Isosceles Triangle related rates problem

In summary, the problem involves a trough with isosceles triangle-shaped ends that is being filled with water at a constant rate. The trough is 9 ft long and its ends are 5 ft across at the top with a height of 1 ft. The task is to determine how fast the water level is rising when the water is 8 inches deep. Using the formula for volume, v=(1/2)bhl, and the given rate of change, dv/dt=14, we can set up an equation to solve for dh/dt. However, we need to express b as a function of h before differentiating. After some attempts, it is determined that b=5h. Plugging this into the equation
  • #1
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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 8 inches deep?

Variables:
b=5 ft
h=1 ft
l=9 ft

Homework Equations


v=(1/2)bhl
dv/dt=14
dh/dt=?
when h=2/3 ft

The Attempt at a Solution


I tried doing it the straightforward way:
v=(1/2)bhl
dv/dt=(1/2)(5)(2/3)(dh/dt)(9)
14=(1/2)(5)(2/3)(dh/dt)(9)
dh/dt=14/15

This is wrong. Then I tried using similar triangles to get the new base to go along with the height of 8 in. For that I got 10/3 ft. So,
v=(1/2)bhl
dv/dt=(1/2)(10/3)(2/3)(dh/dt)(9)
14=(1/2)(10/3)(2/3)(dh/dt)(9)
dh/dt= 7/5

This is also incorrect. I really don't know why. Any help would be greatly appreciated!
 
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  • #2
Welcome to PF!

Hi hks118! Welcome to PF! :smile:

You need to write b as a function of h before you differentiate. :wink:
 
  • #3


tiny-tim said:
Hi hks118! Welcome to PF! :smile:

You need to write b as a function of h before you differentiate. :wink:

Thanks for the reply!

So b=5h correct? but if I plug in my h value, I get 10/3, which was wrong.:confused:
 
  • #4
(just got up :zzz: …)

What was your equation for dv/dt?
 
  • #5
tiny-tim said:
(just got up :zzz: …)

What was your equation for dv/dt?

Well, v=1/2bhl so I figured dv/dt=1/2bh(dh/dt)l. Or do I need db/dt and dl/dt as well?
 
  • #6
hks118 said:
Well, v=1/2bhl so I figured dv/dt=1/2bh(dh/dt)l.

That wouldn't be right even if b was constant. :confused:

Try again. :smile:
 

1. What is an isosceles triangle?

An isosceles triangle is a type of triangle that has two sides of equal length. This means that two of the angles in the triangle are also equal in measure.

2. What is a related rates problem?

A related rates problem is a type of mathematical problem that involves finding the rate at which one quantity changes in relation to another quantity. In the context of an isosceles triangle, this could involve finding the rate at which the length of one side is changing while the length of another side remains constant.

3. How do you solve an isosceles triangle related rates problem?

To solve an isosceles triangle related rates problem, you will need to use the properties of triangles and the concept of related rates. This usually involves setting up an equation with variables representing the changing quantities and using calculus to find the rate of change.

4. What real-world applications use isosceles triangle related rates?

There are many real-world applications that use isosceles triangle related rates, such as physics problems involving motion, engineering problems involving angles and distances, and even financial problems involving the change in value of investments over time.

5. Are there any tips for solving isosceles triangle related rates problems?

Some tips for solving isosceles triangle related rates problems include drawing a clear diagram, carefully labeling the known and unknown quantities, and using the Pythagorean theorem and trigonometric functions to relate the different sides and angles of the triangle. It is also important to carefully set up the equation and correctly apply the concept of related rates.

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