Respuesta :
The depth of the water increasing when the water is 13 feet deep is approximately 1.68 feet per minutes.
We know that the conical water tank has a radius of 12 feet and is 26 feet high.
We also know that water is flowing into the tank at a rate of 30ft³/min. In other words, our derivative of the volume with respect to time t is:
[tex]\frac{dV}{dt} = \frac{10 ft^3}{min}[/tex]
We want to find how fast the depth of the water is increasing when the water is 13 feet deep. So, we want to find dh/ dt.
First, remember that the volume for a cone is given by the formula:
V = 1/3 π r² h
We want to find dh/dt. So, let's take the derivative of both sides with respect to the time t. However, first, let's put the equation in terms of h.
We can see that we have two similar triangles. So, we can write the following proportion:
[tex]\frac{r}{h} = \frac{12}{36}[/tex]
Multiply both sides by h:
⇒ [tex]r = \frac{12}{36} h[/tex]
So, let's substitute this in r:
[tex]V = \frac{1}{3} \pi \frac{12}{36} h^{2} h[/tex]
Square:
[tex]V = \frac{1}{3} \pi \frac{144}{3888} (h^{2} ) h[/tex]
Simplify:
[tex]V = \frac{144}{3888} \pi h^{2}[/tex]
Now, let's take the derivative of both sides with respect to t:
[tex]\frac{d}{dt} (V) = \frac{d}{dt} [\frac{144}{3888} ] \pi h^{3}[/tex]
Simplify:
[tex]\frac{dV}{dt} =\frac{144}{3888} \pi (3h^{2} ) \frac{dh}{dt}[/tex]
We want to find dh/dt when the water is 13 feet deep. So, let's substitute 13 for h. Also, let's substitute 10 for dV/dt. This yields:
[tex]10 = \frac{144}{3888} \pi (3(13^{2} ) \frac{dh}{dt}[/tex]
[tex]10 = \frac{144}{3888} \pi (507) \frac{dh}{dt}[/tex]
[tex]10 = \frac{73008}{3888} \pi \frac{dh}{dt}[/tex]
[tex]38880 = 73008 \pi \frac{dh}{dt}[/tex]
[tex]\frac{dh}{dt} = \frac{38880}{73008} \pi[/tex]
[tex]\frac{dh}{dt} = \frac{38880}{73008} X\frac{22}{7}[/tex]
≈ 1.6737109 feet / min
The water is rising at a rate of approximately 1.68 feet per minute.
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