Isosceles Triangular Prism Related Rates Problem

Click For Summary
SUMMARY

The discussion focuses on solving a related rates problem involving an isosceles triangular prism-shaped trough. The trough is 9 feet long, with a triangular cross-section that is 5 feet wide and 1 foot high. The water is being filled at a rate of 14 ft³/min, and the goal is to determine how fast the water level is rising when it is 7 inches deep. Key equations include the volume formula for the prism and the relationship between the base width (b) and height (h) of the triangle, leading to the conclusion that b = 5h.

PREREQUISITES
  • Understanding of related rates in calculus
  • Familiarity with the volume formula for an isosceles triangular prism
  • Ability to differentiate equations with respect to time
  • Knowledge of basic trigonometric relationships in triangles
NEXT STEPS
  • Study the relationship between base width and height in isosceles triangles
  • Learn how to apply the chain rule in related rates problems
  • Practice solving similar related rates problems involving different shapes
  • Explore the implications of differentiating volume equations in real-world applications
USEFUL FOR

Students studying calculus, particularly those focusing on related rates, as well as educators looking for examples of practical applications of calculus concepts.

maladroit
Messages
42
Reaction score
0

Homework Statement



A trough is 9 ft long and its ends have the shape of isosceles triangles that are 5 ft across at the top and have a height of 1 ft. If the trough is being filled with water at a rate of 14 ft3/min, how fast is the water level rising when the water is 7 inches deep?

I know b, h, l, dv/dt, dl/dt.
I need to first find db/dt then solve for dh/dt

Homework Equations



Volume of Iso. triangular prism= 1/2bh*l
dv/dt=1/2bh(dl/dt)+l(1/2b(dh/dt)+1/2h(db/dt)

I assume that dl/dt=0, so the new equation for the derivative is equal to.. dv/dt=l(1/2b(dh/dt)+1/2h(db/dt)

The Attempt at a Solution


If anyone could just give me some direction where to start in solving for db/dt that would be great!
 
Physics news on Phys.org
Welcome to PF!

Hi maladroit! Welcome to PF! :smile:
maladroit said:
Volume of Iso. triangular prism= 1/2bh*l
dv/dt=1/2bh(dl/dt)+l(1/2b(dh/dt)+1/2h(db/dt)

oooh, that's far too complicated! :cry:

There's a relation between b and h, so write b in terms of h, and then differentiate! :wink:
 
That seems to be just what my problem is... I can't see the relationship between b and h.

I tried setting up a triangle and solving for what b is equal to in terms of h but I am still getting the wrong answer.
I used cos(theta)=(1/2b)/h, differentiated, and solved for what db/dt was equal to.
 
maladroit said:
That seems to be just what my problem is... I can't see the relationship between b and h.

uhh? :confused: … it's a triangle … it's the same all the way up … b = 5h.
 
Once I replaced the db/dt with the 5dh/dt and solved for the problem I got .4817 ft/min but that was incorrect.

The equation I used was...
14=9(1/2*5(dh/dt)+1/2*(7/12)5(dh/dt)
 
maladroit said:
The equation I used was...
14=9(1/2*5(dh/dt)+1/2*(7/12)5(dh/dt)

Sorry, not following that :confused:

can you write it out properly: what is v in terms of h, then what is dv/dt in terms of h, then what is dv/dt when h = 7/12. :smile:
 
Right!
I used the same equation I put before
dv/dt=l(1/2b(dh/dt)+1/2h(db/dt))
and, differentiating b=5h,

dv/dt=l(1/2b(dh/dt)+1/2h(5dh/dt))

Also, I really appreciate your help!
 
hmm … v = (22.5)h2, dv/dt = … ? :wink:
 
oh! I differentiated much too early... thank you so much for your help!
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
7K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
9K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 4 ·
Replies
4
Views
5K
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K