1. ## 2 Trough Questions

First Question:
A filler is feeding a trough at a rate of 1 cubic meter / second. The troughs dimensions are 12m long with ends shaped like an isosceles triangles which measure 2m vertically and 4m wide. How quickly is the height of the feed in the in the trough increasing after 5 minutes?

Second Question:
A grease trough in the shape of an isosceles trapezoidal prism measures 16m long, 3 m across the bottom, 7m across the top and 4m in height. If the trough is filled at a rate of 4 cubic meters / minute, at what rate is the grease increasing when the depth is 2 meters?

If anyone can answer one or both of those I would greatly appreciate it! Thank you!

2. i dunno if its right but i did this for #2

i think i messed up the units but i got ( 1.22 cm/min) as the rate

2m/ 4m/min = y / x --> y = 1/2 x
hence the top is ( 3 + 1/2x + 1/2 x)

volume = (1/2) ( l + b ) h * L

v=(1/2) (3 + 3 + 1/2x + 1/2x) x * 16
v= (1/2) (6 + x) 16x
2 v = 96x + 16x^2

2 dv/dt = 96 + 36x dx/dt
2( 4 ) = 96 + 36 (2) dx/dt

dx/dt = 1.22

im looking at it now this is right but its wrong... cuz the units are off i assumed everything was in metres fix that and adjust my stuff

3. Originally Posted by rock candy
i dunno if its right but i did this for #2

i think i messed up the units but i got ( 1.22 cm/min) as the rate

2m/ 4m/min = y / x --> y = 1/2 x
hence the top is ( 3 + 1/2x + 1/2 x)

volume = (1/2) ( l + b ) h * L

v=(1/2) (3 + 3 + 1/2x + 1/2x) x * 16
v= (1/2) (6 + x) 16x
2 v = 96x + 16x^2

2 dv/dt = 96 + 36x dx/dt
2( 4 ) = 96 + 36 (2) dx/dt

dx/dt = 1.22

im looking at it now this is right but its wrong... cuz the units are off i assumed everything was in metres fix that and adjust my stuff
1.22cm/min = 0.0122m/min

Can anyone verify if they got a similar answer? The units don't matter since they're all in meters anyways. No conversions are needed so you can just work through the numbers and get the right answer.

4. Originally Posted by Sajiko
First Question:
A filler is feeding a trough at a rate of 1 cubic meter / second. The troughs dimensions are 12m long with ends shaped like an isosceles triangles which measure 2m vertically and 4m wide. How quickly is the height of the feed in the in the trough increasing after 5 minutes?

...
After 5 min = 300 s the volume of the feed is: $V(300) = 1\ \frac{m^3}s \cdot 300\ s = 300 m^3$

The volume of the trough is calculated by:

$V_t = \frac12 \cdot w \cdot h \cdot l = \frac12 \cdot 4\ m\cdot 2\ m\cdot 12\ m = 48\ m^3$

Therefore the trough would be filled completely after 48 s.

5. Originally Posted by rock candy
i dunno if its right it is, nearly but i did this for #2

...
2 v = 96x + 16x^2

2 dv/dt = (96 + 32x )dx/dt
...
Originally Posted by Sajiko
1.22cm/min = 0.0122m/min

Can anyone verify Yes
Use the corrected result. You'll get $\dfrac{dx}{dt} = 0.05\ \dfrac{m}{min}$