Rate of change: Height of a conical pile

In summary, the problem involves a conical pile at a gravel yard where sand falls at a rate of 10 cubic feet per minute. The base of the pile is approximately three times the altitude. The question asks for the rate at which the pile is getting taller when it is 15 feet tall. Using implicit differentiation, the equation 2/9πh^3 is derived and differentiated to solve for dh/dt. After correcting a few mistakes, the correct answer of 8/405π ft per minute is obtained.
  • #1
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The problem states: sand falls onto a conical pile at a gravel yard at a rare of 10 cubic feet per minute. The base of the pile is approximately three times the altitude. How fast is the pile getting taller when the pile is 15 feet tall?Volume = πr² h/3

dV = 10 h = 15
dt

When I use implicit differentiation, I get

d [V] = π/3(r²dh/dt + 2rh dr/dt)
dt
solving for dh/dt I get dV/dt(3/πr² - 2rh dr/dt

The answer they give is 8/405πm however, without knowing what the dr/dt is, I'm not sure how to solve this. The closest I cpme is 8/15π, bur r=t this is without a quantity for dr/dt.
 
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  • #2
The base of the pile is approximately three times the altitude.
 
  • #3
I have that. r Approximatelly equals 1.5h, however, I do not know how to find the rate of change for the radius with respect to time (dr/dt) in order to plug that into my equation and solve for the change in height with respect to time t (dh/dt). Since both radius and height change as the volume changes, how do I find the dr/dt to solve for dh/dt?
 
  • #4
BFPerkins said:
I have that.
Then use it! replace r with 1.5h so that you have only one variable.

If [itex]V= (1/3)\pi r^2 h[/itex] and r= 1.5h then [itex]V= (1/3)\pi (1.5)^2h^3[/itex]. Differentiate that!
 
  • #5
Thanks! I see wehere I made my mistake. I went with [tex]2/9\pi\[/tex]r^3

Iwas eliminatimg the wrong variable.

I appreciate the help.
 
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  • #6
Since I am making the height as a function of the radius, I don't think I need to squlare the 1.5. Instead, if i just write [(2/3)πh^3]/3, or (2/9)πh^3, I can differentiate thar and hopefully I will get thte corect answer.
 
  • #7
By any chance, has any0ne else worked this problem out and come with an answer of 8/405[itex]pi[/itex]?

When I put the height as a function of the radius, I get h = 2/3[itex]r^2[/itex]

so the volume = [itex]2/9\pi h^3[/itex]

when I differentiate both sides of the equation with respect to t, I get [itex]V\prime = 2/9\pi h^2 h\prime[/itex]

solving for [itex]h\prime[/itex] I get [itex]h\prime = 3V\prime/2\pi h^2[/itex]

When I plug in the values for h and [itex]V\prime[/itex] I get an answer of [itex]1/15\pi[/itex]

Ant suggestions on where I went wrong?
 
  • #8
I don't understand where h = 2/3 r^2 came from.

h = 2/3 r (not squared). But r = 3/2 h is more relevant.

Volume = 1/3 pi r^2 h = 1/3 pi 9/4 h^2 h = 3/4 pi h^3

... which leads to 8/405 pi
 
  • #9
BFPerkins said:
By any chance, has any0ne else worked this problem out and come with an answer of 8/405[itex]pi[/itex]?

When I put the height as a function of the radius, I get h = 2/3[itex]r^2[/itex]
Several posts back you said "I have that. r Approximatelly equals 1.5h"
How did r become r2? Was that a typo? If r= (3/2)h then r2= (9/4) h^2.

so the volume = [itex]2/9\pi h^3[/itex]
No, V= (1/3)\pi r^2 h= (1/3)\pi (9/4)h^2 h= (3/4)\pi h^3
when I differentiate both sides of the equation with respect to t, I get [itex]V\prime = 2/9\pi h^2 h\prime[/itex]
No, [itex]V'= (3/4)\pi (3h^2 h')= (9/4)\pi h^2 h'[/itex]
solving for [itex]h\prime[/itex] I get [itex]h\prime = 3V\prime/2\pi h^2[/itex]
h'= (4V')/(9\pi h^2)[/itex].

When I plug in the values for h and [itex]V\prime[/itex] I get an answer of [itex]1/15\pi[/itex]

Ant suggestions on where I went wrong?
V'= 10, h= 15 so [itex]h'= (40)/(9\pi (225))= (40)/(2025 \pi)= 8/(405\pi) ft per minute is the correct answer for dh/dt.
 
  • #10
thanks for the help. Abput 15 minutes ago. i saw where i made my mistake and finally came up with the corredt answer.
 

1. What is the formula for calculating the rate of change of the height of a conical pile?

The formula for calculating the rate of change of the height of a conical pile is: dH/dt = r, where dH/dt represents the rate of change, r represents the radius of the base of the cone, and t represents time.

2. How can the rate of change of the height of a conical pile be used in real-life scenarios?

The rate of change of the height of a conical pile can be used to model various physical processes, such as the melting of a snow cone, the filling of a funnel, or the growth of a sandcastle. It can also be applied in engineering and construction projects to determine the rate at which a pile of material will settle or to calculate the rate of erosion in natural structures.

3. How does the shape of the conical pile affect its rate of change?

The shape of the conical pile does not affect its rate of change, as long as the base radius remains constant. This is because the rate of change is dependent on the radius, not the overall shape of the pile.

4. Can the rate of change of the height of a conical pile be negative?

Yes, the rate of change of the height of a conical pile can be negative if the pile is decreasing in height. This could occur if the pile is being eroded or if material is being removed from the top.

5. How accurate is the rate of change formula for calculating the height of a conical pile?

The rate of change formula provides an accurate estimate of the height of a conical pile as long as the pile is approximately cone-shaped and the base radius remains constant over time. However, factors such as compaction, settling, and erosion may affect the actual height of the pile in real-life scenarios.

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