Well, another test. This unit hasn't been so bad for me, I understand most of it. Hopefully I do well in the test. I do have a couple of problems with differentiating implicitly but it's nothing practice cannot solve. Otherwise, I'm good to go.
Good luck.
Showing posts with label Applications of Derivatives. Show all posts
Showing posts with label Applications of Derivatives. Show all posts
Tuesday, May 20, 2008
Monday, May 19, 2008
BOBBOBBOB!
What can I say? Well for one, this unit was simple the second time around =). I really think that I'm going to do well on this next test and yeah, hopefully I don't stumble on something that I haven't yet seen or tried to do. As always, the solution is practice! Hmm, I think the best place to start review would be to go through the past slides, and then see if I have any muddiest points.
This bob was short and straight to the point. I hope everyone remembers to bob and and and! Good luck on the test Wednesday I believe!
This bob was short and straight to the point. I hope everyone remembers to bob and and and! Good luck on the test Wednesday I believe!
Friday, May 16, 2008
Wednesday, May 14, 2008
Monday, May 12, 2008
Thursday, May 8, 2008
Wednesday, May 7, 2008
More optimization problems
Hello everyone this is the scribe post for May 6, sorry it is late but I had to write the AP calc exam and I was not able to post it earlier. Well in Tuesday's class we only did one question, but there were two others for homework. The first question was about a man who was driving his car through the desert, and wanted to know at what distance he should drive through the desert and the paved road to get to the city B 5km south from city A. He can drive 15mph in the desert and 35mph on the paved road, he is 5 km east of city A. So the maximum distance the man could travel is 5km in the desert and 5km of pavement to get to city B (The diagram is on the slides). So that would be unrealistic, so in order to get the minimum amount of time the car must cut time through the desert and pavement alike. So the pavement equation will become 5km subtract x, x the amount of distance that is going to be cut by driving through the desert. Then the amount of distance driving through the desert is found by a^2 + b^2 = c^2. Where a and b is 5-x. So c is (25+x^2)^.5 (square root is ^.5).
As you know Rate, Distance, and Time is all related. Time is Distance over rate. So the total time is (5-x)/35 + ((25+x^2)^.5)/15 =t(x). Solving for the derivative (solved on the slides) , t'(x) =
7x-3((25+x^2)^0.5)/105((25-x^2)^0.5). Now solving for the zero's you get 40x^2-225. So the roots are -15/2((10)^0.5) and 15/2((10)^0.5). We reject the negative value because its not possible in the context of the question. So the man should go and head for the point x = 15/2((10)^0.5), as that point is the minimum value of the First Derivative Function.
Remember to do the other two questions and the homework given in class, the next scribe is Ethan.
As you know Rate, Distance, and Time is all related. Time is Distance over rate. So the total time is (5-x)/35 + ((25+x^2)^.5)/15 =t(x). Solving for the derivative (solved on the slides) , t'(x) =
7x-3((25+x^2)^0.5)/105((25-x^2)^0.5). Now solving for the zero's you get 40x^2-225. So the roots are -15/2((10)^0.5) and 15/2((10)^0.5). We reject the negative value because its not possible in the context of the question. So the man should go and head for the point x = 15/2((10)^0.5), as that point is the minimum value of the First Derivative Function.
Remember to do the other two questions and the homework given in class, the next scribe is Ethan.
Tuesday, May 6, 2008
Friday, May 2, 2008
Optimization problems
Today, we were doing optimization problems.
In order to describe how we are solving these problems, I will be using an example from the slide 5.
It is the open box question where you have to cut squares from the corners to maximize the volume of the box.
First step is to draw a picture of what you want to do. We are cutting squares from the corners and we label the square's side with x. We know that the side of the entire piece is 24 inches per side. Therefore the equation is (24-2x) for the space between the two corner squares you have to cut. Everything is a square so all sides are the same.
Next we know that the volume of the box is going to be V=L*W*H.
We know that the height(H) is x and each side (L & W） is the same and equal to (24-2x), so the equation is going to be
V(x)= （24-2x)*(24-2x)*x
which is
V(x)=((24-2x)^2) * x
Third step is to simplify the equation and you will get
V(x)=(4x^3) - (96x^2) + 576x.
We take the derivative of that to find the max.
V'(x) =(12x^2) - 192x + 576
V'(x) =12((x^2) - 16x + 48)
The derivative at the max is equal to zero.
So we let V'(x) = 0
0 = 12((x^2) - 16x + 48)
0 = ((x^2) - 16x + 48)
0 = (x - 4) (x - 12)
x = 4, 12
x can't equal to 12 because that would make the side of the box equal to zero.
Therefore x = 4.
The last step is to test it.
___+___4__-___12_____
4 by 4 squares should be cut out to maximize the volume of the box.
The next scribe is Dino.
In order to describe how we are solving these problems, I will be using an example from the slide 5.
It is the open box question where you have to cut squares from the corners to maximize the volume of the box.
First step is to draw a picture of what you want to do. We are cutting squares from the corners and we label the square's side with x. We know that the side of the entire piece is 24 inches per side. Therefore the equation is (24-2x) for the space between the two corner squares you have to cut. Everything is a square so all sides are the same.
Next we know that the volume of the box is going to be V=L*W*H.
We know that the height(H) is x and each side (L & W） is the same and equal to (24-2x), so the equation is going to be
V(x)= （24-2x)*(24-2x)*x
which is
V(x)=((24-2x)^2) * x
Third step is to simplify the equation and you will get
V(x)=(4x^3) - (96x^2) + 576x.
We take the derivative of that to find the max.
V'(x) =(12x^2) - 192x + 576
V'(x) =12((x^2) - 16x + 48)
The derivative at the max is equal to zero.
So we let V'(x) = 0
0 = 12((x^2) - 16x + 48)
0 = ((x^2) - 16x + 48)
0 = (x - 4) (x - 12)
x = 4, 12
x can't equal to 12 because that would make the side of the box equal to zero.
Therefore x = 4.
The last step is to test it.
___+___4__-___12_____
4 by 4 squares should be cut out to maximize the volume of the box.
The next scribe is Dino.
Optimization Problems
Hey everyone this is m@rk and i'm scribing for Wednesday's class. First of we took a quiz on visual calculus about what we've learned from the first and second derivative test. Second , Mr. K introduced a new topic called Optimization Problems. Optimization problems are problems that deal with either finding the maximum or minimum of the function in the problem.
Slide 1
First of all you need to do is draw a diagram to help you visualize the problem. Second, construct the constraint equation. The constraint equation is the equation is something that limits the problem. In this case it is 320=2L+2W. Third, find what you are trying to optimize and construct an optimization equation. In this case you are trying to maximize area, so you need to use the area formula A=LW. Fourth, you need to solve for one of the unknown variables on the constraint equation and then plug it in on the optimization equation. We choose to solve for L in this case and we plug it in on the optimization equation. Fifth, solve for the derivative of our new optimization equation. Sixth, solve for the roots of the derivative of our optimization equation. We found out that the root is located at W=80. Seventh, use the first derivative test and find out whether it is a maximum or a minimum. Lastly, answer the question in a complete sentence.
I'm not going to go over the other question since it is just almost the same problem but instead I'm going to give you an outline how to solve an optimization problem, so after digging up the apcalc blog here it is:
Step 2: Write an equation for it. Use V = for volume and M = for material, A for area, etc. I must insist on using descriptive variables, because in optimization, if you are sloppy, you lose track of what's going on.
Step 3: Try to get the equation into a two variable form, so you can take the derivative.
- Step 3a: To do step 1, you will often have to create a second equation from additional information given in the problem. This may require ingenuity, but it should become natural.
- Step 3b: ISolate one of the two variables in the equation drawn in step 3a.
- Step 3c: In the original equation that you are trying to minimize or maximize, replace the variable you isolated in step 3b.
Step 4: Take the derivative of your two-variable equation.
Step 5: Set the derivative to 0, and solve for the value of the remaining variable.
Step 6: Plus the value of that variable into the first two equations to find all dimensions, including the final goal, such as the amount of material, cost, or volume.
NOTE: This is just an outline so follow it loosely. There is really no chronological way to answer this problems.
Thats it , the next scribe is haiyan.
Slide 1
First of all you need to do is draw a diagram to help you visualize the problem. Second, construct the constraint equation. The constraint equation is the equation is something that limits the problem. In this case it is 320=2L+2W. Third, find what you are trying to optimize and construct an optimization equation. In this case you are trying to maximize area, so you need to use the area formula A=LW. Fourth, you need to solve for one of the unknown variables on the constraint equation and then plug it in on the optimization equation. We choose to solve for L in this case and we plug it in on the optimization equation. Fifth, solve for the derivative of our new optimization equation. Sixth, solve for the roots of the derivative of our optimization equation. We found out that the root is located at W=80. Seventh, use the first derivative test and find out whether it is a maximum or a minimum. Lastly, answer the question in a complete sentence.
I'm not going to go over the other question since it is just almost the same problem but instead I'm going to give you an outline how to solve an optimization problem, so after digging up the apcalc blog here it is:
Step 1: Find what you are trying to maximize or minimize. This will be stated excplicitly (in the question).
Step 2: Write an equation for it. Use V = for volume and M = for material, A for area, etc. I must insist on using descriptive variables, because in optimization, if you are sloppy, you lose track of what's going on.
Step 3: Try to get the equation into a two variable form, so you can take the derivative.
- Step 3a: To do step 1, you will often have to create a second equation from additional information given in the problem. This may require ingenuity, but it should become natural.
- Step 3b: ISolate one of the two variables in the equation drawn in step 3a.
- Step 3c: In the original equation that you are trying to minimize or maximize, replace the variable you isolated in step 3b.
Step 4: Take the derivative of your two-variable equation.
Step 5: Set the derivative to 0, and solve for the value of the remaining variable.
Step 6: Plus the value of that variable into the first two equations to find all dimensions, including the final goal, such as the amount of material, cost, or volume.
NOTE: This is just an outline so follow it loosely. There is really no chronological way to answer this problems.
Thats it , the next scribe is haiyan.
Wednesday, April 30, 2008
Monday, April 28, 2008
The late Thursday scribe.
Yes me again. There actually isn't a lot of information to include on this scribe.
Due to the fact, Mr. K thought it was a lot of information to absorb for the past few classes.
So we....
basically reviewed Tuesday's class.
We got into groups, did the second slide question.
And critical numbers of f, are when f' and f'' are equal to zero.
Next scribe... Kristen.
Sorry it's so short...
Due to the fact, Mr. K thought it was a lot of information to absorb for the past few classes.
So we....
basically reviewed Tuesday's class.
We got into groups, did the second slide question.
And critical numbers of f, are when f' and f'' are equal to zero.
Next scribe... Kristen.
Sorry it's so short...
The Last Tuesday Scribe
Sorry it's so late. Kept forgetting and procrastinating. Yes, it was supposed to be me... but, I wasn't even participating during the class, because of APcalc in the corner of the room. But, I can at least explain the slides.
Slide 1, 2
Picture and the question.
Slide 3,4,5
Velocity is the change of distance over the change in time.
a. The average velocity, is the slope of the line, from the interval 0 to 4 of the given position function.
b. Velocity is zero when, the derivative of x(t) is equal to zero.
c. When the point is moving to the right (positive) meaning, the velocity is positive. (Derivative of x is positive)
d. Same thing as c, but moving left (negative)
e. When the derivative is equal to 3
f. Taking the derivative of the velocity function. Or the second derivative of the position function, called the acceleration function is equal to 3.
q. (Not sure why q, but okay) Draw what x(t) looks like. Red graph is the derivative of x(t)
Slide 6,7
The graph, and line test means, you are trying to find from the derivative, where there may be local minimums or maximums on the parent function.
So, when the derivative has values that crosses the x axis(zero) there is a local maximum or minimum. To determine whether it is a maximum or a minimum, on the left side of the zero, if it is a positive value then crosses the x-axis and then is negative, it is a local maximum. Vice versa, it is a local minimum.
Slide 8
The extreme value theorem, says, on a differentiated function, within a closed interval, there exists a maximum and minimum value on the parent function.
Slide 9 and 10 practice what we just learned.
That's it for that scribe. I'm onto Thursday's scribe now. Ciao.
Slide 1, 2
Picture and the question.
Slide 3,4,5
Velocity is the change of distance over the change in time.
a. The average velocity, is the slope of the line, from the interval 0 to 4 of the given position function.
b. Velocity is zero when, the derivative of x(t) is equal to zero.
c. When the point is moving to the right (positive) meaning, the velocity is positive. (Derivative of x is positive)
d. Same thing as c, but moving left (negative)
e. When the derivative is equal to 3
f. Taking the derivative of the velocity function. Or the second derivative of the position function, called the acceleration function is equal to 3.
q. (Not sure why q, but okay) Draw what x(t) looks like. Red graph is the derivative of x(t)
Slide 6,7
The graph, and line test means, you are trying to find from the derivative, where there may be local minimums or maximums on the parent function.
So, when the derivative has values that crosses the x axis(zero) there is a local maximum or minimum. To determine whether it is a maximum or a minimum, on the left side of the zero, if it is a positive value then crosses the x-axis and then is negative, it is a local maximum. Vice versa, it is a local minimum.
Slide 8
The extreme value theorem, says, on a differentiated function, within a closed interval, there exists a maximum and minimum value on the parent function.
Slide 9 and 10 practice what we just learned.
That's it for that scribe. I'm onto Thursday's scribe now. Ciao.
Thursday, April 24, 2008
Tuesday, April 22, 2008
Monday, April 21, 2008
Application of Derivatives
On Friday, our class started talking about the application of derivatives. To start off, Mr. K showed us this slide:
Mr. K then asked us when was the car moving fastest. The class decided that the car was moving fastest during t2, because the slope of the tangent line was the steepest at that point. During the interval [t1,t2], it is clear that the car is speeding up during that interval because the slope of the tangent lines are getting steeper going from t1 to t2. The car is definitely slowing down on the interval [t2,t3] as we can see that the slope of the tangent lines are decreasing.
Now this was a tough graph to decipher. The AP Calculus students had no problem understanding this one, but to those of us who aren't taking AP, this was kind of difficult. I myself found it quite confusing at first, but after clarification, I understand it. Now, you have to find the derivative of f in this graph to find out where is f ' is positive. First we have to find out where the tangent lines in f are zero. Looking at the function f, we can clearly see that the tangent lines are zero in -2 and +2. This means that f ' will have roots at -2 and +2. Then, we look when the tangent lines of f are approaching zero. There are 2 instances in this graph. The first instance is from (-∞, -2]. The slope from that domain is increasing, since it's coming from a negative value to zero. But that still doesn't solve our problem since f ' is still negative at that point. Then, from (-2, 2) in f, we can see that the slope is increasing until (0,0) and it starts to decrease until it reaches (2,3), where it becomes zero. At this point, f ' is positive as the slopes of the tangent lines from (-2, -1) to (0,0) are increasing from zero and reaches its maximum at (0,0), and then it starts to decrease from (0,0) until it reaches (2,3) where the slope is zero. This then solves the first question which asks where is f ' positive, which is from (-2, o) to (2,0).
Now to find out where f '' is positive. From the graph, we can see that f is some kind of a cubic function, so we know that its derivative will be a quadratic function. Now, the derivative of a quadratic function is a straight line, so now we know that f '' is a straight line. Looking at the shape of f ' (say it is -2x2), it is definitely a negative quadratic function because it has a maximum, not a minimum. We then have to find out the derivative of f '' which is represented as -2x2 (However, it IS NOT -2x2, it's just a representation). Using the power rule, we can determine that the derivative of f ', which is f '', is -4x. From this we can see that the function is positive at quadrant 2.
The critical numbers (which is another term for the roots) of f are -2 and 2.
This is a little difficult...
For a), f is increasing wherever f ' is positive.
For b), f(0) is negative.
Well, that was all we talked about last class. Mr. K tried to squeeze in one more slide but we ran out of time.
The next scribe is Kristin.
Mr. K then asked us when was the car moving fastest. The class decided that the car was moving fastest during t2, because the slope of the tangent line was the steepest at that point. During the interval [t1,t2], it is clear that the car is speeding up during that interval because the slope of the tangent lines are getting steeper going from t1 to t2. The car is definitely slowing down on the interval [t2,t3] as we can see that the slope of the tangent lines are decreasing.
Now this was a tough graph to decipher. The AP Calculus students had no problem understanding this one, but to those of us who aren't taking AP, this was kind of difficult. I myself found it quite confusing at first, but after clarification, I understand it. Now, you have to find the derivative of f in this graph to find out where is f ' is positive. First we have to find out where the tangent lines in f are zero. Looking at the function f, we can clearly see that the tangent lines are zero in -2 and +2. This means that f ' will have roots at -2 and +2. Then, we look when the tangent lines of f are approaching zero. There are 2 instances in this graph. The first instance is from (-∞, -2]. The slope from that domain is increasing, since it's coming from a negative value to zero. But that still doesn't solve our problem since f ' is still negative at that point. Then, from (-2, 2) in f, we can see that the slope is increasing until (0,0) and it starts to decrease until it reaches (2,3), where it becomes zero. At this point, f ' is positive as the slopes of the tangent lines from (-2, -1) to (0,0) are increasing from zero and reaches its maximum at (0,0), and then it starts to decrease from (0,0) until it reaches (2,3) where the slope is zero. This then solves the first question which asks where is f ' positive, which is from (-2, o) to (2,0).
Now to find out where f '' is positive. From the graph, we can see that f is some kind of a cubic function, so we know that its derivative will be a quadratic function. Now, the derivative of a quadratic function is a straight line, so now we know that f '' is a straight line. Looking at the shape of f ' (say it is -2x2), it is definitely a negative quadratic function because it has a maximum, not a minimum. We then have to find out the derivative of f '' which is represented as -2x2 (However, it IS NOT -2x2, it's just a representation). Using the power rule, we can determine that the derivative of f ', which is f '', is -4x. From this we can see that the function is positive at quadrant 2.
The critical numbers (which is another term for the roots) of f are -2 and 2.
This is a little difficult...
For a), f is increasing wherever f ' is positive.
For b), f(0) is negative.
Well, that was all we talked about last class. Mr. K tried to squeeze in one more slide but we ran out of time.
The next scribe is Kristin.
Friday, April 18, 2008
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