Respuesta :
The rate in mm per hour at which the height of the water is changing when the height of the water is 2mm is 55.2
Given data,
A partially filled inverted cone contains water. The inverted cone has a 16 mm radius. The inverted cone is 18 mm in height. if the water's volume is growing at a pace of 548 cubic millimeters per hour.
The rate in mm per hour at which the height of the water is changing when the height of the water is 2 mm;
r/h = 16/18
= 8/9
r = 8h/9
The volume of the cone is;
v = [tex]\frac{1}{3}\pi r^2h[/tex]
v = [tex]\frac{1}{3}\pi (8h/9)^2h[/tex]
v = [tex]\frac{64}{243}\pi h^3[/tex]
Now, differentiating concerning x;
[tex]\frac{dv}{dt}[/tex]= [tex]\frac{d}{dt}[/tex] [tex]\frac{64}{243}\pi h^3[/tex]
[tex]\frac{dv}{dt}[/tex]= 64π/243*[tex]\frac{d}{dt}[/tex] h³
[tex]\frac{dv}{dt}[/tex]= [tex]\frac{64}{81}\pi h^2\frac{dh}{dt}[/tex]
We have;
[tex]\frac{dv}{dt}[/tex]= 548 and h = 2mm
[tex]\frac{dh}{dt} = \frac{548*81}{256\pi }[/tex]
[tex]\frac{dh}{dt}[/tex] = 55.19
[tex]\frac{dh}{dt}[/tex] = 55.2mm per hour.
Hence, the rate in mm per hour at which the height of the water is changing when the height of the water is 2mm is 55.2
To learn more about volume click here:
brainly.com/question/1578538
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