4. (20 points: $5+5$) A biologist is studying the growth of a certain species of bacteria. The population of the bacteria, $P$, varies directly with the square of the time, $t$, in hours since the start of the observation.
a. Write a power function representing the verbal statement (include $k$ as the constant of variation).
b. If the population of the bacteria is 2000 after 4 hours, find the constant of variation $k$.
c. Write the power function that models the population of the bacteria for any time $t$.
d. Using the power function, determine the population of the bacteria after 6 hours.
e. If the biologist wants to know when the population will reach 4500, how many hours will it take?
5. $(20$ points: $5+5+5+5)$ Solve the given equations algebraically. Show your work for full credit.
a. $\quad 10 x^3=5120$
b. $5 g^{-2}=25$
c. $\frac{3}{x^4}=12$
d. $10 x^3-5120=0$
Answer (1)
Establishes the power function as P = k t 2 .
Determines the constant of variation k to be 125.
Models the bacteria population with the function P = 125 t 2 .
Calculates the population after 6 hours as 4500 and the time to reach a population of 4500 as 6 hours.
Solves the algebraic equations to find x = 8 , g = 5 5 , x = 2 2 , and x = 8 .
Explanation
Problem Overview Let's break down this problem step by step. We're given a scenario about bacteria growth and some algebraic equations to solve. We'll tackle each part systematically.
Writing the Power Function a. The problem states that the population, P , varies directly with the square of the time, t . This translates to the equation: P = k t 2 where k is the constant of variation.
Finding the Constant of Variation b. We're given that P = 2000 when t = 4 . Let's substitute these values into our equation to find k :
2000 = k ( 4 2 ) 2000 = 16 k k = 16 2000 = 125
Writing the Specific Power Function c. Now that we know k = 125 , we can write the specific power function for this bacteria population: P = 125 t 2
Calculating Population at t=6 d. To find the population after 6 hours, we substitute t = 6 into our power function: P = 125 ( 6 2 ) P = 125 ( 36 ) = 4500
Calculating Time to Reach Population of 4500 e. To find when the population will reach 4500, we set P = 4500 and solve for t :
4500 = 125 t 2 t 2 = 125 4500 = 36 t = 36 = 6 So, it will take 6 hours for the population to reach 4500.
Solving Equation 5a Now, let's solve the algebraic equations: a. 10 x 3 = 5120 x 3 = 10 5120 = 512 x = 3 512 = 8
Solving Equation 5b b. 5 g − 2 = 25 g 2 5 = 25 g 2 = 25 5 = 5 1 g = 5 1 = 5 1 = 5 5 ≈ 0.447
Solving Equation 5c c. x 4 3 = 12 x 4 = 12 3 = 4 1 x = 4 4 1 = 2 1 = 2 1 = 2 2 ≈ 0.707
Solving Equation 5d d. 10 x 3 − 5120 = 0 10 x 3 = 5120 x 3 = 10 5120 = 512 x = 3 512 = 8
Examples
Understanding direct variation and power functions is crucial in various fields. For instance, in physics, the kinetic energy (KE) of an object is directly proportional to the square of its velocity (v), expressed as KE = (1/2)mv^2, where m is the mass. Similarly, in finance, the return on an investment can sometimes be modeled as a power function of the risk taken. These models help predict outcomes and manage resources effectively by understanding how one variable changes in relation to another's power.
