Math Final Project Group 2 Group Member: YDplusSB, Hyunhwa, J-tron foteen Materials:
A measuring cylinder A right circular cone A plastic bag A glass A timer 3 people….
Purpose:
The purpose of our final project is to help us understand calculus espeically the related rate problems better by applying the math into a real world situation. Throughout the process of solving the problem, we could also learn how to edit a video, how to distribute work, and how to work together as a group.
Scenario: There is a right circular cone with a height of 11.3cm, a radius of 4cm. Initially, there is 15ml amount of water in the cone with a height of 4.85cm. As we unlock the bottom of the cone, the water in the cone will drip with a flow rate of 1.875ml/sec. At the same time, some water is poured into the cone at a flow rate of 0.275ml/sec.
Question: What’s the rate of change of the radius after 3sec with the volume of 10.45cm3 and the height of 4.226cm?
Little Video:
The actually video quality is low. We use a white backgroup to highlight the actual experiment which consists of a right circular cone, a plastic bag, one glass, and a measuring cylinder.
Related Rates Co-Authors: Blitzen, Dammitimmad, and mc Casper.
Materials: Bowling Ball
Marble Video Camera Smooth floor (with minimal friction) Paper and pen to do the work Black Box (calculator) Grey Box (brain)
Purpose: For our final project we were asked to create a related rates problem within the real world using a program to draw the actual data from our experiment. Using the video analysis with a program called logger pro we had all of the data needed to solvethe problem at hand.
The Problem: Our experiment consisted of rolling a bowling ball and a marble away from each other at a 90 degree angle. We executed this experiment on a smooth and level concrete floor indoors. We can assume that both balls were rolling at approximately the same speed (see next paragraph below). Our goal is to find the rate of change of that the distance between the two balls is increasing with respect to time.
The great lengths of assuring a constant velocity: Time frame of experiment was roughly 2 seconds Very smooth concrete floor used in experiment No interference after the initial push off Air resistance is negligible due to being indoors and no wind
Figure 1: Video of actual experiment!
Data: Graph of our data: You can see from our graph above that the velocity of the two objects is within minor human error, constant. The only part of the lines that are not constant is the beginning portion of the graph due to the ball's being stationary at the beginning of the video. The slight curve that is also apparent before the constant velocity is reached is due to the initial push from us accelerating the balls and bringing them up to speed.
Using the linear regression on our program (called logger pro), we got the equations for the velocity and displacement of each object individually.
Graphs with the linear regression equations.(Because the linear equations are difficult to read, you can go to our WikiWork Problem to see the equations of these lines more clearly.) http://crmscalc2010.pbworks.com/WikiWork-Problem-2
General Overview How To Solve: After obtaining the data and the linear regressions in loggerpro, we found the velocities using the data begotten from loggerpro. We used the X and Y components of object's velocity to find their total velocity. Next we chose a random point somewhere from the middle of the graph. To find the rate of change of the distance between the two objects we used the derivative of Pythagorean formula (see wiki). We substituted in the displacements for both objects into the formula and their velocities to find dz/dt (the related rate of the distance between the two objects with respect to time).
We choose to repeat this process for three separate points to see how accurate and consistent our answers were.
After completeing our project we determined that there was very little human error in our experiment because all three final answers were only one tenth different from one another. ***See WikiWork problem for full explination of calculations.***
Final Problem (one last time): What is the rate at which the distance between the two objects is increasing? The velocity of the bowling ball is 37.957 inches per second. The velocity of the marble is 43.52 inches per second.
***Please see our WikiWork page to see the calculations and solution to this problem*** http://crmscalc2010.pbworks.com/WikiWork-Problem-2
OK, so the first thing we did in class today was observe an obese sumo wrestler plunging to his death on a pair
of skis he probably uses as tooth picks.
This was an intro, in some way or another, into a new topic that we will
be studying for the rest of the quarter: Related Rates.I will get to exactly what those are in a bit, but first a
quick review.
(fill in the blanks)
Calculus is the mathematics of _______, and everything changes with respect to ______.
Algebra is the study of the relationship between _______.
Calculus is the study of the relationship between the _______ __ _______ of the variables (echem… derivative)
Related Rates:
Related rates are, (as far as we are concerned), how multiple derivatives are related. These variable derivatives include...
dx/dt= instantaneous rate of change of horizontal displacement with respect to time (or as we will be looking at it for the rest of this scribe post, horizontal velocity )
dy/dt= instantaneous rate of change ofvertical displacement with respect to time (or as we will be looking at it for the rest of this scribe post, vertical velocity)
Take, for instance, a cone... (just bear with me, I did this all in Paint)
Say you were to fill that cone with some agua...
It becomes evident that the rate at which the height (vertical) of the water increases is directly dependent on the width (horizontal) of the cone. Both the rate of vertical growth and horizontal growth are unique derivatives, and they clearly have a relationship. (Hence the term related rates).
(answers to quiz): Change, time, variables, rate of change.
Next, we took a rubber pig, Porky, and ruthlessly pinned him against a wall using a long pvc pipe, mercilessly using him as a demonstration for our own convenience, completely oblivious to his own mutilated feelings. The situation is simply this: Porky is on top of the pipe, pinned against the wall. We decide to lower Porky by slowly sliding the bottom of the pipe away from the wall at a constant rate until Porky reaches the ground only to go into years of shock therapy.
The rest of class was dedicated to filling out Exploration 4.9A, which Bru so graciously posted.
The same scenario of Porky's descent is posed in the exploration. (In the exploration, the pipe becomes a ladder and the wall becomes a skyscraper). And the million dollar question that we answered in the investigation is...
How is the rate at which Porky is descending related to the rate at which the bottom of the ladder is moving way from the skyscraper?
In order to answer this, we need to find the velocity at which the bottom of the ladder moves away from the skyscraper and the rate at which Porky descends the building face.
The horizontal velocity is given to be 2 ft/s, the initial horizontal distance from the skyscraper is 10 ft, and the length of the ladder is 26 ft.
In order to find the vertical velocity, we need to find the height of porky at one second intervals. We can do this because we know that the bottom leg is expanding by two feet every second, and the hypotenuse stays the same.
If you are having trouble imagining this, here is a crude visual to help you out. Each colored triangle represents Porky and the ladder's position at different seconds.
Now comes the subject responsible for the largest amount of point deductions on our tests...Algebra!!!
We must use the Pythagorean theorem to determine the length of each part of the triangle at each second.
For instance...
if in the first second, the horizontal length is 10 ft, and the length of the ladder is 26 ft (see image above), we can use Pythag to determine the height of Porky.
a=10 ft
b=?
c=26 ft
b=24ft
Now, let's take the first second after the ladder is moved.
Because the ladder is moving at a constant rate of 2 ft/s away from the building, the initial length is 10 ft, and this is the first second, the horizontal length of the ladder from the building will be 12 ft from the skyscraper.
And since we know that the length of the ladder itself is constant, we can substitute the values
a=12 ft
b=?
c=26 ft
into the Pythagorean theorem, which gives us a b value of 23.06 ft.
This train of thought can be applied to the triangle every second until Porky reaches the ground, keeping in mind that the ladder's distance from the skyscraper will increase by two feet every second.
(Image taken from 3rd slide of 2-17)
After calculating all of the a and b distance values we are able to answer the initial question:
The rate at which Porky descends is related to the rate of the ladder's movement by some variable rate.
After we found all of the b values manually, we were asked to find an equation that related the changing a and b values.
For the sake of easiness, a and b can be substituted for x and y, respectively, leaving us with the equation...
To solve the unknown rate of Porky's velocity, we simply implicitly differentiate with respect to time...
This equation relates Porky's velocity and the ladder's velocity. It is called a Differential Equation because it contains a derivative.
So, that concludes Monday's lesson, the HW for Friday is pages 1-3 of the handout Bru gave us in class, Section 4-9A: Related Rates Warm Up.
PS, I kinda lied about being over with learning derivatives, I just really wanted to put a dancing cat in the post.
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