Mathematics: During a rainfall, the depth of water in a rain guage increases at a rate modeled by R(t)=0.5+t cos(πt^3/80), where t is the time in hours since the start of the rainfall and R(t) is measured in centimeters per hour. How much did the depth of water in the rain gauge increase from t=0 to t=3 hours

25% as a fraction is 1/4 and as a decimal it’s 0.251/2 as a percentage is 25% and as a decimal it’s 0.256.66% as a fraction is 333/50 and as a percentage it’s 666%...

During a rainfall, the depth of water in a rain

Ответ:

yasmin599

21.01.2022

Mathematics

$Q \approx 5.333\,cm$

Step-by-step explanation:

The half-hourly rates are:

$R(0) = 0.5\,\frac{cm}{h}$

$R(0.5) =1\,\frac{cm}{h}$

$R(1) = 1.499\,\frac{cm}{h}$

$R(1.5) =1.987\,\frac{cm}{h}$

$R(2) = 2.402\,\frac{cm}{h}$

$R(2.5) = 2.544\,\frac{cm}{h}$

$R(3) = 1.966\,\frac{cm}{h}$

The depth of water can be estimated by numerical integration:

$Q \approx [0.5\,\frac{cm}{h} + 0.5\cdot (1\,\frac{cm}{h} - 0.5\,\frac{cm}{h} )+1\,\frac{cm}{h} + 0.5\cdot (1.499\,\frac{cm}{h} - 1\,\frac{cm}{h} )+1.499\,\frac{cm}{h} + 0.5\cdot (1.987\,\frac{cm}{h} - 1.499\,\frac{cm}{h} ) + 1.987\,\frac{cm}{h} + 0.5\cdot (2.402\,\frac{cm}{h} - 1.987\,\frac{cm}{h} )+2.402\,\frac{cm}{h} + 0.5\cdot (2.544\,\frac{cm}{h} - 2.402\,\frac{cm}{h} )+1.966\,\frac{cm}{h} + 0.5\cdot (2.544\,\frac{cm}{h} - 1.966\,\frac{cm}{h} )] \cdot (0.5\,h)$

$Q \approx 5.333\,cm$

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During a rainfall, the depth of water in a rain guage increases at a rate modeled by R(t)=0.5+t cos(πt^3/80), where t is the time in hours since the start of the rainfall and R(t) is measured in centimeters per hour. How much did the depth of water in the rain gauge increase from t=0 to t=3 hours

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