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Sagot :
The rate of change of the depth of water in the tank when the tank is half
filled can be found using chain rule of differentiation.
When the tank is half filled, the depth of the water is changing at 1.213 ×
10⁻² ft.³/hour.
Reasons:
The given parameter are;
Height of the conical tank, h = 8 feet
Diameter of the conical tank, d = 10 feet
Rate at which water is being pumped into the tank, = 3/5 ft.³/hr.
Required:
The rate at which the depth of the water in the tank is changing when the
tank is half full.
Solution:
The radius of the tank, r = d ÷ 2
∴ r = 10 ft. ÷ 2 = 5 ft.
Using similar triangles, we have;
[tex]\dfrac{r}{h} = \dfrac{5}{8}[/tex]
The volume of the tank is therefore;
[tex]V = \mathbf{\dfrac{1}{3} \cdot \pi \cdot r^2 \cdot h}[/tex]
[tex]r = \dfrac{5}{8} \times h[/tex]
Therefore;
[tex]V = \dfrac{1}{3} \cdot \pi \cdot \left( \dfrac{5}{8} \times h\right)^2 \cdot h = \dfrac{25 \cdot h^3 \cdot \pi}{192}[/tex]
By chain rule of differentiation, we have;
[tex]\dfrac{dV}{dt} = \mathbf{\dfrac{dV}{dh} \cdot \dfrac{dh}{dt}}[/tex]
[tex]\dfrac{dV}{dh}=\dfrac{d}{h} \left( \dfrac{25 \cdot h^3 \cdot \pi}{192} \right) = \mathbf{\dfrac{25 \cdot h^2 \cdot \pi}{64}}[/tex]
[tex]\dfrac{dV}{dt} = \dfrac{3}{5} \ ft.^3/hour[/tex]
Which gives;
[tex]\dfrac{3}{5} = \mathbf{\dfrac{25 \cdot h^2 \cdot \pi}{64} \times \dfrac{dh}{dt}}[/tex]
When the tank is half filled, we have;
[tex]V_{1/2} = \dfrac{1}{2} \times \dfrac{1}{3} \times \pi \times 5^2 \times 8 =\mathbf{ \dfrac{25 \cdot h^3 \cdot \pi}{ 192}}[/tex]
Solving gives;
h³ = 256
h = ∛256
[tex]\dfrac{3}{5} \times \dfrac{64}{25 \cdot h^2 \cdot \pi} = \dfrac{dh}{dt}[/tex]
Which gives;
[tex]\dfrac{dh}{dt} = \dfrac{3}{5} \times \dfrac{64}{25 \cdot (\sqrt[3]{256}) ^2 \cdot \pi} \approx \mathbf{1.213\times 10^{-2}}[/tex]
When the tank is half filled, the depth of the water is changing at 1.213 × 10⁻² ft.³/hour.
Learn more here:
https://brainly.com/question/9168560
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