rate of change problem...

[billbarber]

Hi. I thought I had this problem figured out but my answers aren't making sense logically.

A tank of water in the shape of a cone is leaking water at a constant rate of 2 ft^3/hr. The base radius of the tank is 5 ft and the height of the tank is 14 ft.
a) At what rate is the depth of the water in the tank changing when the depth of the water is 6 ft.?
b) At what rate is the radius of the top of the water in the tank changing when the depth of the water is 6 ft.?

Ok. So to start out I used the Volume formula V=pi(r^2)(h)/3 and multiplied both sides by d/dt. Since we know that dV/dt= 2 ft^3 I plugged that in and the 5 for the radius and solved for dh/dt and got .0764 ft/hr. (that's rounded answer.) I then subtracted the new depth (6 ft) from the original height (14 ft) to get the change in the height, which is 8 ft. I then divided that by .0764 to see how long it took for the height to reach that point and got 104.7120 hrs. (rounded again.) I multiplied that by 2 to find how much the volume had changed in that amount of time, and got 209.4241. (rounded.) I then subtracted that from the volume of the entire cone (366.5191) to get the volume of the remaining water and got 157.0950. I then put this into the Volume formula with the height 6 ft. to try to solve for the radius, and the answer was slightly larger than the original radius, which shouldn't happen as you get closer to the tip of the cone. So I stopped the problem there and haven't tried to solve for the rate of change of that radius yet, since it seemed like I had gone wrong somewhere. Where did I go wrong?

Thanks, any help would be greatly appreciated.

 

[Deleted member 4993]

Hi. I thought I had this problem figured out but my answers aren't making sense logically.

A tank of water in the shape of a cone is leaking water at a constant rate of 2 ft^3/hr. The base radius of the tank is 5 ft and the height of the tank is 14 ft.
a) At what rate is the depth of the water in the tank changing when the depth of the water is 6 ft.?
b) At what rate is the radius of the top of the water in the tank changing when the depth of the water is 6 ft.?

Ok. So to start out I used the Volume formula V=pi(r^2)(h)/3 and multiplied both sides by d/dt. Since we know that dV/dt= 2 ft^3 I plugged that in and the 5 for the radius and solved for dh/dt and got .0764 ft/hr. (that's rounded answer.) I then subtracted the new depth (6 ft) from the original height (14 ft) to get the change in the height, which is 8 ft. I then divided that by .0764 to see how long it took for the height to reach that point and got 104.7120 hrs. (rounded again.) I multiplied that by 2 to find how much the volume had changed in that amount of time, and got 209.4241. (rounded.) I then subtracted that from the volume of the entire cone (366.5191) to get the volume of the remaining water and got 157.0950. I then put this into the Volume formula with the height 6 ft. to try to solve for the radius, and the answer was slightly larger than the original radius, which shouldn't happen as you get closer to the tip of the cone. So I stopped the problem there and haven't tried to solve for the rate of change of that radius yet, since it seemed like I had gone wrong somewhere. Where did I go wrong?

Thanks, any help would be greatly appreciated.

To be able show you where did you go wrong - at least I - need to see your work in greater detail. You have some problem with mathematical term and I am confused about what you are trying to do.

For example you said - "multiplied both sides by d/dt"

That does not make sense - you cannot do that. So I need to see what you really meant by that statement - what did you actually carry out.

First of all you have to start with a volume at a given time - not the full volume.

The volume at a given time, consists of an empty cone at the top of the full cone. Let the height of the water level from the base = h

height of the empty cone = 14-h

radius of the empty cone = (14-h)/14 * 5 = 5 - 5h/14

The volume of water at a given time V = pi/3 [350 - (5 - 5h/14)^2 * (14-h)]

Then differentiate both sides - w.r.t. time (t).

dV/dt = pi/3[-2*(5 - 5h/14) * (-5/14) * (14-h) - (5 - 5h/14)^2 *(-1)] * dh/dt

Now start working from here....

 

[billbarber]

Yeah, I was having trouble trying to find the terminology, differentiate with respect to time is what I meant.

Couple of questions.

I also forgot to mention the diagram shows the cone with the point at the bottom like an ice cream cone not with the point at the top like a pyramid. So the part with water is actually the second cone. Does that affect the equations you gave me?

In the Volume at a given point formula, what is the 350 from?

 

[Deleted member 4993]

Yeah, I was having trouble trying to find the terminology, differentiate with respect to time is what I meant.

Couple of questions.

I also forgot to mention the diagram shows the cone with the point at the bottom like an ice cream cone not with the point at the top like a pyramid. So the part with water is actually the second cone. Does that affect the equations you gave me? Yes the equation will be different - but the principle will be same

In the Volume at a given point formula, what is the 350 from? 5^2 * 14