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Calculus

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Submitted By justladd
Words 983
Pages 4
Justin Ladd
Calc 1
4/12/2011
Mini Project 2

Mini Project 2

For this project I took a different approach. I wanted to try something that was a little more changeling for me and at the same time I wanted a problem that would make me think. I know for some people that this may not seem to be that hard of a problem but for me these types of problem are difficult. I wanted to pick a problem that pertained to my major, which is Mechanical Engineering but that did not work out to well for me on the last project because I ended up treating the problem like an engineering problem and not a calculus problem, so with that being said that’s one reason that I picked this problem. The other reason that I choose this problem is that it seemed interesting to me and it is a story problem. For me personally I struggle with story problems and have difficulty comprehending what the problem is asking me to find. So let me tell what the problem is and then I will explain how I think it is related to the curriculum and a real world phenomenon:
A water trough is 10 m long and a cross-section has the shape of an isosceles trapezoid that is 30 cm wide at the bottom, 80 cm wide at the top, and has height 50 cm. If the trough is being filled with water at the rate of 0.2 m3/min, how fast is the water level rising when the water is 30 cm deep?
This problem is related to the curriculum because it is about linear approximation and differentials (Stewart). We covered this section is 3.9 in our text book and I thought that this was a changeling section. Now for a real life situation well we can use the math and the problem solving skills that are involved in this problem and apply them to many different situations that involve a rate of change for filling a container with a liquid for example maybe a tanker truck filling with milk or a ships hold filling with natural gas the possibilities are limitless. These types of problems ask you to think about what it is you are finding and what remembering information that you should already know like the area of a triangle or the volume of a cylinder.
.25 m For the first part of the problem I will need to figure out what information I am missing and what information I have. So the first thing I am going to do is draw a picture of the cross-section and label it with the information provided in the problem. I also need to convert all the measurements to the same units so I am picking meters since the Volume is in meter, the conversion factor for this is (cm) / 100 = meters this means that there are 100 cm in just 1 meter and I will label it all (a).

a
0.25 m 0.30 m 0.25 m =0.80 m
(a)
r
0.5 m h = 0.50m 10 m h h

0.30 m

Now that I have a picture to look at I need to figure out what the area of a trapezoid is and I am also trying to find the volume. The formula for area of a trapezoid is h * (b₁ + b₂)/2 and for Volume of trapezoidal trough = h * (b₁ + b₂)/2 * L. If you remember form reading the question it says “an isosceles trapezoid” this means that the trapezoid can be broken up into a rectangle and two similar triangles, I had to look this up online. The rectangle has a width of 0.3 m and a height of 0.50 m. The similar triangles have a top that measures 0.25 m and a height of 0.5 m. Now that I have broken the information in the question up I can start to put my equations together. By the similar triangles we get rh=0.25 m0.5 m=12 now I need to rewrite it into r = 12x. Now we can let h be the depth of the water and V be the volume of the water in the trough at any given time t. Looking at the cross-section that I drew we can determine that: V=10(0.3x+2* 12rh)
Now we put in our first equation and we get: V=100.3x+12xx.
This equation can be simplified even further into V=3x+5x2
Now I will need to take the derivative of this equation with respect to t dVdt=3+10x
We were giving the rate of change for the volume in the original equation which was 0.2 cubic meters per minute. So now we need to set this new equation equal to this 0.2=3+10x
Now divide by both sides and then substitute in 0..3 for x this is the water level of the trough 0.23+10x=0.23+100.3=0.03333 this answer is in meters per minute I want to convert it back to cm to I would multiply 0.03333 * 100 and this gives me 3.3333333 which is 103 cm/min when the water level at 30 cm. Now that I have got my equation and I found out how fast the water level is rising when the trough is being filled at a rate of 0.2 cubic meters per minute. I want to show what happens to the water level if you increase the rate of flow by 0.1 and keep the water level the same. To do this I will just make a graph.
As you can see if you increase the rate of water flowing into the trough the volume will increase also.

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