Jon's bathtub exists rectangular and its base exists 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.
A differential equation in mathematics exists an equation that links the derivatives of one or more unknown functions. Applications often involve functions that reflect physical quantities, derivatives that depict the rates at which those values change, and a differential equation that establishes a connection between the three.
If we take a look at a rectangular bathtub, the volume of the bathtub can be expressed as:
Volume (V) = length × breadth × height
where; base = length × breadth = 18ft²
The volume of the rectangular bathtub = 18h ......... (1)
Using differentiation to differentiate 18h with respect to t implicitly, then:
[tex]$\frac{\mathrm{dV}}{\mathrm{dt}}=18 \frac{\mathrm{dh}}{\mathrm{dt}}$[/tex]
When the rate of rising of the volume is 0.2 ft² / min
[tex]$0.2=18 \frac{\mathrm{dh}}{\mathrm{dt}}$[/tex]
substitute the values in the above equation, we get
[tex]$\frac{\mathrm{dh}}{\mathrm{dt}}=\frac{1}{18} \times(0.2)$[/tex]
[tex]$\frac{\mathrm{dh}}{\mathrm{dt}}=0.011 \mathrm{ft} / \mathrm{min}$[/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 exists 0.011 ft/ min.
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