Finding Rate of Change for Water Level in a Trough

[answerseeker]

A trough is 10 ft long and its ends have the shape of isosceles triangles that are 3ft across at the top and have a height of 1ft. if the trough is filled with water at a rate of 12ft^3/min, how fast does the water level rise when the water is 6inches deep?

so far i have: dv/dt= 12ft^3/min and need to find dh/dt.
V= base* height*length ? where length=10 ft.
what i don't understand is how come in my notes, the next step is that b becomes 0.5*b*h??

 

[Integral]

I have no way to help you with what is written in your notes. But...

The expression:

V= b*h*L

Gives the volume of a solid with a rectangular cross section. You have a solid with a triangular cross section so the the expression

V = .5*b*h*L

Would be correct.

can you complete the problem from there?

 

[HallsofIvy]

A trough is 10 ft long and its ends have the shape of isosceles triangles that are 3ft across at the top and have a height of 1ft. if the trough is filled with water at a rate of 12ft^3/min, how fast does the water level rise when the water is 6inches deep?

so far i have: dv/dt= 12ft^3/min and need to find dh/dt.
V= base* height*length ? where length=10 ft.
what i don't understand is how come in my notes, the next step is that b becomes 0.5*b*h??

V= base*height*length for a RECTANGULAR trough. This is TRIANGULAR. The area of a triangle is 0.5*b*h.

 

[answerseeker]

I think I get it now. since it's a triangular prism, you take the area of the triangle and multiply it by the base of the rectangle. I think that's correct now.