration as time increases without bound? (a) It increases without bound (b) It approaches 20 mg/L (c) It approaches 0 mg/L (d) It remains constant at 20 mg/L
Solution:
Ans: (c) Explanation: As t increases, the exponential decay term \(e^{-0.5t}\) approaches 0 faster than the linear term t increases. Therefore, \(\lim_{t \to \infty} 20te^{-0.5t} = 0\). The drug concentration eventually decreases to 0 mg/L as the body metabolizes it.
Section B: Fill in the Blanks
Q9:In an exponential growth model \(y = ab^t\) where b > 1, the value b is called the __________ .
Solution:
Ans: growth factor Explanation: The growth factor b determines how rapidly the quantity increases over each time interval. When b > 1, the model represents exponential growth; when 0 < b="">< 1,="" it="" represents="" exponential="">
Q10:For the quadratic model \(y = ax^2 + bx + c\), the x-coordinate of the vertex is given by the formula __________ .
Solution:
Ans: \(x = -\frac{b}{2a}\) Explanation: This vertex formula provides the x-coordinate where the parabola reaches its maximum or minimum value. The y-coordinate is found by substituting this x-value back into the original equation.
Q11:In a logistic growth model \(P(t) = \frac{L}{1 + ae^{-kt}}\), the value L represents the __________ .
Solution:
Ans: carrying capacity Explanation: The carrying capacity is the maximum population that the environment can sustain. As time approaches infinity, the population P(t) approaches L, representing the upper limit of growth.
Q12:A model that can be written in the form \(y = mx + b\) where m and b are constants represents __________ growth or decay.
Solution:
Ans: linear Explanation:Linear models show constant rate of change, represented by the slope m. This is fundamentally different from exponential models where the rate of change itself changes over time.
Q13:If an investment of $5000 grows to $8000 in 4 years with continuous compounding at rate r, the relationship is modeled by \(8000 = 5000e^{4r}\). Solving for r requires using the __________ function.
Solution:
Ans: natural logarithm (or ln) Explanation: To solve for r, we divide both sides by 5000 to get \(1.6 = e^{4r}\), then take the natural logarithm of both sides: \(\ln(1.6) = 4r\), giving \(r = \frac{\ln(1.6)}{4}\).
Q14:The general form of a quadratic function is \(f(x) = ax^2 + bx + c\). When a < 0,="" the="" parabola="" opens="" __________="" and="" has="" a="" maximum="">
Solution:
Ans: downward Explanation: When the leading coefficient a is negative, the parabola opens downward, creating a maximum point at the vertex. When a > 0, the parabola opens upward with a minimum point.
Section C: Word Problems
Q15:The number of subscribers to a streaming service is modeled by \(S(t) = 12000(1.15)^t\), where t is the number of years since 2020. Find the number of subscribers in 2020 and predict the number of subscribers in 2025.
Solution:
Ans: For 2020, t = 0: \(S(0) = 12000(1.15)^0 = 12000(1) = 12000\) subscribers
For 2025, t = 5: \(S(5) = 12000(1.15)^5\) \(S(5) = 12000(2.0113571875)\) \(S(5) ≈ 24136\) subscribers
Final Answer: In 2020 there were 12,000 subscribers; in 2025 there will be approximately 24,136 subscribers.
Q16:A coffee shop's daily profit P(x) in dollars from selling x hundred cups of coffee is modeled by \(P(x) = -5x^2 + 60x - 100\). Find the number of cups that should be sold to maximize profit, and determine the maximum profit.
Solution:
Ans: Find the vertex using \(x = -\frac{b}{2a}\): \(x = -\frac{60}{2(-5)} = -\frac{60}{-10} = 6\) hundred cups = 600 cups
Final Answer: The coffee shop should sell 600 cups to maximize profit at $80.
Q17:The temperature of a heated object cooling in a room is modeled by \(T(t) = 72 + 128e^{-0.05t}\), where T is temperature in °F and t is time in minutes. Find the initial temperature of the object and the temperature after 20 minutes.
Solution:
Ans: Initial temperature at t = 0: \(T(0) = 72 + 128e^{-0.05(0)}\) \(T(0) = 72 + 128e^0\) \(T(0) = 72 + 128(1) = 200°\)F
Temperature at t = 20: \(T(20) = 72 + 128e^{-0.05(20)}\) \(T(20) = 72 + 128e^{-1}\) \(T(20) = 72 + 128(0.3679)\) \(T(20) = 72 + 47.09\) \(T(20) ≈ 119.1°\)F
Final Answer: Initial temperature is 200°F; after 20 minutes the temperature is approximately 119.1°F.
Q18:A construction company determines that the cost C(x) in thousands of dollars to build x houses is given by \(C(x) = 2x^2 + 10x + 50\). If they sell each house for $80,000, write a profit function P(x) and determine how many houses they must build to break even.
Final Answer: The company breaks even when building approximately 1 house or 34 houses. Profit occurs between these values.
Q19:A wildlife preserve models its deer population with \(P(t) = \frac{800}{1 + 15e^{-0.4t}}\), where t is years after introduction. Find the initial deer population and the population after 5 years. What is the carrying capacity?
Solution:
Ans: Initial population at t = 0: \(P(0) = \frac{800}{1 + 15e^{-0.4(0)}}\) \(P(0) = \frac{800}{1 + 15e^0}\) \(P(0) = \frac{800}{1 + 15} = \frac{800}{16} = 50\) deer
Final Answer: Initial population is 50 deer; after 5 years there are approximately 264 deer; carrying capacity is 800 deer.
Q20:A medication's concentration in the bloodstream is modeled by \(C(t) = 15te^{-0.3t}\), where C is in mg/L and t is in hours. Find the concentration after 2 hours and determine when the concentration reaches its maximum value.
Solution:
Ans: Concentration at t = 2: \(C(2) = 15(2)e^{-0.3(2)}\) \(C(2) = 30e^{-0.6}\) \(C(2) = 30(0.5488)\) \(C(2) ≈ 16.46\) mg/L
To find maximum, take derivative and set equal to zero: \(C'(t) = 15e^{-0.3t} + 15t(-0.3)e^{-0.3t}\) \(C'(t) = 15e^{-0.3t}(1 - 0.3t)\)
Set C'(t) = 0: Since \(e^{-0.3t}\) is never zero: \(1 - 0.3t = 0\) \(0.3t = 1\) \(t = \frac{1}{0.3} = \frac{10}{3} ≈ 3.33\) hours
Final Answer: After 2 hours the concentration is approximately 16.46 mg/L; maximum concentration occurs at approximately 3.33 hours.
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