Answer:
[tex]Q \approx 5.333\,cm[/tex]
Step-by-step explanation:
The half-hourly rates are:
[tex]R(0) = 0.5\,\frac{cm}{h}[/tex]
[tex]R(0.5) =1\,\frac{cm}{h}[/tex]
[tex]R(1) = 1.499\,\frac{cm}{h}[/tex]
[tex]R(1.5) =1.987\,\frac{cm}{h}[/tex]
[tex]R(2) = 2.402\,\frac{cm}{h}[/tex]
[tex]R(2.5) = 2.544\,\frac{cm}{h}[/tex]
[tex]R(3) = 1.966\,\frac{cm}{h}[/tex]
The depth of water can be estimated by numerical integration:
[tex]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)[/tex]
[tex]Q \approx 5.333\,cm[/tex]
During a rainfall, the depth of water in a rain gauge increases. The increase in depth of water from 0 to 3 hours is 1.4658 cm.
Given information:
During a rainfall, the depth of water in a rain gauge increases at a rate modeled by [tex]R(t)=0.5+t \cos(\dfrac{\pi t^3}{80})[/tex].
here t is the time in hours since the start of the rainfall and R(t) is measured in centimeters per hour.
It is required to find the increase in depth of water from t=0 to t=3 hours.
At t=0, the depth of water will be,
[tex]R(t)=0.5+t \cos(\dfrac{\pi t^3}{80})\\R(0)=0.5+0 \cos(\dfrac{\pi\times 0^3}{80})\\R(0)=0.5\rm\; cm[/tex]
At t=3, the change depth of water will be,
[tex]R(t)=0.5+t \cos(\dfrac{\pi t^3}{80})\\R(3)=0.5+3 \cos(\dfrac{\pi\times 3^3}{80})\\R(3)=1.9658\rm\; cm[/tex]
So, the change in depth of water from 0 to 3 hours is,
[tex]R(3)-R(0)=1.9658-0.5\\=1.4658\rm\;cm[/tex]
Therefore, the increase in depth of water from 0 to 3 hours is 1.4658 cm.
For more details about rate of change, refer to the link:
https://brainly.com/question/8223651
