contestada

During a rainfall, the depth of water in a rain gauge increases at a rate modeled by r(t)=0.5+tcos(πt380) , 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?

Respuesta :

lhmarianateixeira

During a rainfall, the depth of water in a rain gauge increases  1.4658 cm.

What is a cosine function?

Cosine is a trigonometric function, used in a right triangle to define the ratio of the side adjacent to and the hypotenuse of this triangle.

During a rainfall, the depth of water in a rain gauge increases at a rate modeled by:

[tex]R(t)=0.5+tcos(\frac{\pi t^3}{80})[/tex]

Where:

  • t is the time in hours since the start of the rainfall
  • R(t) is measured in centimeters per hour.

So, when t=0, the depth of water will be:

[tex]R(t)=0.5+tcos(\frac{\pi t^3}{80})\\R(0)=0.5+0cos(\frac{\pi 0^3}{80})\\R(0)=0.5 cm[/tex]

So, when t=3, the depth of water will be:

[tex]R(t)=0.5+tcos(\frac{\pi t^3}{80})\\R(3)=0.5+3cos(\frac{\pi 3^3}{80})\\R(3)=1.9658 cm[/tex]

So the difference will be:

[tex]R(3)-R(0)=1.4658 cm[/tex]

See more about rate of change at brainly.com/question/8223651

Otras preguntas