Water Trough (Related Rates)

In summary, we can use the formula V=(1/3)(pi)(h/2)^2(h) to find the volume of a cone with height h and radius r. When the height is increasing at a rate of 0.5 m/s, the volume is also increasing at a rate of 981.25m^3/s.
  • #1
Draggu
102
0
a) 1. Homework Statement
A water trough is 10m long, and a cross section has the shape of an isosceles triangle that is 1m across at the top and 50cm high. The trough is being filled with water at a rate of 0.4m^3/min. How fast is the water level rising when the water is 40cm deep?

b) As a volcano erupts, pouring lava over its slope, it maintains the shape of a cone, with height twice as large as the radius of the base. If the height is increasing at a rate of 0.5 m/s, and all the lava stays on the slopes, at what rate is the lava pouring out of the volcano when the volcano is 50m high?

Homework Equations


h=height
w=width

The Attempt at a Solution


a)
dV/dt = 0.4m^3/min
V=(1/2)hw(10)
=5hw

w/1=h/0.5
w=2h

V=5hw = 5h(2h) = 10h^2

dV/dt=20hh'

0.4 = 20(0.4)h'
0.05=h'

I am almost sure it is correct but I am just looking for a confirmation. I will add units of course later.

b)

h' = 0.5m
r=(h/2)
h=50
v'=?

I think we are searching for the rate the volume decreases..so it would be 981.25m^3/s.
 
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  • #2
Draggu said:
a) 1. Homework Statement
A water trough is 10m long, and a cross section has the shape of an isosceles triangle that is 1m across at the top and 50cm high. The trough is being filled with water at a rate of 0.4m^3/min. How fast is the water level rising when the water is 40cm deep?

b) As a volcano erupts, pouring lava over its slope, it maintains the shape of a cone, with height twice as large as the radius of the base. If the height is increasing at a rate of 0.5 m/s, and all the lava stays on the slopes, at what rate is the lava pouring out of the volcano when the volcano is 50m high?



Homework Equations


h=height
w=width


The Attempt at a Solution


a)
dV/dt = 0.4m^3/min
V=(1/2)hw(10)
=5hw

w/1=h/0.5
w=2h

V=5hw = 5h(2h) = 10h^2

dV/dt=20hh'

0.4 = 20(0.4)h'
0.05=h'

I am almost sure it is correct but I am just looking for a confirmation. I will add units of course later.

b)

h' = 0.5m
r=(h/2)
h=50
v'=?

I think we are searching for the rate the volume decreases..so it would be 981.25m^3/s.

Can someone please look this over and tell me if I am doing it correctly? I have a test tomorrow.
 
  • #3
Hi Darggu

yep i think I'm getting the same as you, I find it easier to go through working by leaving numbers out until the end so..

[tex] w(h) = 2.h [/tex]

[tex] A(h) = \frac{hw(h)}{2} = h^2 [/tex]

[tex] V(h) = A(h).L = L.h^2 [/tex]

[tex] \frac{dV(h)}{dt} = 2hL\frac{dh}{dt} [/tex]

[tex] \frac{dh}{dt} = \frac{dV}{dt}}\frac{1}{2hL} [/tex]

i'm not sure what you mean for the 2nd one, you want to do the same method as the first, realte volumtric rate of change to rate of chenge of height

what is the volume of a cone?

Also i don't think the volume is decreasing...
 
  • #4
Here is my work.

V=(1/3)(pi)(r^2)h
= (1/3)(pi)(h/2)(h)
= (1/12)(pi)(h^3)

V' = (1/4)(pi)(h^2)h'
=(1/4)(pi)(50^2)(0.5)
=981.25

You are right, the volume is not decreasing, it's just shooting that much out every second, or?
 
  • #5
Draggu said:
Here is my work.

V=(1/3)(pi)(r^2)h
= (1/3)(pi)(h/2)(h)
do you mean
= (1/3)(pi)(h/2)^2(h) ok got it in next line...
Draggu said:
= (1/12)(pi)(h^3)

V' = (1/4)(pi)(h^2)h'
=(1/4)(pi)(50^2)(0.5)
=981.25

You are right, the volume is not decreasing, it's just shooting that much out every second, or?

yeah i think you've got it, looking good
 

1. How do you calculate the rate of change in a water trough?

The rate of change in a water trough can be calculated using the related rates formula, which is based on the principles of calculus. The formula is: dV/dt = A * dh/dt, where dV/dt represents the rate of change of the volume of water in the trough, A represents the cross-sectional area of the trough, and dh/dt represents the rate of change of the water's height in the trough.

2. What factors affect the rate of change in a water trough?

The rate of change in a water trough can be affected by several factors, including the volume of water in the trough, the shape and size of the trough, the rate at which water is being added or removed, and the rate at which the water's height is changing. These factors can be represented in the related rates formula and used to calculate the rate of change.

3. How is the related rates formula used in real-life situations?

The related rates formula is commonly used in real-life situations involving rates of change, such as when filling or draining a water trough. It can also be applied to other scenarios, such as the changing volume of a balloon or the growth rate of a population. The formula helps to determine how different variables affect each other and how they change over time.

4. Can the related rates formula be used to predict future changes in a water trough?

Yes, the related rates formula can be used to predict future changes in a water trough. By plugging in different values for the variables in the formula, the rate of change can be calculated for different scenarios. This can help to predict how the water level in the trough will change over time and how long it will take to reach a certain volume or height.

5. How does the shape of a water trough affect the rate of change?

The shape of a water trough can affect the rate of change in two main ways. First, the cross-sectional area of the trough (represented by the variable A in the related rates formula) will vary depending on the shape of the trough. This will impact the rate of change of the water's volume. Second, the shape of the trough may also affect the rate at which the water's height changes, as different shapes may result in different rates of water flow in and out of the trough.

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