Answer:
The water level in the bath tub is rising at a rate of 0.0111 ft/s
Explanation:
Volume of the bath tub = (Area of base) × (height)
Area of base = 18 ft² (constant)
Height = h (variable)
V = 18h
(dV/dt) = 18 (dh/dt)
If (dV/dt) = 0.2 ft³/s
0.2 = 18 (dh/dt)
(dh/dt) = (0.2/18)
(dh/dt) = 0.0111 ft/s
Hope this Helps!!!
Jon's bathtub is rectangular and its base is 18 ft2. The water level rising if Jon is filling the tub at a rate of 0.2 ft3/min is 0.011 ft/min.
If we take a look at a rectangular bathtub, the volume of the bathtub can be expressed as:
Volume (V) = length × breadth × height
where;
∴
Using differentiation to differentiate 18h with respect to t implicitly, then:
[tex]\mathbf{\dfrac{dV}{dt} = 18\dfrac{dh}{dt}}[/tex]
when the rate of rising of the volume is 0.2 ft³/min
Then;
[tex]\mathbf{0.2 = 18 \dfrac{dh}{dt}}[/tex]
[tex]\mathbf{ \dfrac{dh}{dt} = \dfrac{1}{18}\times (0.2) }[/tex]
[tex]\mathbf{ \dfrac{dh}{dt} = 0.01 1 \ ft/min}[/tex]
Therefore, we can conclude that the rate at which the water level rises if Jon is filling the tub at 0.2 ft3/min is 0.011 ft/min.
Learn more about the rectangular shape here:
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