Today, Mr. Hansen started out by asking the class a few questions about calculus to see what we already knew. These included "What does the word 'calculus' mean?" and "Who uses calculus?" From that we gathered that calculus was invented by Sir Isaac Newton in the mid 1600's. This was made more famous because Halley used calculus to calculate precisely the path of a comet. Today calculus is used by all sorts of mathematicians, physicists, engineers, and scientists. The word literally translates to "pebble."
In class Mr. Hansen also tossed a tennis ball up and caught it, asking us to graph time and the ball's height in feet above the floor. The ball slowly decelerated until it reached the apex of its climb and then accelerated back down. Mr. Hansen caught the ball lower than where he had tossed it from. This was shown in many of our graphs. Next, Mr. Hansen also bounced the ball, asking us to graph this is well. This time, there was a point on the graph at which the ball's height decrease stopped suddenly where the ball hit the floor. The ball then accelerated off the floor but slowly decelerated as it climbed up again and Mr. Hansen caught it.
We started off today's class with a bit of review from the last class as well as some history that Mr. Hansen forgot from yesterday's class. Archimedes was an ancient greek who made lots of calculus-related discoveries, which were only repeated when Newton made incredible calculus contributions about 1500 to 2000 years later.
We spent the rest of the class talking about the activity from the day before and the concepts behind it. The activity was about trying to calculate the exact velocity at a specific moment in time of a falling paper. The class decided that we cannot find the velocity at a specific point in time, because it is just a point. We need two points to calculate velocity -- velocity measures change, if there is no change, there is no velocity. Mr. Hansen also explained that average velocity is equal to distance travelled divided by elapsed time, or delta s (distance) divided by delta t (time). On a graph, the average velocity is represented by m sec, or the slope of the secant line of the interval (a secant line is a line connects points in a specific time interval), so long as it is a time v. distance graph. We can mathematically show that we cannot calculate velocity at a specific point because velocity is equal to delta s/delta t -- distance travelled/elapsed time. If no distance was travelled and no time elapsed, as would be the case at a specific point, the expression would be 0/0, which is undefined.
However, we found that we can get close to the velocity at a certain instant by reducing the time interval with the instant included. With the points given, the average velocity of the interval [9/60, 10/60] would be the most accurate. Despite the fact that by keeping the time interval small, the accuracy increases, that would be a more accurate average velocity than that of the interval [9/60, 10/60]. This is because in the former interval, there is a part that has a velocity less than the velocity at 10/60 and a part that has a greater velocity, thus putting the average somewhere close to that of 10/60, whereas in the latter, 10/60 is at the end of the range of the interval, resulting in a less accurate velocity.
As the time elapsed approaches zero, we can get a more and more accurate calculation of the velocity at an instant.
At the end of class, we related the two worlds in which we had been working for this activity: the world with physical objects and the world with mathematical abstractions (graphs). Average velocity in the physical world is essentially the same as m sec (slope of the secant line, as long as the graph is time v. distance). Also, in the physical world, as time elapsed approaches zero, we get closer to the velocity of the instant. Similarly, in the graph world, as the change in t approaches zero, we get closer to m tan (slope of the tangent line, a line that touches the graph only at one point, so that its slope would be equal to about that of the velocity at that instant.
Monday August 25th Today in class we focused on using a second method to find Instantaneous Velocity. We started by briefly looking in to the difference between speed and velocity, speed being the Magnitude of Velocity (Total distance/Elapsed Time), but Mr. Hansen said we would wait a little before delving into this further...
As I alluded to earlier, the majority of the class time was devoted to learning another way to find Vinst (Instantaneous Velocity). While we already know the "chart method" it is far less precise and much more tedious than the so-called Delta Process, an algebraic method for finding Mtan (Vinst). The Delta Process relays on finding a fixed point near the tangent line on a graph, and then using limits to move the point infinitesimally close to the fixed point, generating a much more precise Tangant line. A key concept for this process is "h", a very small quantity that is added to the fixed point's x-value, which gets closer and closer to the fixed point. In class we used the example of y=4-x^2 with the fixed point (1,3) to learn this new process. The first step is to create a second point that is close to the fixed point and generate a secant line. This is where "h" comes in, as we used the point (1+h, 4 - [1+h]^2) to generate the secant line. The X-coordinate is the same as the fixed point's X, with "h" added. This x-coordinate (1+h in this case) is then plugged in to the original equation (y=4-x^2) to find what the Y-coordinate would be at x=1+h. After finding both coordinates, we were able to find the slope of the secant line between the fixed point and our second point. (Slope = change in y/change in x). After some plain old algebraic simplifying, we calculated Msec (slope of secant line) to be { (h[-2-h])/h }. To find Mtan, we simply moved the second point closer to the fixed point. By doing this, we move "h" closer to 0 seeing as "h" is the coordinate difference for the two points. In the end, we generated the following equation : Mtan = Limit (h -> 0) [h/h(-2-h)] = -2
Today in class we talked about tangent lines in geometry and calculus. "In geometry, a tangent line is a line that intersects a circle at exactly one point." Calculus is a bit different... This is because in Calculus, the tangent sometimes goes through more than one point. In Calculus, a tangent line to point P is "the line through P whose sl0pe equals the limit of the slopes of secant PQ as Q approaches P along the graph. Basically, its the delta process!! m=lim(h->0) {f(x+h)-f(x)/ h}
We then practiced the delta process on a worksheet, finding the slopes of tangents, and also the slopes of perpendicular lines to the tangents. The back of the paper was review, finding the average velocity from a chart and asking tough questions. Mr. Hansen hinted that the tough question (#5) will be close to something for the quiz on Thursday. So watch out!
Students were given a position - time graph, and they then had to walk in the motion detector in a way that would create the prescribed position time graph. Students had to think carefully about acceleration and its impact on the position time graph. We then discussed the history of calculus; specifically, we discussed the fact that Newton was unable to clearly define the concept of "limit". The students started working on an investigation of how different functions behave near points where they are undefined.
Today's class started off with a quiz. After said quiz, we discussed the various limits of the functions on the limit worksheet. The central point of the discussion was whether we could be sure of what the y-value would be on certain graphs as they approached the point at which they were undefined.
In the end, we determined that we could only be truly sure on the first graph (it would be very close to 5). In the second graph, though it is easy to determine the y-value of any x, it is impossible to make an accurate guess of a number close to 2 as we do not know which side the limit is approaching from. For the third graph, the only thing we can say with certainty is that y will become very, very large as it approaches the limit. No limit was found for the final graph.
addition to 8-28-08 comment We also discussed three different reasons that the limit of f(x), as x approaches c, might fail to exist: (i) a jump (b) f(x) growing without bound (a vertical asymptote) (c) infinitely rapid oscillation of the f(x) values
Friday in class we had a reading quiz over Thursday night's homework. Then we went over the answers.
We found another way to explain the limit of a function as x approaches a. We decided that we look at the y-coordinates for x-values CLOSE to a, but we ignore x=a. Then we looked at two examples.
The first example was f(x)= (x^2-9) / (x-3), where x≠3. Where x=3, f(x)=2. We drew the graph for this function and determined that as x approaches 3 for the limit of this function, the y-value equals 6. There was an open circle on the graph at the point (3,6), but there was also a point disconnected from the line at (3,2). This brought up a good point: we should never consider the y-value to be 2, because the LIMIT means we are getting CLOSE to x=3 but never x=3.
The second example was the limit of 3x/x as x approaches 0. We determined that the y-value is 3, and the graph is a horizontal line with an open circle at (0,3).
We also talked about air resistance and velocity in cars, and whether you should use the air conditioner versus the windows when driving on the highway. Mr. Hansen said that the amount of drag when you have the windows open uses the same amount of gas as the air conditioner, so really the only way to conserve fuel would be to sit in your hot car, perspiring.
Then we did a worksheet that Mr. Hansen handed out, and then we got our quizzes from Thursday back.
Today we incorporated a real life situation into our Calculus class by discussing Hurricane Gustav. We thought of ways that calculus would apply when predicting the course of the hurricane, such as instantaneous velocity. Because the hurricane is constantly in motion, and changing, it is crucial to apply things such as forces of the wind, the heat of the water, and previous speed of the storm to predict where it will land, and how big it is.
We then tied together the relationship between instantaneous velocity and limits. Instantaneous velocity cannot be calculated normally because it can never equal a certain value. As it gets closer and closer to that value, there is either an asymptote or a hole in the graph. We can only understand instantaneous velocity through limits.
We then did a couple examples using the Algebra of Limits. We learned how to evaluate a 0/0 limit by factoring or by using conjugate radicals.
The theorem in the book was review in class: if f(x) = g(x) for all all values except x ≠a This is possible for the two functions to have the same limit because we don't consider "a", we consider everything except for "a."
Example 1:
First we discovered if you substitute the number, and you get 0/0 then it is indeterminate. If it is possible to factor the numerator or denominator, do so and see if anything simplifies.
Example 2:
lim √(x+3) - √(3) / x x->0
First we sub in 0 for x and you get 0/0. This is indeterminate so you must multiply by the conjugate radical. This is the value of the numerator, but the signs are opposite. You must multiply this value to the numerator and denominator because we don't want to mess up the function. Technically all it equals is 1, so it will not alter the function. In this case you would multiply by √(x+3) + √(3) / √(x+3) + √(3). The math ends up easy because of the opposite addition and subtraction signs. The numerator eventually simplifies to x. As for the denominator, it is smart not to multiply it out because normally something from the numerator and denominator will cancel out. This limit ends up equaling 1/ (2*√(3))
We next talked about sing the Laws of limits. It would probably too complication to type out, so refer to the book for specifics. So far we know that if the limit has an equation that is cubed, find the limit first and then cube the limit. Next we learned that the limit of a product is equal to the product of each limit.
Finally, we finished off class with a worksheet that applied these laws, and also used algebra of limits, but we did not finish all of it!
On Wednesday, Mr. Hansen reminded us that we were having a quiz friday on a continuation of limits. He then proceded to ask anyone if they had any questions on the homework. The most confusing problem that day was problem 19 on page 115. the limit as x approaches seven of the square root of x plus two MINUS three all over x minus seven. Vishnu walked up to the board after Mr. Hansen told him to do the problem, where he proceded to solve the problem using conjuga te binomial (I think that's what it is called). after that we found that the limit was one sixth. After the homework was over, we were given a worksheet that focused on limits, especially the squeeze method on the front side, and on the back, evaluating limits for normal old functions and piece-wise functions. This was the day we learned that we should put a little positive sign on the zero if its positive (and vice-versa) to show if the function is going to not be undefined, just being divided by some very small positive or negative number.
Today in class we started by answering troubling questions on the homework. We deemed homework questions number 7 and 26 the hardest. For number 7, we found that many different graphs could possibly work as long as they followed the given asymptotes. For number 26 we found that splitting up the function into two different limit equations doesn't always give us accurate answers. We decided instead to find what the bound of sin pi/x is and multiply that by the sqrt of(x^3 +x^2) to find the bound of the entire function. We then replaced 0 for x (x was approaching 0) and determined the function had to be between 0 and 0, thus leaving the only possible answer for the limit as x approaches 0 to be 0.
For the remainder of class we learned how to "use the delta process to find a slope formula that works at every point on a given function." On the graph we came up with our fixed point of "x" and "the function of x." We determined a nearby moving point: (x+h, the function of x+h). We then followed the same process of finding the slope and getting rid of the "h" on the bottom. To find the mtan we found the limit as h approaches 0 and this became our derivative function.
Friday, Sept. 12 Yesterday we worked in groups and found the derived function formulas for y=x^3, y=x^4 and y=x^5. Students then conjectured that if N is a positive integer, the derivative formula for f(x) = x^ N will always be f'(x) = N x^(N-1). Today in class we proved this conjecture, working together as a class. The work is also shown on the embedded slideshow. Fixed point (x, x^N) Moving point (x+h, (x+h)^N) msec = [(x+h)^N-x^N]/ h =[x^N + Nx^(N-1)h + terms with h^2, h^3,...,h^N - x^N] / h
the x^N terms subtract out and we then factor out h to get (h/h)[Nx^(N-1)+ terms with h, h^2, ...,h^(N-1)
we evaluate the limit as h-->0 (h/h) = 1 all terms that have h, h^2,...h^(N-1) will approach ZERO We are left with NX^(N-1)= f'(x)
Monday, sept. 15th We worked on test review for half of the class (since this weekend was homecoming). We then discussed some other derivative formulas: (1) dc / dx = 0 if c is a constant. We discussed this geometrically [the function y = c is a horizontal line whose slope is 0 at each point] and algebraically: Fixed point (x,c) moving point (x+h, c) y' = lim (as h->0) of [ c - c] / h = lim (as h->0) of 0 / h =0
(2) We also discussed the derivative of a constant multiple of a power function, f(x) = Kx^n We discussed how the factor of K stretches the graph vertically so that increases all of the slopes by a factor of K, and we also used the delta process and noticed how the K factors out: lim(h->0) [K(x+h)^n - K x^n] / h = lim(h->0) K [{(x+h)^n-x^n}/h] = K lim(h->0) [(x+h)^n - x^n]/h =K f'(x)
Today, we started off with some review, looking back at the motion detector activity that we did earlier in teh year. We looked at a graph (possibly a cubic function) that showed s(t) vs. t, or distance from the detector vs. time. A tangent line was drawn on the graph, which we said indicated the instantaneous velocity at that point. It was negative, which is important because velocity is a vector, meaning that both direction and magnitude matter.
Then we looked at another graph of a cubic function that showed v(t) vs. t, or velocity in ft/s vs. time in seconds. A similar tangent line was drawn on the graph, but because it was a velocity vs. time graph, we said that the tangent indicated accelleration because it was the rate of change in the velocity, or the feet/second/second.
Then we put an equation to the graphs, and tried to determine the acceleration using the equation. We found the derivative of the position vs. time graph to find the equation that would signify velocity, and then took the derivative of that to find acceleraation.
Then we applied this to a different situation, one where a balloon's volume of air was changing. The slope of the tangent line in this case was equal to the instantenous rate at which the air was leaving the balloon. A secant line was also drawn on the graph, and it signified the average rate that air was leaving the balloon at for a given time period. Then we worked on a worksheet about derivatives of polynomials.
Today we opened class by discussing the previous night's homework. Mr. Hansen checked to make sure everyone had done it, and we then went over some of the more difficult problems. After that we began a new discussion of tangent lines, specifically if a parabola could be tangent to another parabola. For this to be possible, the graphs would have to intersect once, and the slopes would have to be the same at that point. Then we did the problem: If f(x)=x^3+Ax+B is tangent to g(x)=Cx+x^2 at the point (3,15) what are the values of A,B and C. We first solved 15=C(3)+(3)^2 to find that C=2 We then tried to solve 15=(3)^3+A(3)+B but we could only get it down to -12=3A+B, which has infinitely many solutions. To solve this, we realized we could take f'(x) and g'(x), seeing they would still equal themselves at x=3. We then derived the two equations and got f'(x)=3x^2+A and g'(x)=C+2x. Seeing f(3)=g(3), f'(3) also = g'(3) so
27+a=c +6 a=c-21 a=2-21 a=-19
Now that we had a, we plugged it in to 15=(3)^3+(-19)(3)+B to get 15=27-57+B 15=-30+ B 45=B
After learning this method, we spent the rest of the class working on a packet Mr. Hansen handed out.
Today in class we talked about how to show displacement and distance. One cannot simply subtract f(a)-f(b) on a function,to find the distance, because often times, if the object went forward and then backward, (for example, we used a ball going up, then down a ramp) the value of f(a)-f(b) will be negetive. To make sure that you don't get a negetive value, you need to put absolute value around the subtraction.
Today's class began with a review of the tougher problems on the homework the night previous.
The lesson proper began with a review of how to find the derivative of a function such as the square root of x, which can be accomplished by considering it to be x^1/2 rather than a square root. The problem that was presented to us was to find, accurately, a point on the square root function without using a calculator. The specific value of x we were to evaluate was 9.16. We took a known point close to it (9,3), and used that point with the derivative function to find the slope of the tangent line. Using the slope and the point, we obtained the point/slope form of the derivative function. Plugging 9.16 into this "linearization" of y(x) at x=9, we determined the square root of 9.16 to be approximately 3.026. In general, the closer the point in question is to the known point, the more accurate the value will be.
Wednesday we took a quiz for the first half of class.
Then we reviewed the definition of linearization, which is when we use the tangent line to estimate values for a non-polynomial function.
We also reviewed what the symbol dy/dx represents, which is the slope of the tangent line and also the derivative of y. dy and dx are called differentials, and dy = the change in y along the tangent line, while dx = the change in x along the tangent line.
On the graph of a curve, the difference between the x-value a and (a+h) is h, which also equals ∆x or dx. For ∆y, the distance does not equal dy, which determines the difference between the slope of the tangent line and the curve of the graph.
Then we used the same example as on Tuesday, estimating √9.16 on the graph of y=√x. We said dy/dx=1/(2√x) and then we multiplied both sides by dx. Then we subbed in 9 for x (because it is closest to 9.16) and got dy=0.1667dx. Then we got dy=0.026 when we made dx=0.16, because that is the difference between 9 and 9.16.
We then applied this linearization principle to A=πr^2. We found that dA=2π(5)(0.02)≈∆A, when the radius is 5 and when we are estimating for 5.02.
Today Mr. Hansen started out by handing back people's quizzes who had missed a day. Then we began working on the packet we were given awhile ago with the graphs. We worked on question 4, which required us to find the equation for the linearization of the function y=x^3-4x-1. Then we would use this to find the xintcercept of the function. so the first thing to do was find y' which was 3x^2-4. Then we substituted 2 into the derivative to get 8, which is the slope of the linearization of the function. Then we plugged in 2 into the initial function to get the y coordinate. Thus, we have a point in the slope. So, we write the equation Y+1=8(X-2) and equal this to 0 to find xintercept. thus we get the x intercept to equal 17/8. After this, we wrote down Newton's Method, which was to find the xintercepts of functions using tangent lines. The first step was to start out with an x intercept estimate. With this we would find Y' or mtan. After that we would find the y value of the initail equation and then write the equation with of the tangent line using the x estimate to find an even better x estimate. After that, we have X2 and use this over again.
Today, we talked a lot about differentiability and continuity. We used a specific piecewise function as an example to highlight an idea that will be explained later in this post. Our task was to make a road that was descending parabolically and the next part of the road (which was increasing linearly) meet smoothly at a specific point. In order to do this, we had to make the piecewise function continuous and differentiable at that specific point. We solved for the two missing variables by combining the equation making them continuous and differentiable.
The main idea of this class was:
If f(x) is differentiable at x = c, then f(x) is also continuous at x = c.
If f(x) is not continuous at x = c, then f(x) is not differentiable at x = c
21 comments:
Today, Mr. Hansen started out by asking the class a few questions about calculus to see what we already knew. These included "What does the word 'calculus' mean?" and "Who uses calculus?" From that we gathered that calculus was invented by Sir Isaac Newton in the mid 1600's. This was made more famous because Halley used calculus to calculate precisely the path of a comet. Today calculus is used by all sorts of mathematicians, physicists, engineers, and scientists. The word literally translates to "pebble."
In class Mr. Hansen also tossed a tennis ball up and caught it, asking us to graph time and the ball's height in feet above the floor. The ball slowly decelerated until it reached the apex of its climb and then accelerated back down. Mr. Hansen caught the ball lower than where he had tossed it from. This was shown in many of our graphs. Next, Mr. Hansen also bounced the ball, asking us to graph this is well. This time, there was a point on the graph at which the ball's height decrease stopped suddenly where the ball hit the floor. The ball then accelerated off the floor but slowly decelerated as it climbed up again and Mr. Hansen caught it.
We started off today's class with a bit of review from the last class as well as some history that Mr. Hansen forgot from yesterday's class. Archimedes was an ancient greek who made lots of calculus-related discoveries, which were only repeated when Newton made incredible calculus contributions about 1500 to 2000 years later.
We spent the rest of the class talking about the activity from the day before and the concepts behind it. The activity was about trying to calculate the exact velocity at a specific moment in time of a falling paper. The class decided that we cannot find the velocity at a specific point in time, because it is just a point. We need two points to calculate velocity -- velocity measures change, if there is no change, there is no velocity. Mr. Hansen also explained that average velocity is equal to distance travelled divided by elapsed time, or delta s (distance) divided by delta t (time). On a graph, the average velocity is represented by m sec, or the slope of the secant line of the interval (a secant line is a line connects points in a specific time interval), so long as it is a time v. distance graph. We can mathematically show that we cannot calculate velocity at a specific point because velocity is equal to delta s/delta t -- distance travelled/elapsed time. If no distance was travelled and no time elapsed, as would be the case at a specific point, the expression would be 0/0, which is undefined.
However, we found that we can get close to the velocity at a certain instant by reducing the time interval with the instant included. With the points given, the average velocity of the interval [9/60, 10/60] would be the most accurate. Despite the fact that by keeping the time interval small, the accuracy increases, that would be a more accurate average velocity than that of the interval [9/60, 10/60]. This is because in the former interval, there is a part that has a velocity less than the velocity at 10/60 and a part that has a greater velocity, thus putting the average somewhere close to that of 10/60, whereas in the latter, 10/60 is at the end of the range of the interval, resulting in a less accurate velocity.
As the time elapsed approaches zero, we can get a more and more accurate calculation of the velocity at an instant.
At the end of class, we related the two worlds in which we had been working for this activity: the world with physical objects and the world with mathematical abstractions (graphs). Average velocity in the physical world is essentially the same as m sec (slope of the secant line, as long as the graph is time v. distance). Also, in the physical world, as time elapsed approaches zero, we get closer to the velocity of the instant. Similarly, in the graph world, as the change in t approaches zero, we get closer to m tan (slope of the tangent line, a line that touches the graph only at one point, so that its slope would be equal to about that of the velocity at that instant.
Monday August 25th
Today in class we focused on using a second method to find Instantaneous Velocity. We started by briefly looking in to the difference between speed and velocity, speed being the Magnitude of Velocity (Total distance/Elapsed Time), but Mr. Hansen said we would wait a little before delving into this further...
As I alluded to earlier, the majority of the class time was devoted to learning another way to find Vinst (Instantaneous Velocity). While we already know the "chart method" it is far less precise and much more tedious than the so-called Delta Process, an algebraic method for finding Mtan (Vinst). The Delta Process relays on finding a fixed point near the tangent line on a graph, and then using limits to move the point infinitesimally close to the fixed point, generating a much more precise Tangant line. A key concept for this process is "h", a very small quantity that is added to the fixed point's x-value, which gets closer and closer to the fixed point. In class we used the example of y=4-x^2 with the fixed point (1,3) to learn this new process. The first step is to create a second point that is close to the fixed point and generate a secant line. This is where "h" comes in, as we used the point (1+h, 4 - [1+h]^2) to generate the secant line. The X-coordinate is the same as the fixed point's X, with "h" added. This x-coordinate (1+h in this case) is then plugged in to the original equation (y=4-x^2) to find what the Y-coordinate would be at x=1+h. After finding both coordinates, we were able to find the slope of the secant line between the fixed point and our second point. (Slope = change in y/change in x). After some plain old algebraic simplifying, we calculated Msec (slope of secant line) to be { (h[-2-h])/h }. To find Mtan, we simply moved the second point closer to the fixed point. By doing this, we move "h" closer to 0 seeing as "h" is the coordinate difference for the two points. In the end, we generated the following equation :
Mtan = Limit (h -> 0) [h/h(-2-h)] = -2
Today in class we talked about tangent lines in geometry and calculus. "In geometry, a tangent line is a line that intersects a circle at exactly one point."
Calculus is a bit different...
This is because in Calculus, the tangent sometimes goes through more than one point.
In Calculus, a tangent line to point P is "the line through P whose sl0pe equals the limit of the slopes of secant PQ as Q approaches P along the graph. Basically, its the delta process!!
m=lim(h->0) {f(x+h)-f(x)/ h}
We then practiced the delta process on a worksheet, finding the slopes of tangents, and also the slopes of perpendicular lines to the tangents. The back of the paper was review, finding the average velocity from a chart and asking tough questions. Mr. Hansen hinted that the tough question (#5) will be close to something for the quiz on Thursday. So watch out!
Students were given a position - time graph, and they then had to walk
in the motion detector in a way that would create the prescribed position time graph. Students had to think carefully about acceleration and its impact on the position time graph.
We then discussed the history of calculus; specifically, we discussed the fact that Newton was unable to clearly define the concept of "limit". The students started working on an investigation of how different functions behave near points where they are undefined.
Today's class started off with a quiz. After said quiz, we discussed the various limits of the functions on the limit worksheet. The central point of the discussion was whether we could be sure of what the y-value would be on certain graphs as they approached the point at which they were undefined.
In the end, we determined that we could only be truly sure on the first graph (it would be very close to 5). In the second graph, though it is easy to determine the y-value of any x, it is impossible to make an accurate guess of a number close to 2 as we do not know which side the limit is approaching from. For the third graph, the only thing we can say with certainty is that y will become very, very large as it approaches the limit. No limit was found for the final graph.
addition to 8-28-08 comment
We also discussed three different reasons that the limit of f(x), as x approaches c, might fail to exist:
(i) a jump
(b) f(x) growing without bound (a vertical asymptote)
(c) infinitely rapid oscillation of the f(x) values
Friday in class we had a reading quiz over Thursday night's homework. Then we went over the answers.
We found another way to explain the limit of a function as x approaches a. We decided that we look at the y-coordinates for x-values CLOSE to a, but we ignore x=a. Then we looked at two examples.
The first example was f(x)= (x^2-9) / (x-3), where x≠3. Where x=3, f(x)=2. We drew the graph for this function and determined that as x approaches 3 for the limit of this function, the y-value equals 6. There was an open circle on the graph at the point (3,6), but there was also a point disconnected from the line at (3,2). This brought up a good point: we should never consider the y-value to be 2, because the LIMIT means we are getting CLOSE to x=3 but never x=3.
The second example was the limit of 3x/x as x approaches 0. We determined that the y-value is 3, and the graph is a horizontal line with an open circle at (0,3).
We also talked about air resistance and velocity in cars, and whether you should use the air conditioner versus the windows when driving on the highway. Mr. Hansen said that the amount of drag when you have the windows open uses the same amount of gas as the air conditioner, so really the only way to conserve fuel would be to sit in your hot car, perspiring.
Then we did a worksheet that Mr. Hansen handed out, and then we got our quizzes from Thursday back.
Today we incorporated a real life situation into our Calculus class by discussing Hurricane Gustav. We thought of ways that calculus would apply when predicting the course of the hurricane, such as instantaneous velocity. Because the hurricane is constantly in motion, and changing, it is crucial to apply things such as forces of the wind, the heat of the water, and previous speed of the storm to predict where it will land, and how big it is.
We then tied together the relationship between instantaneous velocity and limits. Instantaneous velocity cannot be calculated normally because it can never equal a certain value. As it gets closer and closer to that value, there is either an asymptote or a hole in the graph. We can only understand instantaneous velocity through limits.
We then did a couple examples using the Algebra of Limits. We learned how to evaluate a 0/0 limit by factoring or by using conjugate radicals.
The theorem in the book was review in class:
if f(x) = g(x) for all all values except x ≠a
This is possible for the two functions to have the same limit because we don't consider "a", we consider everything except for "a."
Example 1:
First we discovered if you substitute the number, and you get 0/0 then it is indeterminate. If it is possible to factor the numerator or denominator, do so and see if anything simplifies.
Example 2:
lim √(x+3) - √(3) / x
x->0
First we sub in 0 for x and you get 0/0. This is indeterminate so you must multiply by the conjugate radical. This is the value of the numerator, but the signs are opposite. You must multiply this value to the numerator and denominator because we don't want to mess up the function. Technically all it equals is 1, so it will not alter the function. In this case you would multiply by √(x+3) + √(3) / √(x+3) + √(3). The math ends up easy because of the opposite addition and subtraction signs. The numerator eventually simplifies to x. As for the denominator, it is smart not to multiply it out because normally something from the numerator and denominator will cancel out. This limit ends up equaling 1/ (2*√(3))
We next talked about sing the Laws of limits. It would probably too complication to type out, so refer to the book for specifics. So far we know that if the limit has an equation that is cubed, find the limit first and then cube the limit. Next we learned that the limit of a product is equal to the product of each limit.
Finally, we finished off class with a worksheet that applied these laws, and also used algebra of limits, but we did not finish all of it!
On Wednesday, Mr. Hansen reminded us that we were having a quiz friday on a continuation of limits. He then proceded to ask anyone if they had any questions on the homework. The most confusing problem that day was problem 19 on page 115. the limit as x approaches seven of the square root of x plus two MINUS three all over x minus seven. Vishnu walked up to the board after Mr. Hansen told him to do the problem, where he proceded to solve the problem using conjuga te binomial (I think that's what it is called). after that we found that the limit was one sixth. After the homework was over, we were given a worksheet that focused on limits, especially the squeeze method on the front side, and on the back, evaluating limits for normal old functions and piece-wise functions. This was the day we learned that we should put a little positive sign on the zero if its positive (and vice-versa) to show if the function is going to not be undefined, just being divided by some very small positive or negative number.
Today in class we started by answering troubling questions on the homework. We deemed homework questions number 7 and 26 the hardest. For number 7, we found that many different graphs could possibly work as long as they followed the given asymptotes. For number 26 we found that splitting up the function into two different limit equations doesn't always give us accurate answers. We decided instead to find what the bound of sin pi/x is and multiply that by the sqrt of(x^3 +x^2) to find the bound of the entire function. We then replaced 0 for x (x was approaching 0) and determined the function had to be between 0 and 0, thus leaving the only possible answer for the limit as x approaches 0 to be 0.
For the remainder of class we learned how to "use the delta process to find a slope formula that works at every point on a given function." On the graph we came up with our fixed point of "x" and "the function of x." We determined a nearby moving point: (x+h, the function of x+h). We then followed the same process of finding the slope and getting rid of the "h" on the bottom. To find the mtan we found the limit as h approaches 0 and this became our derivative function.
Friday, Sept. 12
Yesterday we worked in groups and found the derived function formulas for y=x^3, y=x^4 and y=x^5. Students then conjectured that if N is a positive integer, the derivative formula for f(x) = x^ N will always be f'(x) = N x^(N-1).
Today in class we proved this conjecture, working together as a class. The work is also shown on the embedded slideshow.
Fixed point (x, x^N)
Moving point (x+h, (x+h)^N)
msec = [(x+h)^N-x^N]/ h
=[x^N + Nx^(N-1)h + terms with h^2, h^3,...,h^N - x^N] / h
the x^N terms subtract out and we then factor out h to get
(h/h)[Nx^(N-1)+ terms with h, h^2, ...,h^(N-1)
we evaluate the limit as h-->0
(h/h) = 1
all terms that have h, h^2,...h^(N-1) will approach ZERO
We are left with
NX^(N-1)= f'(x)
Monday, sept. 15th
We worked on test review for half of the class (since this weekend was homecoming). We then discussed some other derivative formulas:
(1) dc / dx = 0 if c is a constant.
We discussed this geometrically [the function y = c is a horizontal line whose slope is 0 at each point] and algebraically: Fixed point (x,c)
moving point (x+h, c)
y' = lim (as h->0) of [ c - c] / h
= lim (as h->0) of 0 / h
=0
(2) We also discussed the derivative of a constant multiple of a power function, f(x) = Kx^n
We discussed how the factor of K
stretches the graph vertically so that increases all of the slopes by a factor of K, and we also used the delta process and noticed how the K factors out:
lim(h->0) [K(x+h)^n - K x^n] / h
= lim(h->0) K [{(x+h)^n-x^n}/h]
= K lim(h->0) [(x+h)^n - x^n]/h
=K f'(x)
Thursday, September 18
Today, we started off with some review, looking back at the motion detector activity that we did earlier in teh year. We looked at a graph (possibly a cubic function) that showed s(t) vs. t, or distance from the detector vs. time. A tangent line was drawn on the graph, which we said indicated the instantaneous velocity at that point. It was negative, which is important because velocity is a vector, meaning that both direction and magnitude matter.
Then we looked at another graph of a cubic function that showed v(t) vs. t, or velocity in ft/s vs. time in seconds. A similar tangent line was drawn on the graph, but because it was a velocity vs. time graph, we said that the tangent indicated accelleration because it was the rate of change in the velocity, or the feet/second/second.
Then we put an equation to the graphs, and tried to determine the acceleration using the equation. We found the derivative of the position vs. time graph to find the equation that would signify velocity, and then took the derivative of that to find acceleraation.
Then we applied this to a different situation, one where a balloon's volume of air was changing. The slope of the tangent line in this case was equal to the instantenous rate at which the air was leaving the balloon. A secant line was also drawn on the graph, and it signified the average rate that air was leaving the balloon at for a given time period. Then we worked on a worksheet about derivatives of polynomials.
Today we opened class by discussing the previous night's homework. Mr. Hansen checked to make sure everyone had done it, and we then went over some of the more difficult problems. After that we began a new discussion of tangent lines, specifically if a parabola could be tangent to another parabola. For this to be possible, the graphs would have to intersect once, and the slopes would have to be the same at that point. Then we did the problem:
If f(x)=x^3+Ax+B is tangent to
g(x)=Cx+x^2
at the point (3,15) what are the values of A,B and C.
We first solved 15=C(3)+(3)^2 to find that C=2
We then tried to solve 15=(3)^3+A(3)+B but we could only get it down to -12=3A+B, which has infinitely many solutions. To solve this, we realized we could take f'(x) and g'(x), seeing they would still equal themselves at x=3. We then derived the two equations and got f'(x)=3x^2+A and g'(x)=C+2x. Seeing f(3)=g(3), f'(3) also = g'(3) so
27+a=c +6
a=c-21
a=2-21
a=-19
Now that we had a, we plugged it in to
15=(3)^3+(-19)(3)+B to get
15=27-57+B
15=-30+ B
45=B
After learning this method, we spent the rest of the class working on a packet Mr. Hansen handed out.
Today in class we talked about how to show displacement and distance. One cannot simply subtract f(a)-f(b) on a function,to find the distance, because often times, if the object went forward and then backward, (for example, we used a ball going up, then down a ramp) the value of f(a)-f(b) will be negetive. To make sure that you don't get a negetive value, you need to put absolute value around the subtraction.
Today's class began with a review of the tougher problems on the homework the night previous.
The lesson proper began with a review of how to find the derivative of a function such as the square root of x, which can be accomplished by considering it to be x^1/2 rather than a square root. The problem that was presented to us was to find, accurately, a point on the square root function without using a calculator. The specific value of x we were to evaluate was 9.16. We took a known point close to it (9,3), and used that point with the derivative function to find the slope of the tangent line. Using the slope and the point, we obtained the point/slope form of the derivative function. Plugging 9.16 into this "linearization" of y(x) at x=9, we determined the square root of 9.16 to be approximately 3.026. In general, the closer the point in question is to the known point, the more accurate the value will be.
Wednesday we took a quiz for the first half of class.
Then we reviewed the definition of linearization, which is when we use the tangent line to estimate values for a non-polynomial function.
We also reviewed what the symbol dy/dx represents, which is the slope of the tangent line and also the derivative of y. dy and dx are called differentials, and dy = the change in y along the tangent line, while dx = the change in x along the tangent line.
On the graph of a curve, the difference between the x-value a and (a+h) is h, which also equals ∆x or dx. For ∆y, the distance does not equal dy, which determines the difference between the slope of the tangent line and the curve of the graph.
Then we used the same example as on Tuesday, estimating √9.16 on the graph of y=√x. We said dy/dx=1/(2√x) and then we multiplied both sides by dx. Then we subbed in 9 for x (because it is closest to 9.16) and got dy=0.1667dx. Then we got dy=0.026 when we made dx=0.16, because that is the difference between 9 and 9.16.
We then applied this linearization principle to A=πr^2. We found that dA=2π(5)(0.02)≈∆A, when the radius is 5 and when we are estimating for 5.02.
Today Mr. Hansen started out by handing back people's quizzes who had missed a day. Then we began working on the packet we were given awhile ago with the graphs. We worked on question 4, which required us to find the equation for the linearization of the function y=x^3-4x-1. Then we would use this to find the xintcercept of the function. so the first thing to do was find y' which was 3x^2-4. Then we substituted 2 into the derivative to get 8, which is the slope of the linearization of the function. Then we plugged in 2 into the initial function to get the y coordinate. Thus, we have a point in the slope. So, we write the equation Y+1=8(X-2) and equal this to 0 to find xintercept. thus we get the x intercept to equal 17/8. After this, we wrote down Newton's Method, which was to find the xintercepts of functions using tangent lines. The first step was to start out with an x intercept estimate. With this we would find Y' or mtan. After that we would find the y value of the initail equation and then write the equation with of the tangent line using the x estimate to find an even better x estimate. After that, we have X2 and use this over again.
Wednesday, October 22, 2008
Today, we talked a lot about differentiability and continuity. We used a specific piecewise function as an example to highlight an idea that will be explained later in this post. Our task was to make a road that was descending parabolically and the next part of the road (which was increasing linearly) meet smoothly at a specific point. In order to do this, we had to make the piecewise function continuous and differentiable at that specific point. We solved for the two missing variables by combining the equation making them continuous and differentiable.
The main idea of this class was:
If f(x) is differentiable at x = c, then f(x) is also continuous at x = c.
If f(x) is not continuous at x = c, then f(x) is not differentiable at x = c
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