Question

the following model represents the total spent on education funding, in millions of dollars, in a state from 2000 to 2010, where the time t = 0 represents the year 2000: e(t) = 33 ln(2935 − 269t) find the rate at which the total spent on education funding in that state was changing in the year 2007. (round your answer to the nearest integer.) the total spent on education funding was? by? million dollars per year in the year 2007.

Explanation:

Step1: Identify the value of t for 2007

Since \( t = 0 \) represents 2000, for the year 2007, \( t=2007 - 2000=7 \).

Step2: Find the derivative of \( E(t) \)

Given \( E(t)=33\ln(2935 - 269t) \), we use the chain rule. The derivative of \( \ln(u) \) is \( \frac{1}{u}u' \), where \( u = 2935-269t \) and \( u'=- 269 \). So,
\( E'(t)=33\times\frac{-269}{2935 - 269t}=\frac{-33\times269}{2935 - 269t} \)

Step3: Substitute \( t = 7 \) into \( E'(t) \)

First, calculate the denominator: \( 2935-269\times7=2935 - 1883 = 1052 \)
Then, calculate the numerator: \( - 33\times269=-8877 \)
So, \( E'(7)=\frac{-8877}{1052}\approx - 8.44 \)
But we are interested in the rate of change in terms of magnitude (since the question is about the rate at which it was changing, and we can take the absolute value for the rate of change in millions of dollars per year). So, \( |E'(7)|\approx8 \) (rounded to the nearest integer). Wait, wait, let's recalculate:

Wait, \( 33\times269 = 33\times(270 - 1)=33\times270-33=8910 - 33 = 8877 \)
Denominator: \( 2935-269\times7=2935 - 1883 = 1052 \)
\( \frac{8877}{1052}\approx8.44 \), so when rounded to the nearest integer, it is 8? Wait, no, wait, maybe I made a mistake. Wait, the function is \( E(t)=33\ln(2935 - 269t) \), so the derivative is \( E'(t)=33\times\frac{-269}{2935 - 269t} \). The negative sign indicates a decreasing rate, but the question says "the rate at which the total spent on education funding was changing", so we can consider the absolute value for the magnitude. Wait, but let's check the calculation again.

Wait, \( t = 7 \), so \( 2935-269\times7=2935 - 1883 = 1052 \)
\( E'(t)=\frac{33\times(- 269)}{2935 - 269t} \)
So \( E'(7)=\frac{33\times(-269)}{1052}=\frac{-8877}{1052}\approx - 8.44 \). So the rate of change is approximately - 8.44 million dollars per year. Rounding to the nearest integer, we get - 8, but since the question is about the rate at which it was changing (the magnitude or the value), but maybe the negative sign indicates a decrease. But the problem says "the rate at which the total spent on education funding was changing", so we can present the value as approximately - 8 or 8 (if we consider the speed of change). But let's check the calculation again. Wait, maybe I messed up the derivative. Let's re - derive:

The derivative of \( \ln(u) \) with respect to \( t \) is \( \frac{u'}{u} \). Here, \( u = 2935-269t \), so \( u'=-269 \). Then \( E(t)=33\ln(u) \), so \( E'(t)=33\times\frac{u'}{u}=33\times\frac{-269}{2935 - 269t} \). So that part is correct.

Now, \( 33\times269 = 8877 \), \( 2935-269\times7 = 2935 - 1883 = 1052 \). Then \( \frac{8877}{1052}\approx8.44 \), so \( E'(7)\approx - 8.44 \). Rounding to the nearest integer, we get - 8. But maybe the problem expects the absolute value? Wait, the question says "the rate at which the total spent on education funding was changing", so the rate can be negative (indicating a decrease) or positive (increase). Since \( 2935-269t \) is the argument of the log, and as \( t \) increases, \( 2935 - 269t \) decreases, so \( \ln(2935 - 269t) \) decreases, so \( E(t) \) decreases, so the rate of change is negative. But when we round - 8.44 to the nearest integer, we get - 8? Wait, no, - 8.44 is closer to - 8 than to - 9? Wait, - 8.44 is 0.44 units away from - 8 and 0.56 units away from - 9, so it rounds to - 8. But maybe the problem has a typo or I misread the function. Wait, the function is \( E(t)=33\ln(2935 - 269t) \)? Let me check the numbers again. 269*7 = 1883, 2935 - 1883 = 1052. 33*269: 30*269 = 8070, 3*269 = 807, so 8070+807 = 8877. 8877/1052≈8.44. So the rate of change is approximately - 8 million dollars per year (rounded to the nearest integer).

Answer:

\(-8\) (or \(8\) if we consider the magnitude of the rate of change)