Lava erupts from volcano...Not all lava stays on slopes....

[wind]

Lava erupts from a volcano at a constant rate of 20 000 m^3/s (cubic meters per second). The volcano maintains the shape of a cone, with the height being 50% greater than the radius of the base. However, not all the lava stays in the slopes of the volcano. Some of the lava flows on the flat ground beyond the slopes to make a circular pattern. The height of the volcano is increasing at a rate of 0.1 m/s when the height is 700 m.

a) At what rate is lava flowing beyond the slopes of the volcano when the height is 700 m?

b) If the thickness of the circular pattern of lava flow approaching nearby villages when the height of the volcano is 700m and the radius of the circular lava pattern is 500m?

dv/dt = 20 000 m3/s

h = r (0.50)
r = 0.50 / h
hd/ht = 0.1
dl/dt= ? when h = 700

v = pi r^2 h / 3
v = pi (0.50/h) h / 3
v = pi (0.50)/ 3

This question is confusing. Can someone explain it to me? Thanks
______________________________________________
Edited by stapel -- Reason for edit: spelling, punctuation, capitalization, etc

 

[stapel]

b) If the thickness of the circular pattern of lava flow approaching nearby villages when the height of the volcano is 700m and the radius of the circular lava pattern is 500m?

I'm sorry, but the above is not a complete sentence: If the thickness is [some omitted amount] when the height is 700 meters and the radius is 500 meters, then... what?

Thank you.

Eliz.

 

[wind]

Sorry

If the thickness of the circular pattern of lava beyond the volcano is 20cm, how fast is the lava flow approaching nearby villages when the height of the volcano is 700m and the radius of the circular lava pattern is 500m?

 

[stapel]

dv/dt = 20 000 m3/s

h = r (0.50)
r = 0.50 / h
hd/ht = 0.1
dl/dt= ? when h = 700

v = pi r^2 h / 3
v = pi (0.50/h) h / 3
v = pi (0.50)/ 3

To which part of the exercise does the above apply? What is your reasoning?

Please be specific. Thank you.

Eliz.

 

[wind]

^

to part a

Givin

dv/dt = 20 000 m3/s (constant)

h = r (0.50)
r = 0.50 / h
dh/dt = 0.1
dl/dt= ? when h = 700

let l rep lava

dv/ht
v = pi r^2 h / 3
v = pi (0.50/h) h / 3
v = pi (0.50)/ 3

...I think this is wrong

so now there is the circle

v= pi r^2 h
d/dt v = d/dt ( pi r^2 h)
d/dt v = d/dt ( pi 2r h)

but when it says " when the height is 700m" are they talking about the volcano or the circle.

I don't understand this questionm

 

[galactus]

but when it says " when the height is 700m" are they talking about the volcano or the circle.

I don't understand this questionm

They're talking about the height of the volcano.

It would seem you need to find the rate of change of the diameter of the

lava flow. The diameter and the height of the volcano are the same.

When the height is 700' the diameter of the volcano is also 700'.

Since dh/dt=1/10 m/s, then dD/dt=1/10 m/s.

It would appear that part 'a' is making this observation.

For part b, you have a washer(annulus). Find the rate of change of the outside radius. The center is the volcano.

The inside radius is 350 when the volcano is 700' tall. Therefore, the

outside ring is 150' wide(500-350). You have a cylinder in the outside ring of height 0.20 m.

Hope this helps get you started.