Solved on Sep 12, 2023

Calculate the average fuel consumption rate from 6 AM to 3 PM given the formula $f(t) = 0.9t^2 - 0.2t^{0.4} + 10$ , where $t$ is the time in AM and $f(t)$ is the fuel consumption in barrels.

STEP 1

Assumptions1. The factory operates from6 AM to6 PM. . The fuel consumption is given by the function $f(t)=0.9 t^{}-0. t^{0.4}+10$ , where $t$ is the time in AM and $f(t)$ is the number of barrels of fuel oil.
3. We are asked to find the average rate of consumption from6 AM to3 PM.

STEP 2

The average rate of consumption over a time interval is given by the total consumption over that interval divided by the length of the interval. In this case, the total consumption is given by the integral of the consumption function over the interval, and the length of the interval is the difference in time.
$Average\, rate = \frac{1}{b - a} \int_{a}^{b} f(t) dt$ where $a$ and $b$ are the start and end times respectively.

STEP 3

Now, plug in the given values for the start time (6 AM) and end time (3 PM). Note that the time is measured in hours after midnight, so6 AM is6 hours after midnight and3 PM is15 hours after midnight.
$Average\, rate = \frac{1}{15 -6} \int_{6}^{15} f(t) dt$

STEP 4

Substitute the given function $f(t)$ into the integral.
$Average\, rate = \frac{1}{15 -6} \int_{6}^{15} (0.9 t^{2}-0.2 t^{0.4}+10) dt$

STEP 5

Calculate the integral.
$Average\, rate = \frac{1}{15 -} \left[0.3 t^{3} -0.5 t^{0.} +10t \right]_{}^{15}$

STEP 6

Evaluate the integral at the limits of integration.
$Average\, rate = \frac{1}{15 -6} \left[ (0.3 \cdot15^{3} -0.5 \cdot15^{0.6} +10 \cdot15) - (0.3 \cdot6^{3} -0.5 \cdot6^{0.6} +10 \cdot6) \right]$

STEP 7

Calculate the value of the expression.
$Average\, rate = \frac{1}{9} \left[ (0.3 \cdot3375 -0.5 \cdot7.37 +150) - (0.3 \cdot216 -0.5 \cdot1.47 +60) \right]$

STEP 8

implify the expression to find the average rate of consumption.
$Average\, rate = \frac{1}{} \left[ (1012.5 -3.69 +150) - (64.8 -0.74 +60) \right]$

STEP 9

Calculate the average rate of consumption.
$Average\, rate = \frac{}{9} \left[1159 -124.54 \right] = \frac{}{9} \cdot1034.46 \approx114.94$ The average rate of consumption from6 AM to3 PM is approximately114.94 barrels of fuel oil per hour.

Was this helpful?

Calculate the average fuel consumption rate from 6 AM to 3 PM given the formula $f(t) = 0.9t^2 - 0.2t^{0.4} + 10$ , where $t$ is the time in AM and $f(t)$ is the fuel consumption in barrels.

STEP 1

Assumptions1. The factory operates from6 AM to6 PM. . The fuel consumption is given by the function $f(t)=0.9 t^{}-0. t^{0.4}+10$ , where $t$ is the time in AM and $f(t)$ is the number of barrels of fuel oil.
3. We are asked to find the average rate of consumption from6 AM to3 PM.

STEP 2

STEP 3

Now, plug in the given values for the start time (6 AM) and end time (3 PM). Note that the time is measured in hours after midnight, so6 AM is6 hours after midnight and3 PM is15 hours after midnight.
$Average\, rate = \frac{1}{15 -6} \int_{6}^{15} f(t) dt$

STEP 4

Substitute the given function $f(t)$ into the integral.
$Average\, rate = \frac{1}{15 -6} \int_{6}^{15} (0.9 t^{2}-0.2 t^{0.4}+10) dt$

STEP 5

Calculate the integral.
$Average\, rate = \frac{1}{15 -} \left[0.3 t^{3} -0.5 t^{0.} +10t \right]_{}^{15}$

STEP 6

Evaluate the integral at the limits of integration.
$Average\, rate = \frac{1}{15 -6} \left[ (0.3 \cdot15^{3} -0.5 \cdot15^{0.6} +10 \cdot15) - (0.3 \cdot6^{3} -0.5 \cdot6^{0.6} +10 \cdot6) \right]$

STEP 7

Calculate the value of the expression.
$Average\, rate = \frac{1}{9} \left[ (0.3 \cdot3375 -0.5 \cdot7.37 +150) - (0.3 \cdot216 -0.5 \cdot1.47 +60) \right]$

STEP 8

implify the expression to find the average rate of consumption.
$Average\, rate = \frac{1}{} \left[ (1012.5 -3.69 +150) - (64.8 -0.74 +60) \right]$

STEP 9

Calculate the average rate of consumption.
$Average\, rate = \frac{}{9} \left[1159 -124.54 \right] = \frac{}{9} \cdot1034.46 \approx114.94$ The average rate of consumption from6 AM to3 PM is approximately114.94 barrels of fuel oil per hour.

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Calculate the average fuel consumption rate from 6 AM to 3 PM given the formula f(t)=0.9t2−0.2t0.4+10f(t) = 0.9t^2 - 0.2t^{0.4} + 10f(t)=0.9t2−0.2t0.4+10, where ttt is the time in AM and f(t)f(t)f(t) is the fuel consumption in barrels.

STEP 1

STEP 2

STEP 3

STEP 4

Substitute the given function f(t)f(t)f(t) into the integral.Average rate=115−6∫615(0.9t2−0.2t0.4+10)dtAverage\, rate = \frac{1}{15 -6} \int_{6}^{15} (0.9 t^{2}-0.2 t^{0.4}+10) dtAveragerate=15−61​∫615​(0.9t2−0.2t0.4+10)dt

STEP 5

Calculate the integral.Average rate=115−[0.3t3−0.5t0.+10t]15Average\, rate = \frac{1}{15 -} \left[0.3 t^{3} -0.5 t^{0.} +10t \right]_{}^{15}Averagerate=15−1​[0.3t3−0.5t0.+10t]15​

STEP 6

STEP 7

STEP 8

implify the expression to find the average rate of consumption.Average rate=1[(1012.5−3.69+150)−(64.8−0.74+60)]Average\, rate = \frac{1}{} \left[ (1012.5 -3.69 +150) - (64.8 -0.74 +60) \right]Averagerate=1​[(1012.5−3.69+150)−(64.8−0.74+60)]

STEP 9

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Calculate the average fuel consumption rate from 6 AM to 3 PM given the formula $f(t) = 0.9t^2 - 0.2t^{0.4} + 10$ , where $t$ is the time in AM and $f(t)$ is the fuel consumption in barrels.

Substitute the given function $f(t)$ into the integral.
$Average\, rate = \frac{1}{15 -6} \int_{6}^{15} (0.9 t^{2}-0.2 t^{0.4}+10) dt$

Calculate the integral.
$Average\, rate = \frac{1}{15 -} \left[0.3 t^{3} -0.5 t^{0.} +10t \right]_{}^{15}$

implify the expression to find the average rate of consumption.
$Average\, rate = \frac{1}{} \left[ (1012.5 -3.69 +150) - (64.8 -0.74 +60) \right]$