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Water Trough (Related Rates)

by Draggu
Tags: rates, trough, water
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Draggu
#1
Mar9-09, 03:13 PM
P: 102
a) 1. The problem statement, all variables and given/known data
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?



2. Relevant equations
h=height
w=width


3. 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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Draggu
#2
Mar9-09, 09:00 PM
P: 102
Quote Quote by Draggu View Post
a) 1. The problem statement, all variables and given/known data
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?



2. Relevant equations
h=height
w=width


3. 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.
lanedance
#3
Mar9-09, 09:45 PM
HW Helper
P: 3,307
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...

Draggu
#4
Mar9-09, 10:02 PM
P: 102
Water Trough (Related Rates)

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?
lanedance
#5
Mar9-09, 11:49 PM
HW Helper
P: 3,307
Quote Quote by Draggu View Post
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...
Quote Quote by Draggu View Post
= (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


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