Question

Problem Statement: A cone-shaped water tank is being filled with water. The tank has a height of 10 feet and a diameter of 5 feet at the top. If the water is being pumped into the tank at a rate of 3 cubic feet per minute, how fast is the water level rising when the water is 5 feet deep? Answer Key: Labeled Drawing: You can create a simple drawing of a cone with the given dimensions. Label the height ( (h) ), radius ( { }^{circledR} ), and the water level ( (mathrm{y}) ). Formulas: 1. Volume of a cone: [ V=frac{1}{3} pi r^{2} h ] 2. Given that the cone is similar at all levels, we have [ frac{r}{h}=frac{R}{H} ] where ( mathrm{R} ) is the radius of the tank and ( mathrm{H} ) is the height of the tank. Complete Solution: 1. From the similarity of the triangles, we have [ r=frac{R}{H} h=frac{1}{2} h ] 2. Substituting ( r ) into the volume formula, we get [ V=frac{1}{3} pileft(frac{1}{2} h ight)^{2} h=frac{pi}{12} h^{3} ] 3. Differentiating both sides with respect to time ( t ), we get [ frac{d V}{d t}=frac{pi}{4} h^{2} frac{d h}{d t} ] 4. We know that ( mathrm{dV} / mathrm{dt}=3 ) cubic feet per minute. 5. Substituting ( mathrm{h}=5 ) feet and ( mathrm{dV} / mathrm{dt}=3 ) cubic feet per minute into the equation, we can solve for ( mathrm{dh} / mathrm{dt} ) [ frac{d h}{d t}=frac{3}{frac{pi}{4} cdot 5^{2}}=0.152 mathrm{ft} / mathrm{min} ] So, when the water is 5 feet deep, the water level is rising at a rate of approximately 0.152 feet per minute.

          Problem Statement: A cone-shaped water tank is being filled with water. The tank has a height of 10 feet and a diameter of 5 feet at the top. If the water is being pumped into the tank at a rate of 3 cubic feet per minute, how fast is the water level rising when the water is 5 feet deep?

Answer Key:
Labeled Drawing: You can create a simple drawing of a cone with the given dimensions. Label the height ( (h) ), radius ( { }^{circledR} ), and the water level ( (mathrm{y}) ).

Formulas:
1. Volume of a cone:
[
V=frac{1}{3} pi r^{2} h
]
2. Given that the cone is similar at all levels, we have
[
frac{r}{h}=frac{R}{H}
]
where ( mathrm{R} ) is the radius of the tank and ( mathrm{H} ) is the height of the tank.
Complete Solution:
1. From the similarity of the triangles, we have
[
r=frac{R}{H} h=frac{1}{2} h
]
2. Substituting ( r ) into the volume formula, we get
[
V=frac{1}{3} pileft(frac{1}{2} h
ight)^{2} h=frac{pi}{12} h^{3}
]
3. Differentiating both sides with respect to time ( t ), we get
[
frac{d V}{d t}=frac{pi}{4} h^{2} frac{d h}{d t}
]
4. We know that ( mathrm{dV} / mathrm{dt}=3 ) cubic feet per minute.
5. Substituting ( mathrm{h}=5 ) feet and ( mathrm{dV} / mathrm{dt}=3 ) cubic feet per minute into the equation, we can solve for ( mathrm{dh} / mathrm{dt} )
[
frac{d h}{d t}=frac{3}{frac{pi}{4} cdot 5^{2}}=0.152 mathrm{ft} / mathrm{min}
]

So, when the water is 5 feet deep, the water level is rising at a rate of approximately 0.152 feet per minute.

$Problem Statement: A cone-shaped water tank is being filled with water. The tank has a height of 10 feet and a diameter of 5 feet at the top. If the water is being pumped into the tank at a rate of 3 cubic feet per minute, how fast is the water level rising when the water is 5 feet deep? Answer Key: Labeled Drawing: You can create a simple drawing of a cone with the given dimensions. Label the height ( (h) ), radius ( ^circledR ), and the water level ( (mathrmy) ). Formulas: 1. Volume of a cone: [ V=frac13 pi r^2 h ] 2. Given that the cone is similar at all levels, we have [ fracrh=fracRH ] where ( mathrmR ) is the radius of the tank and ( mathrmH ) is the height of the tank. Complete Solution: 1. From the similarity of the triangles, we have [ r=fracRH h=frac12 h ] 2. Substituting ( r ) into the volume formula, we get [ V=frac13 pileft(frac12 h ight)^2 h=fracpi12 h^3 ] 3. Differentiating both sides with respect to time ( t ), we get [ fracd Vd t=fracpi4 h^2 fracd hd t ] 4. We know that ( mathrmdV / mathrmdt=3 ) cubic feet per minute. 5. Substituting ( mathrmh=5 ) feet and ( mathrmdV / mathrmdt=3 ) cubic feet per minute into the equation, we can solve for ( mathrmdh / mathrmdt ) [ fracd hd t=frac3fracpi4 cdot 5^2=0.152 mathrmft / mathrmmin ] So, when the water is 5 feet deep, the water level is rising at a rate of approximately 0.152 feet per minute.$

Added by Demon N.

Question

Please give Ace some feedback