Problem with related rates already found the answer, but with one part

In summary, the water level is rising at a rate of 3 square feet per second when the height is 1/2 foot.
  • #1
coolkid800
6
0
Hey guys can anyone please help find the answer for A, I just can't figure out what they want. I mean I was still able to figure out the answer to B, but I just kept getting the answer to A wrong and I only have one try left. Please help!

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A water trough is 15 feet long, and its cross section is an equilateral triangle with sides 3 feet long. Water is pumped into the trough at a rate of 10 cubic feet per second. How fast is the water level rising when the depth of the water is 1/2 foot?

( Hint: First, what is the height h of an equilateral triangle of side length s? Next, what is the area of an equilateral triangle in terms of the side length s? Then write the area in terms of h. The volume of the water in the trough at time t is the product of the cross-sectional area with water and the length of the trough. )

a) What is the height h of an equilateral triangle of side length s?

h = ____ ft.

Note: Answers I said were (3*sqrt(3))/2 and 1/sqrt(3) I thought all I had to do was the pythagorean theorem...but apparently its not the answer

b) The water level is rising at a rate of __(2*sqrt(3))/3 ft./sec.__
 
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  • #2
A perpendicular dropped from one vertex of an equilateral triangle bisects the opposite side. The altitude is one leg of a right triangle with hypotenuse of length s and other leg of length s/2. If we call that height x, then we have (Pythagorean theorem) s2= x2+ s2/4. Then x2= s2- s2/4= 3s2/4 and x= [itex]\sqrt{3}s/2[/itex] feet. Did you forget the "s"? And what is the other answer? (a) only asks for one answer.

When the distance from the base of the trough to water level, measured along the slant side, is s, then the height is [itex]sqrt{3}s/2[/itex] and the "base" of the triangle is s so the area is (1/2)base*height= [itex]\sqrt{3}s^2/4[/itex] square feet so the volume of water in the trough is [itex]V= 3\sqrt{3}s^2/4[/itex]cubic feet. [itex]dV/dt= 3\sqrt{3}s/2 ds/dt[/itex]. You know know that dV/dt= 10 cubic feet per second so you can find ds/dt immediately. But that is not the rate at which the water level is rising! You know that [itex]h= \sqrt{3}s/2[/itex] so [itex]dh/dt= \sqrt{3}/2 ds/dt[/itex].
 
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  • #3
Yeah, there's only one answer, the 2 answers I said were answers I tried to see if they were right.

I have a question, so S is supposed to be a variable included in the answer or I cannot plug in a number for it to get an exact numerical answer? And what is "sup" supposed to mean?
 
  • #4
Yes, s is the distance from the base of the trough to the water level measured along the slant side rather than straight up (the height). Since the water is rising, that is a variable. The problem asks you to find how fast the height is increasing when the height is 1/2 foot. You will have to use [itex]h= \sqrt{3}s/2[/itex] to find s when h= 1/2 and put that into the formula.

The "sup" was supposed to be the html tag for "superscript" but I forgot the "/" to end it: s[ sup ]2[ /sup ] without the spaces is s2.
 

Related to Problem with related rates already found the answer, but with one part

1. What is the concept of related rates in science?

The concept of related rates in science involves finding the rate of change of one quantity with respect to another quantity, where both quantities are changing over time. This is often done by using the chain rule in calculus to relate the rates of change of the two quantities.

2. How do you approach solving a problem with related rates?

To solve a problem with related rates, it is important to clearly identify and define the two changing quantities, determine how they are related, and then use the chain rule to find the rate of change of one quantity with respect to the other. It is also helpful to draw a diagram and label all given information.

3. Can you give an example of a problem with related rates?

One example of a problem with related rates is finding the rate at which the area of a circle is changing with respect to time, given the rate at which the radius is changing. This can be solved by using the formula for the area of a circle and the chain rule to relate the two rates of change.

4. What is the importance of understanding related rates in science?

Understanding related rates is important in many scientific fields, such as physics, chemistry, and engineering. It allows scientists to analyze and predict the behavior of systems where quantities are changing over time, and is essential in solving real-world problems and making accurate calculations.

5. How do you know if you have found the correct answer to a problem with related rates?

When solving a problem with related rates, it is important to check that the units of the final answer make sense and are consistent with the given information. It is also helpful to double check all calculations and make sure the answer is reasonable based on the given values and the context of the problem.

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