Related Rates, why doesn't my solution work?

In summary: ZIn summary, the conversation discusses a problem involving piling sand in the shape of a cone, where the cone's height and diameter are always equal. The given information is that the height is increasing at a rate of 5 ft/min when the height is 10 ft, and the goal is to find the rate at which the sand is pouring onto the pile. The solution involves finding the volume of the cone and using implicit differentiation to find the rate of change of the radius. However, there is an error in the initial formula for the volume, as the cone's height and diameter being equal means that the volume should be (2/3)*pi*r^3 instead of (1/3)*pi*r^3. This mistake leads
  • #1
Tclack
37
0
Conical water tank, vertex down. Radius of 10ft, 24 ft high. Water flows into tank at 20 cubic ft/ min. How fast is the depth of the water increasing at 16ft?

r=10ft
y=16ft
dV/dt=20 cubic ft/min
dy/dt=?

So here was my attempt:

V=1/3 (pi r^2) y
I treated r as a constant so I could find dv/dt in terms of dy/dt

dV/dt= 1/3 pi r^2 dy/dt
dy/dt= (dV/dt)3/(pi r^2)

I found t by taking similar triangles
10/24=r/16 so r = 10*16/24 = 20/3

I plugged in the data and got 27/20pi

The answer is: 9/20pi

Here's the actual solution:
they changed r in terms of y
r=(2/3)y

V=(1/3) (pi r^2) y substituting r gives V=(1/3) pi(2/3)^2 (y^3)=(4pi/27) y^3
dV/dt= (4pi/27)(3y^2)dy/dt

dy/dt= dV/dt * 27/[4pi(3y^2)]
plugging in 16ft for y yields: 9/20pi, the right answer

What went wrong with my solution? Both solutions seem mutually correct. They're not, but I fail to recognize what's wrong with it.
 
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  • #2
Your solution assumes r is a constant. It's not. As the level of the water decreases r changes as well as y.
 
  • #3
That makes sense. I think given a similar problem, I would make the same mistake.

So, next time I get a similar problem, I'll do implicit differentiation, and get the term dr/dt, I would look at my givens and see that I have no info on dr/dt, then I would go back and get rid of r (by substituting it with y) so I don't have to deal with it...

I appreciate it, thanks.
 
  • #4
Another question: I'm working on a problem involving piling sand in the shape of a cone. The cone's height and diameter are always equal. The information gave that it's height is increasing at a rate of 5 ft /min at the instant when it's 10ft, I'm supposed to find the rate at which the sand is pouring onto the pile.
V=1/3 pi r^3
so dV/dt= pi (r^2)dr/dt
When I worked through it, I was stumped as to what to put down for dr/dt

it's DIAMETER is increasing at 5 ft/min, If at that moment, it were to stay at 5 ft/min, a minute later the new diameter would be 15ft. the radius went from 5ft to 7.5 ft. So It appears that the radius is increasing at 2.5 ft/min.

When I plugged in that information, I got the wrong answer. The books answer matches if you assume dr/dt=5ft/min

What's wrong with me now?
 
  • #5
Tclack said:
Another question: I'm working on a problem involving piling sand in the shape of a cone. The cone's height and diameter are always equal. The information gave that it's height is increasing at a rate of 5 ft /min at the instant when it's 10ft, I'm supposed to find the rate at which the sand is pouring onto the pile.
V=1/3 pi r^3
so dV/dt= pi (r^2)dr/dt
When I worked through it, I was stumped as to what to put down for dr/dt

it's DIAMETER is increasing at 5 ft/min, If at that moment, it were to stay at 5 ft/min, a minute later the new diameter would be 15ft. the radius went from 5ft to 7.5 ft. So It appears that the radius is increasing at 2.5 ft/min.

When I plugged in that information, I got the wrong answer. The books answer matches if you assume dr/dt=5ft/min

What's wrong with me now?

The volume is V=(1/3)*pi*r^2*h. If h is equal the diameter, then h=2r. That makes the volume (2/3)*pi*r^3. Not (1/3)*pi*r^3. I think what's wrong with you is that you may need some sleep. You just aren't thinking about this very clearly.
 
  • #6
Yeah I think you're right, I worked the problem again given your new information and I got the wrong answer...because in whatever screwed up mathematical system I'm using, 5^3 =625.

And I have another mystery, but I have a feeling it's not the mystery I think it is. I should look at it tomorrow after I've lain in bed with my eyes shut for 8 hours.

Thanks for your insight. ZZZ
 

1. Why do we use related rates in science?

Related rates are used in science to determine how two or more variables are changing with respect to each other. This is important in understanding the relationships between different quantities and how they affect each other in real-world situations.

2. How do I know when to use related rates in a problem?

You can identify a related rates problem when the rates of change of two or more quantities are given and you are asked to find the rate of change of another quantity that is related to them. This often involves using the chain rule and setting up an equation that relates the different rates of change.

3. Why is my solution to a related rates problem not working?

There are a few common reasons why a solution to a related rates problem may not work. One possibility is that the equation or setup is incorrect, so it is important to carefully read and understand the problem before attempting to solve it. Another common mistake is not properly differentiating the equation using the chain rule.

4. What are some tips for solving related rates problems?

Some tips for solving related rates problems include carefully reading and understanding the problem, drawing a diagram to visualize the situation, identifying the given rates of change and the rate you are looking for, and setting up an equation that relates these rates using the chain rule. It may also be helpful to use units and label all variables in the equation.

5. Can I use related rates in real-life situations?

Yes, related rates can be used to solve real-life problems in fields such as physics, engineering, and economics. For example, related rates can be used to determine the rate at which a balloon is inflating, the rate at which a population is growing, or the rate at which a chemical reaction is occurring. Understanding related rates can also help in making informed decisions in everyday life, such as determining the most efficient route to take when driving to a destination.

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