The given height of the cylinder of 1.5 m, and radius of 1 m, and the rate
of dripping of 110 cm³/s gives the following values.
1) The rate of change of the oil's radius when the radius is 0.5 m is r' ≈ 9.34 × 10⁻⁵ m/s
2) The rate of change of the oil's height when the height is 20 cm is h' ≈ 1.97 × 10⁻³ m/s
3) The rate the oil radius is changing when the radius is 10 cm is approximately 0.175 m/s
The given parameters are;
Height of the tank, h = 1.5 m
Radius of the tank, r = 1 m
Rate at which the oil is dripping from the tank = 110 cm³/s = 0.00011 m³/s
[tex]1) \hspace{0.15 cm}V = \frac{1}{3} \cdot \pi \cdot r^2 \cdot h[/tex]
From the shape of the tank, we have;
[tex]\dfrac{h}{r} = \dfrac{1.5}{1}[/tex]
Which gives;
h = 1.5·r
[tex]V = \mathbf{\frac{1}{3} \cdot \pi \cdot r^2 \cdot (1.5 \cdot r)}[/tex]
[tex]\dfrac{d}{dr} V =\dfrac{d}{dr} \left( \dfrac{1}{3} \cdot \pi \cdot r^2 \cdot (1.5 \cdot r)\right) = \dfrac{3}{2} \cdot \pi \cdot r^2[/tex]
[tex]\dfrac{dV}{dt} = \dfrac{dV}{dr} \times \dfrac{dr}{dt}[/tex]
[tex]\dfrac{dr}{dt} = \mathbf{\dfrac{\dfrac{dV}{dt} }{\dfrac{dV}{dr} }}[/tex]
[tex]\dfrac{dV}{dt} = 0.00011[/tex]
Which gives;
[tex]\dfrac{dr}{dt} = \mathbf{ \dfrac{0.00011 }{\dfrac{3}{2} \cdot \pi \cdot r^2}}[/tex]
When r = 0.5 m, we have;
[tex]\dfrac{dr}{dt} = \dfrac{0.00011 }{\dfrac{3}{2} \times\pi \times 0.5^2} \approx 9.34 \times 10^{-5}[/tex]
The rate of change of the oil's radius when the radius is 0.5 m is r' ≈ 9.34 × 10⁻⁵ m/s
2) When the height is 20 cm, we have;
h = 1.5·r
[tex]r = \dfrac{h}{1.5}[/tex]
[tex]V = \mathbf{\frac{1}{3} \cdot \pi \cdot \left(\dfrac{h}{1.5} \right) ^2 \cdot h}[/tex]
r = 20 cm ÷ 1.5 = [tex]13.\overline3[/tex] cm = [tex]0.1\overline3[/tex] m
Which gives;
[tex]\dfrac{dr}{dt} = \dfrac{0.00011 }{\dfrac{3}{2} \times\pi \times 0.1 \overline{3}^2} \approx \mathbf{1.313 \times 10^{-3}}[/tex]
[tex]\dfrac{d}{dh} V = \dfrac{d}{dh} \left(\dfrac{4}{27} \cdot \pi \cdot h^3 \right) = \dfrac{4 \cdot \pi \cdot h^2}{9}[/tex]
[tex]\dfrac{dV}{dt} = \dfrac{dV}{dh} \times \dfrac{dh}{dt}[/tex]
[tex]\dfrac{dh}{dt} = \dfrac{\dfrac{dV}{dt} }{\dfrac{dV}{dh} }[/tex]
[tex]\dfrac{dh}{dt} = \mathbf{\dfrac{0.00011}{\dfrac{4 \cdot \pi \cdot h^2}{9}}}[/tex]
When the height is 20 cm = 0.2 m, we have;
[tex]\dfrac{dh}{dt} = \dfrac{0.00011}{\dfrac{4 \times \pi \times 0.2^2}{9}} \approx \mathbf{1.97 \times 10^{-3}}[/tex]
The rate of change of the oil's height when the height is 20 cm is h' ≈ 1.97 × 10⁻³ m/s
3) The volume of the slick, V = π·r²·h
Where;
h = The height of the slick = 0.1 cm = 0.001 m
Therefore;
V = 0.001·π·r²
[tex]\dfrac{dV}{dr} = \mathbf{ 0.002 \cdot \pi \cdot r}[/tex]
[tex]\dfrac{dr}{dt} = \mathbf{\dfrac{0.00011 }{0.002 \cdot \pi \cdot r}}[/tex]
When the radius is 10 cm = 0.1 m, we have;
[tex]\dfrac{dr}{dt} = \dfrac{0.00011 }{0.002 \times \pi \times 0.1} \approx \mathbf{0.175}[/tex]
The rate the oil radius is changing when the radius is 10 cm is approximately 0.175 m
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