A software company is monitoring the rate, in lines of code per minute, at which one of its automated programs writes code. The rate can be modeled by the function where is the number of minutes since the program started running and and are constants. At the moment the program started (), the rate was 450 lines per minute, and after 20 minutes, the rate had decreased to 300 lines per minute. According to the model, which of the following is closest to the rate, in lines per minute, that the program will approach as becomes very large?

Solution available with step-by-step explanation

SAT Nonlinear Functions Question 129 | Free SAT Prep | Aniko

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Question 129·Hard·Nonlinear Functions

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A software company is monitoring the rate, in lines of code per minute, at which one of its automated programs writes code. The rate $R$ can be modeled by the function

R(t) = a e^{-0.03t} + b,

where $t$ is the number of minutes since the program started running and $a$ and $b$ are constants. At the moment the program started ( $t = 0$ ), the rate was 450 lines per minute, and after 20 minutes, the rate had decreased to 300 lines per minute.

According to the model, which of the following is closest to the rate, in lines per minute, that the program will approach as $t$ becomes very large?

For exponential models of the form $A e^{kt} + C$ , immediately recognize that as $t$ grows, the exponential part goes to 0 (when $k < 0$ ), so the function approaches the constant term $C$ . On SAT questions like this, translate any given data points into equations, solve the small system for the constants (using quick subtraction to eliminate one variable and a calculator for exponentials if allowed), and then match the resulting long-term value to the closest answer choice.

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Focus on what happens as time gets very large

Look at the term $e^{-0.03t}$ in the function. As $t$ becomes very large, does this term grow or shrink, and what does that do to $R(t)$ ?

Translate the given information into equations

Use $R(0) = 450$ and $R(20) = 300$ to write two equations involving $a$ and $b$ . Plug $t = 0$ and $t = 20$ into the function.

Solve the system to find the constant term

Once you have the two equations in $a$ and $b$ , subtract one from the other to eliminate $b$ and solve for $a$ . Then substitute back to find $b$ , and compare that value with the answer choices.

Desmos Guide

Use Desmos to solve for the constants

In one line, enter a = 150/(1 - e^(-0.6)), and in the next line enter b = 450 - a. Desmos will display numerical values for $a$ and $b$ that satisfy the two given conditions.

Define the rate function with those constants

Add a new line and type R(t) = a*e^(-0.03*t) + b. Then create a table for $R$ (click the gear icon and choose “Table”) and include values like $t = 0$ , $t = 20$ , and some much larger $t$ values (for example, $t = 200$ or $t = 500$ ) to see how the rate changes.

Read off the long-term rate

Look at the $R(t)$ values in the table for very large $t$ . They will get very close to a constant number; this number is the value of $b$ , the rate the program approaches as time increases. Choose the answer option that is closest to that number.

Step-by-step Explanation

Identify the long-term behavior of the model

The rate is modeled by

R(t) = a e^{-0.03t} + b

As $t$ gets very large, the factor $e^{-0.03t}$ becomes extremely small and approaches 0, so the term $a e^{-0.03t}$ fades away.

That means $R(t)$ approaches the constant value $b$ for large $t$ . So the question is really asking us to find $b$ .

Use the initial rate to get the first equation

At $t = 0$ , the rate was 450 lines per minute. Substitute $t = 0$ into the model:

R(0) = a e^{0} + b = a + b

We are told $R(0) = 450$ , so

a + b = 450

This is our first equation relating $a$ and $b$ .

Use the rate at 20 minutes to get the second equation

After 20 minutes, the rate was 300 lines per minute. Substitute $t = 20$ into the model:

R(20) = a e^{-0.03 \cdot 20} + b = a e^{-0.6} + b

We are told $R(20) = 300$ , so

a e^{-0.6} + b = 300

Now we have a system of two equations in $a$ and $b$ :

\begin{aligned} a + b &= 450 \\ a e^{-0.6} + b &= 300 \end{aligned}

Solve for $b$ and match it to the closest answer choice

Subtract the second equation from the first to eliminate $b$ :

(a + b) - (a e^{-0.6} + b) = 450 - 300

This simplifies to

a - a e^{-0.6} = 150

Factor out $a$ :

a(1 - e^{-0.6}) = 150 \quad \Rightarrow \quad a = \frac{150}{1 - e^{-0.6}}

Now use $a + b = 450$ to solve for $b$ :

b = 450 - a = 450 - \frac{150}{1 - e^{-0.6}}

Using a calculator, $e^{-0.6} \approx 0.55$ , so

a \approx \frac{150}{1 - 0.55} = \frac{150}{0.45} \approx 333, \quad b \approx 450 - 333 = 117

So the model says the rate approaches about 117 lines per minute as $t$ becomes very large. Among the choices given, the closest value is 120, so that is the correct answer.

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Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation