Mathematics, 26.03.2020 20:19, TylieW
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
Answers: 1
![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)](/tpl/images/0565/6308/3e06b.png)
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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(...
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