Related Rates 2.0

What are Related Rates?

http://www.donself.com/images/confused-baby.bmp

A related rate problem has a few qualities:

a. 2 changing quantities

b. Hopefully enough information about those quantities

c. And (as with most Calculus problems) asks you to figure out the rate at which on is changing

There are a few tricks to helping you solve these most complex problems:

1. DRAW A PICTURE!

You might think your too cool for school... but guess what? Your not.

Drawing a picture gives you time to organize and think, which could help with a breakthrough.

2. Identify the quantities

Determine and label, which quantities can change (variables) and which are constants?

Again, keeping these organized is your road to success.

3. Algebra, Algebra, Algebra...

Write your equation... remember what you learned in Algebra? You should!

4. Calculus... your new friend

Use your newly sharpened Calculus skills to differentiate the equation. This should be easy as pie... right?

5. Find the rates of change

Dissect the problem. Find out what quantities could be rates of change (I'm talking about units here!)

6. Substitute!

Here comes the fun. Now you can work with your sweet equation to answer the problem.

Wait! Don't forget... this was a word problem if I remember correctly. Which means......?

Answer in a complete sentence, with units of course.

This is a good site to practice... in case you feel you need to work on your skills.

I know that this is the kind of problems we were doing last week, but just a little play by play to help anyone who is still confused out.

Don't forget the take home quiz assigned in class... all about Related Rates.

And we have a new final project about Related Rates coming up, so start brainstorming and coming up with fun creative ideas.

The next scribe will be Blitzen.

http://www.timtim.com/public/images/drawings/large/Skier.gif

-Skirdude-

Scribe Post Feb 19th

Related Rates!!

Followed by last scribe post, in this post we will talk about the extension of Porky Problem.

This post will follow the exploration done in class: Section 4.9 A: Related Rates Warm Up

(http://www.ehow.com/how-does_4696774_credit-card-companies-use-calculus.html)

Review!

Calculus is the mathematics of change, and everything changes with respect to time.

In algebra we study the relationship between variables.

In calculus we study the relationship between the rate of change of variables.

There are infinite amount of scenarios that can be produced.

- A bathtub is filling at the rate of 2gal/min

- A balloon is deflating at a rate of 3 cubic centimeters/second

- A football is kicked with an initial upward velocity of 30yd/s

Lets translate these scenarios into mathematical models!

all theses scenarios are rates volume changing respect to time.

Now let's play with some real related rates problems!

Make sure you have the exploration in front of you!

Rippling Water Problem

- This problem deals with the radius of the circle, and the area.

Therefore we should write out the equation involving two different variables.

Let A be Area

Now we are using Implicit Differentiation

to express the rate of change of the area of the circle with respect to time t.

Process starts with stating that this equation will be derived with respect to variable.

Use chain rule to implicitly differentiate the equation.

This is the hardest part where everyone is having hard time, or at least some of us!!

It is simple, let's look at again.

- This is our original equation

each of variables, in this case A and r, is a function composed with t.

Since the letter r itself is a function, outside function is squared, and inside function is r

Using chain rule,

derive the outside, leave inside alone

- We get the result.

now we have to multiply the derivative of inside function, which is r'

However, since the function r is the function of t, we can also write out the derivative of r in different form.

It's the change in numerator over change in denominator.

This is why there is additional fraction representing derivative after each implicit differentiations.

This is how to do the implicit differentiation relating two different rates.

Rest of the question and answers are posted in Bru's post of Friday's class.

If there is any additional question, please let me know.

-Hyunhwa-

:0 Our next Scribe Post will be by Skirdude

Scribe Post 2-17

Great news everybody.....

….WE’RE FINISHED LEARNING NEW DERIVATIVES!!! (sorta) WOOHOO!!!!!

http://media.photobucket.com/image/haleluya%20cartoon/ottoneb/animated136.gif


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 of vertical 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=24 ft

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.

The next scribe post will be Hyunhwa!