Volume and Related Rates: Solving Homework Problems with Triangle Base

[Qube]

Homework Statement

http://i.minus.com/jz5YbMhv91p6K.png

Homework Equations

Volume is the area of the base (or cross section) times height.

The Attempt at a Solution

The base is a triangle. The area of a triangle is (0.5)(base length)(height). In this case that's 0.5*20*15, or 10 times 15, or 150.

V = 150(h).

We differentiate volume with respect to time, since the problem gives us dV/dt and expects us to find dh/dt.

dV/dt = 150(dh/dt).

24 = 150(dh/dt)

The change in height with respect to time is 24/150 inches/second.

This is an unfamiliar problem and I am unsure if I proceeded correctly. Am I right? It seems as if the height variable disappeared; that h = 8 as stated in the problem is extraneous information and that the rate of height increase is constant. Intuitively, this feels correct. But is it?

 

[LCKurtz]

Homework Statement

http://i.minus.com/jz5YbMhv91p6K.png

Homework Equations

Volume is the area of the base (or cross section) times height.

The Attempt at a Solution

The base is a triangle. The area of a triangle is (0.5)(base length)(height). In this case that's 0.5*20*15, or 10 times 15, or 150.

V = 150(h).

We differentiate volume with respect to time, since the problem gives us dV/dt and expects us to find dh/dt.

dV/dt = 150(dh/dt).

24 = 150(dh/dt)

The change in height with respect to time is 24/150 inches/second.

This is an unfamiliar problem and I am unsure if I proceeded correctly. Am I right? It seems as if the height variable disappeared; that h = 8 as stated in the problem is extraneous information and that the rate of height increase is constant. Intuitively, this feels correct. But is it?

No. Draw a picture of your trough with the grain at depth $y$ where $y$ is between $0$ and $h=15$. Write a formula for the volume of the grain when its depth is $y$. You should be able to get $V$ as a function of $y$ alone. Then differentiate that equation with respect to time to get started.

 

[Qube]

Alright, I think I got it now. What I had in mind was something like a triangular prism standing up on its triangular base with the height as the length of the side.

This is my work. I used similar triangles to derive a formula relating b to h so I wouldn't be doing multivar. calculus.

https://scontent-b-mia.xx.fbcdn.net/hphotos-prn2/v/1452080_10200999313166755_1652918791_n.jpg?oh=69422d2b2355ac1a310473f4c2c1814e&oe=5279C052

 

[LCKurtz]

Alright, I think I got it now. What I had in mind was something like a triangular prism standing up on its triangular base with the height as the length of the side.

That isn't what you want. A trough has its wide opening at the top and a point at the bottom. So the two vertical ends of the trough are the triangles standing on their point with the 20 inch side on top and parallel to the ground. Let's see your work.

[Edit] Apparently you added the picture after your original post. I will look at it. OK I looked and it looks good.

 

[Qube]

That isn't what you want. A trough has its wide opening at the top and a point at the bottom. So the two vertical ends of the trough are the triangles standing on their point with the 20 inch side on top and parallel to the ground. Let's see your work.

[Edit] Apparently you added the picture after your original post. I will look at it.

Yeah ... I guess English is important to the interpretation of math! Thanks again. :)