How Fast Does Water Rise in a Triangular Trough When Filled?

[Draggu]

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.

 

[Draggu]

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.

 

[lanedance]

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..

$$w(h) = 2.h$$

$$A(h) = \frac{hw(h)}{2} = h^2$$

$$V(h) = A(h).L = L.h^2$$

$$\frac{dV(h)}{dt} = 2hL\frac{dh}{dt}$$

$$\frac{dh}{dt} = \frac{dV}{dt}}\frac{1}{2hL}$$

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]

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]

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...

= (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