Today was the first day of school. We discussed a few things around the portal. We also discussed when calculus was developed, which was the late 1600s. Newton and Leibaiz invented it to help on physic questions and motions. We also discussed who use it in today's society. Scientists (bio, chem, physics and more), engineers, meteorologists and more use calculus to help them out. The literal meaning of the word calculus in Latin means stone or pebble. We also did experiments with a tennis ball and tried to plot how the graph would look. We discussed this and then got onto the worksheet. We started answering a few questions and learned about average velocity. Average velocity is distance traveled/ elasped time. We did the worksheet and had some left to do for homework. -Chris
Today we went over the Calculus Ball Activity handout and discovered that average velocity increases as time goes on. In addition, we discovered that average velocity, which is equal to the change in distance over the change in time, in the physical world, is equal to the slope of a secant on a distance/time graph in the math world. Also, as time approaches zero, instantaneous velocity can be equated, which is equal to the slope of a tangent. On the handout, we tried out various ways to find instantaneous velocity. On of these ways was to find the average of the average velocities of the time intervals before and after the point you are trying to find.
After finding the problems associated with the graphing and chart methods for finding instantaneous velocity, which is that they are not precise enough, we learned the algebraic method or delta process for finding the instantaneous velocity. In the algebraic method, after picking a point to solve for, you make (x+h, f(x+h)) the other point. After finding the equation for the slope of the secant line, you then solve for the equation as h approaches the limit 0. Ex. Find M tan @ (1,3) for the equation 4-x^2 Find the x and y values for 1+h... (1+h, 4-(1+h)^2) (1+h, 4-h^2-2h-1) (1+h, -h^2-2h+3) Find the secant... ((-h^2-2h+3)-3)/((1+h)-1) (-h^2-2h)/h (h/h)(-h-2) Then solve for the limit as h approaches zero answer is -2 which is also M tan of point (1,3) Afterwards we discussed problem 49 on page 24. To write the equation for the graph you have to write down separate equations for each section of the graph, for instance, in the interval -1 (is less than) x (is less than) 2, the equation is y=x+1, and for the interval 2 (is less than) x (is less than) 4 the equation is y=-(3/2)x+6 On a sidenote, on this blog site, every time you use the (less than) symbol, you may get a message that your HTML cannot be accepted
Today we talked about the calculus definition of a tangent line. The definition for geometry, "a line that intersects a circle in exactly one point" won't work for the circumstances we discussed in class. The circumstances were a an irregular curve that would make the tangent line intersect at more than one point, a linear graph, and a tangent line that did not capture the slope of the curve. So the definition of a tangent line for calculus becomes: "Given a fixed point P on the graph with Q a nearby point also on the graph. the 'tangent line to the graph at P' is defined to be the line through P whose slope equals the limit of the slopes of secants PQ as Q approaches P along the graph." This definition actually is just a verbal version of the equation we read about on p.140 of our textbook. m= lim f(x) - f(a)/x - a x->a Then we reviewed the delta process and more on the sheet that Mr. Hansen passed out to us. --Amanda
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.
On Friday we took a reading quiz and then we went over the answers.
We then talked about how to find f(x) in a different way, in which we look at the x values as they approach a but not when x=a.
I believe the example we used was y=(x^2-9)/(x-3) and x cannot equal 3 and that at x=3, f(x)=2. And as we got x-values closer and closer to 3, the y-values got closer to 6. but when we got to x=3, there was an open circle because that is where the function is undefined, though there still was a point disconnected from the line which was (3,2)
I think the next example was 3x/x and therefore when x=0, the function is undefined, but as the graph approaches 0, the y values are staying consistent at 3. So this function was just a horizontal line with an open circle at (0,3).
Then we started a worksheet in class and got our quizzes back from Thursday and went over them briefly
In class today we went over a theorem, if f(x) = g(x) for all points except where x=a then lim x approaching a for f(x) lim x approaching a g(x) provided they both exist. The way that we proved this was by factoring. For example when lim x->2 (x^2- 8x+15)/ x-5 the top would factor into (x-3)(x-5)/ x-5, then you would plug two in and get 2 as the final answer. We have to do this because we cannot input 5 in the original equation because that would make it undefined, as 0/0 case an indeterminate form. Also, we can use conjugate radicals, 1/1 to get rid of radicals on the top.
In Class today we continued our discussion on “Evaluating limits without a graph.” Example 1: lim┬(x→5^+ )〖(x+2)/(x-5)〗
When we plugged 5 in for x we found that the answer was undefined but NOT indeterminate. As x approaches 5 from the right, the y values are increasing without bound. The y values are positive because x > 5 and so we are dividing by a very small number as the x values get closer and closer to 5. So in this example, the answer is + ∞. On the graph of this function there is a vertical asymptote at x = 3.
The limit of a function will only equal positive or negative infinity when we plug the value that x is approaching into the function and get (any number ≠ 0) / 0.
Today we reviewed our homework questions, and discussed a new topic: Continuity of a Function. Many students had trouble with homework problem #21 on page 115 which was evaluate the limit(unless it does not exist) of lim x-> -4 (1/4 + 1/x)/(4 + x). We multiplied the numerator and the denominator by 4x/4x, the congeget radical. We factored the numerator, but left the denominator unfactored because that made the simplifying process more obvious. We discovered that the lim x-> -4 (1/4 + 1/x)/(4 + x) = -1/16 We also had a handout about Continuity. We defined Continuity as “A function f is continuous at a number c if all three statements below are true: 1) f(c) is defined(c is in the domain of f) 2) lim x->c f(x) exists 3) lim x->c f(x) = f(c)” We also tried to define continuity with our own words and we came up with, unbroken with no vertical asymptotes, puncture points, or jumps. We have our second quiz tomorrow so make sure to STUDY!
On friday, we went over the greatest integer function and the stair step function. The greatest integer functions and the staisrstep functions both assigns each real number to an integer that is equal to or less than the real number. For example, if the x value is 1.1 than the y value is 1. If the x value is 2 than the y value is 2. If the x value is 2.9 than the y value is 2.
On Monday we went over the quiz. We talked about splitting limits and doing one then they other for a compound limit. We talked about #2d and the differences between positive infinity and negative infinity. We talked about #2g and how an absolute value can be a piecewise. We talked about a hurricane and the how each part determines the path and as long as it is continuous it will only be off by a little but if it isn't it could be way off; continuity determines the reliability of a prediction.
Continuity Day 2 Wkst. -determine which are continuous, which are not, and why -Began intermediate value property
is f(c) is continuous at x=c then... 1) f(c) must be defined 2) lim x->c f(x) exists 3) limmm x-> c f(x)=f(c)
intermediate value: all the y-values have to hit a point on the graph, the graph doesn't have to be continuous
if f is continuous it had the intermediate property but not necessarily the other way around
Horizontal asymptotes graphs CAN cross horizontal asymptotes such as in the graph of the point on a guitar string, they cannot cross vertical asymptotes
algebra techniques related to limits as x-> infinity
lim x->infinity 5/x^2 =0 (a constant over a very large number is equal to zero)
ex: lim x->infinity (6+8x-5x^2)/(2x^2-3x+1) you want to look for the highest power of x in the denominator and multiply by "one", in this case since x^2 is the highest value, you would multiply by (1/x^2)/(1/x^2)
you would then get [(6/x^2)+(8/x)-5]/[2-(3/x)+(1/x^2)] this would simplify to (0+0-5)/(2-0+0) simplifying to -5/2 which is the limit as x-> infinity
we then worked on a handout call IVP and lim x->infinty f(x) + review
Last night in homework we found the slope of the tangent line of y=x^2 – 4x using the delta process. Today we learned how to use the delta process to find the equation of the line. So first if we can find the tangent’s slope using a graph and table first that will show how the actual equation to use as a check later in the process. So the graph looks like a regular parabola with x-intercepts at x=0 and x=4. Then we found different values for the slope of the tangent line at different x values. x mtan 1 -2 2 0 3 2 4 4 So from the graph and from these values we found the lines equation to be mtan= 2x – 4. Then, we can use our new process to find this equation using the delta process. So we decide the fixed point would be (x, x^2 – 4x) and the moving point would be ((x+h), (x+h)^2 – 4(x+h)). So to set up the equation it would be msec= [(x+h)^2—4(x+h)] – (x^2—4x) / (x+h) – x So after tons of algebra you would get msec=h/h (2x + h – 4). So then we can say that mtan= lim h/h (2x + h – 4) = 2x – 4 h → 0 The equation that we developed is called a derived formula or a derivative function since we took steps to derive it. In class we did a practice problem so here it is: Find the derivative function for y=x^3 – 9x *answer is 3x^2 – 9
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)
Today we reviewed the shortcuts: if f(x)=x^N then f'(x)=Nx^n-1 if f(x)=Kx^N then f'(x)=KNx^N-1 if f(x)= ax^2+bx^2+c then f'(x)=2ax+b
Mr. Hansen passed out a worksheet that focused on the derivative of functions. We completed the quick review and a few of the discussion questions. Mr. Hansen passed out tests back and we spent the remainder of the class on our test questions.
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)
Friday in class we talked about tangency. First we went over the question that Mr. Hansen posed to us on Thursday’s homework, how can we tell if a particle’s speed is increasing or decreasing? First, we decided that speed is equal to |v(t)| or the magnitude of the velocity. Then we could tell that if the v(t) and a(t) have the same sign (+ or – ) the speed is increasing, and if opposite signs it is decreasing. The book states that as “a particle is being pushed in the same way it is moving , speed is increasing.” Then we need to review two facts: 1. On the graph of f, the value of f’ tells me the slope of the tangent line. 2. If s(t) is a position function, s’(t) tells me the instantaneous velocity. Then, from looking at pictures and deciding which one illustrated “tangency” we must really define what “tangency” means. So we decided that “tangency” means: 1. The lines/curves must intersect 2. The lines/curves must have the same slope where/when they intersect Then after defining this we used this definition and our knowledge of derivatives to solve practice problems on our sheet handed out to us by Mr. Hansen.
Well today we began the day with a review of the homework. Then Mr. Hansen told us not to use a certain word phrasing; the derivative is the tangent line. Which is incorrect because the derivative is the equation of the tangent line. We had a quiz on the reading. The reading was about using tangent lines as estimates to more complicated functions. We used an example of square root of x. We needed to find square root of 9.16 so we found the tangent to a close easy to find out point. We used 9 because its square root is just 3. This meant that the solution would be close to 3 just slightly above. Tomorrow we have a half period quiz so im off to study.
On September 22, we first practiced using the delta process on the function of x^(1/2), to find this value we multiplied the the value [(x+h)^(1/2)-x^(1/2)]/h by the conjugate radical. Afterwards, we practiced finding the point of intersection for tangent lines to a curve, to do this, we found the equations for both tangent lines and then used these equations to find the point of intersection. Finally, we practiced finding the coefficients for curves tangent to each other. For a curve to be tangent to another, they have to share a point, and at that point, their derivatives have to be equal. Mr. Hansen said that sometimes, if we are stuck on a math problem, we should draw a picture.
Quick comment about the phrase "The derivative is the tangent line". The output values of the derivative of function f are equal to the slopes of tangent lines to f; we still do NOT have the equation of the tangent line until we use this slope together with the point of tangency to write an equation for the line tangent to f
On September 30th, 2008 we covered a lot of topics. First we went over homework problems from the night prior. We went over page 325 #3 and page 252 #15. We reviewed the differentials and more. After reviewing the homework, the class continued onto the lab for the day. We were handed a sheet and we started to do the sheet in pairs. On the first side we had a parabola, and we found the tangent line at certain points. We then plotted the ordered pairs (x, mtan) on the axes on the sheet. On the following sheet we had a graph and we had to graph the derivative of it. The line of y prime was under the x-axis only when the slope of y was negative. When the line of y prime was above the x-axis the slope of y was positive. We then tried estimating this on the next page and continued to test this theorem.
On October 1st, we reviewed how the values of the derivitive of a function are equal to the slope of the function. Also, we stated these three fact; first, when the graph of the function is becoming larger and has a positive slope, the derivitive of the function is above the x-axis. Second, when the graph of the function is decreasing and has a negative slope, the graph of the derivative is below the x-axis. Lastly, when the slope of the function is 0, the graph of the derivitive hits the x-axis. We also talked about what a terrace point is. A terrace point is a point on the graph where the funciton lingers near the x-axis before and after it hits the axis. An example of this occurs at the origin of the graph of y=x^3.
On Monday October 3rd, We worket on two worksheets. One of the worksheets helped us review the grapghs of functions and their derivatives in 2 ways. We had to sketch the graph of the derivative from looking at graph of the original function. We also had to look at a graph which had an original function, it's first derivative, and it's second derivative, and then we had to identify which is which. Also, on this worksheet, we had to use the equations for osition, velocity and acceleration. The original equation is position, the first derivative is velocity, and the second derivative is acceleration. On the other worksheet, we worked on finding anti-derivatives. To find an Anti-Derivative, you add 1 to the exponent of the derivative and divide the coefficient by the new exponent.
Today we reviewed: the values of f' equal the slope on f today we graphed f from the f' graph Every time the f' graph hits the x-axis there is either a relative maximum, relative minimum or a terrace point Also, if the graph of f' is below the x-axis, then the slopes on f is decreasing, and if the points on f' are above the x-axis, then the slopes on f is increasing Mr. Hansen handed out a worksheet and we did three practice problems
Well, today we established that our goal for the next part of this unit is to figure out the derivatives for sin and cos. So, to begin Mr. Hansen passed us out a piece of paper, which we used as our notes today. We reviewed trig ideas and indeterminate limits first. Then we used a graph of what looks like sin x and drew its derivative. From drawing this graph we could estimate that the derivative of sin x seems to be cos x. But we cannot determine this for sure yet. Then we used our calculator to estimate the limit as x approaches 0 of sin x divided by x. This came out to equal one. Then to end we used a unit circle and the questions on Mr. Hansen’s sheet to prove that this limit has to equal one. So beginning here we start to discover the derivatives of trigonometric functions.
Today Oct. 21, we went over the reasons why some functions are not differentiable. There are three ways, when the graph has a corner/clasp, when there is a vertical asymptote or when the graph/function is not continuous. We did a couple examples to prove this. Then we worked on 3 and 4 from the worksheet and saw how parabolas, quadratics, cubics, etc. when zoomed in upon enough you will see that it becomes a line. But when you do this for a graph/function without a differentiable and there is a corner then zooming further and further in does nothing. We then talked about statements, their converse and the contra-positive for the statement. When one of the first two is true then the contra-postive must also be true. Then we went of a homework problem to close class.
Today we had our quiz over differentiability and then we started talking about concave and inflection points. Concave up is when the tangent lines are below the curve and concave down is when the tangent lines are above the curve. We then found out how to find the inflection point, which is by finding the derivative of the original function and when that function reaches a relative minimum or maximum, and then to find that exact point, you have to set the second derivative equal to zero. The x-coordinate you get from that is the inflection point.
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Today was the first day of school. We discussed a few things around the portal. We also discussed when calculus was developed, which was the late 1600s. Newton and Leibaiz invented it to help on physic questions and motions. We also discussed who use it in today's society. Scientists (bio, chem, physics and more), engineers, meteorologists and more use calculus to help them out. The literal meaning of the word calculus in Latin means stone or pebble. We also did experiments with a tennis ball and tried to plot how the graph would look. We discussed this and then got onto the worksheet. We started answering a few questions and learned about average velocity. Average velocity is distance traveled/ elasped time. We did the worksheet and had some left to do for homework.
-Chris
Today we went over the Calculus Ball Activity handout and discovered that average velocity increases as time goes on. In addition, we discovered that average velocity, which is equal to the change in distance over the change in time, in the physical world, is equal to the slope of a secant on a distance/time graph in the math world. Also, as time approaches zero, instantaneous velocity can be equated, which is equal to the slope of a tangent. On the handout, we tried out various ways to find instantaneous velocity. On of these ways was to find the average of the average velocities of the time intervals before and after the point you are trying to find.
After finding the problems associated with the graphing and chart methods for finding instantaneous velocity, which is that they are not precise enough, we learned the algebraic method or delta process for finding the instantaneous velocity. In the algebraic method, after picking a point to solve for, you make (x+h, f(x+h)) the other point. After finding the equation for the slope of the secant line, you then solve for the equation as h approaches the limit 0.
Ex. Find M tan @ (1,3) for the equation 4-x^2
Find the x and y values for 1+h...
(1+h, 4-(1+h)^2)
(1+h, 4-h^2-2h-1)
(1+h, -h^2-2h+3)
Find the secant...
((-h^2-2h+3)-3)/((1+h)-1)
(-h^2-2h)/h
(h/h)(-h-2)
Then solve for the limit as h approaches zero
answer is -2 which is also M tan of point (1,3)
Afterwards we discussed problem 49 on page 24.
To write the equation for the graph you have to write down separate equations for each section of the graph, for instance, in the interval -1 (is less than) x (is less than) 2, the equation is y=x+1, and for the interval 2 (is less than) x (is less than) 4 the equation is y=-(3/2)x+6
On a sidenote, on this blog site, every time you use the (less than) symbol, you may get a message that your HTML cannot be accepted
Today we talked about the calculus definition of a tangent line. The definition for geometry, "a line that intersects a circle in exactly one point" won't work for the circumstances we discussed in class. The circumstances were a an irregular curve that would make the tangent line intersect at more than one point, a linear graph, and a tangent line that did not capture the slope of the curve. So the definition of a tangent line for calculus becomes: "Given a fixed point P on the graph with Q a nearby point also on the graph. the 'tangent line to the graph at P' is defined to be the line through P whose slope equals the limit of the slopes of secants PQ as Q approaches P along the graph." This definition actually is just a verbal version of the equation we read about on p.140 of our textbook. m= lim f(x) - f(a)/x - a
x->a
Then we reviewed the delta process and more on the sheet that Mr. Hansen passed out to us.
--Amanda
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.
On Friday we took a reading quiz and then we went over the answers.
We then talked about how to find f(x) in a different way, in which we look at the x values as they approach a but not when x=a.
I believe the example we used was y=(x^2-9)/(x-3) and x cannot equal 3 and that at x=3, f(x)=2. And as we got x-values closer and closer to 3, the y-values got closer to 6. but when we got to x=3, there was an open circle because that is where the function is undefined, though there still was a point disconnected from the line which was (3,2)
I think the next example was 3x/x and therefore when x=0, the function is undefined, but as the graph approaches 0, the y values are staying consistent at 3. So this function was just a horizontal line with an open circle at (0,3).
Then we started a worksheet in class and got our quizzes back from Thursday and went over them briefly
In class today we went over a theorem, if f(x) = g(x) for all points except where x=a then lim x approaching a for f(x) lim x approaching a g(x) provided they both exist. The way that we proved this was by factoring. For example when lim x->2 (x^2- 8x+15)/ x-5 the top would factor into (x-3)(x-5)/ x-5, then you would plug two in and get 2 as the final answer. We have to do this because we cannot input 5 in the original equation because that would make it undefined, as 0/0 case an indeterminate form. Also, we can use conjugate radicals, 1/1 to get rid of radicals on the top.
In Class today we continued our discussion on “Evaluating limits without a graph.”
Example 1:
lim┬(x→5^+ )〖(x+2)/(x-5)〗
When we plugged 5 in for x we found that the answer was undefined but NOT indeterminate. As x approaches 5 from the right, the y values are increasing without bound. The y values are positive because x > 5 and so we are dividing by a very small number as the x values get closer and closer to 5. So in this example, the answer is + ∞. On the graph of this function there is a vertical asymptote at x = 3.
The limit of a function will only equal positive or negative infinity when we plug the value that x is approaching into the function and get (any number ≠ 0) / 0.
Today we reviewed our homework questions, and discussed a new topic: Continuity of a Function. Many students had trouble with homework problem #21 on page 115 which was evaluate the limit(unless it does not exist) of
lim x-> -4 (1/4 + 1/x)/(4 + x).
We multiplied the numerator and the denominator by 4x/4x, the congeget radical. We factored the numerator, but left the denominator unfactored because that made the simplifying process more obvious. We discovered that the
lim x-> -4 (1/4 + 1/x)/(4 + x) = -1/16
We also had a handout about Continuity. We defined Continuity as “A function f is continuous at a number c if all three statements below are true:
1) f(c) is defined(c is in the domain of f)
2) lim x->c f(x) exists
3) lim x->c f(x) = f(c)”
We also tried to define continuity with our own words and we came up with, unbroken with no vertical asymptotes, puncture points, or jumps. We have our second quiz tomorrow so make sure to STUDY!
On friday, we went over the greatest integer function and the stair step function. The greatest integer functions and the staisrstep functions both assigns each real number to an integer that is equal to or less than the real number. For example, if the x value is 1.1 than the y value is 1. If the x value is 2 than the y value is 2. If the x value is 2.9 than the y value is 2.
On Monday we went over the quiz. We talked about splitting limits and doing one then they other for a compound limit. We talked about #2d and the differences between positive infinity and negative infinity. We talked about #2g and how an absolute value can be a piecewise. We talked about a hurricane and the how each part determines the path and as long as it is continuous it will only be off by a little but if it isn't it could be way off; continuity determines the reliability of a prediction.
Continuity Day 2 Wkst.
-determine which are continuous, which are not, and why
-Began intermediate value property
-Charles
9/9/08
is f(c) is continuous at x=c then...
1) f(c) must be defined
2) lim x->c f(x) exists
3) limmm x-> c f(x)=f(c)
intermediate value: all the y-values have to hit a point on the graph, the graph doesn't have to be continuous
if f is continuous it had the intermediate property but not necessarily the other way around
Horizontal asymptotes
graphs CAN cross horizontal asymptotes such as in the graph of the point on a guitar string, they cannot cross vertical asymptotes
algebra techniques related to limits as x-> infinity
lim x->infinity 5/x^2 =0 (a constant over a very large number is equal to zero)
ex: lim x->infinity (6+8x-5x^2)/(2x^2-3x+1)
you want to look for the highest power of x in the denominator and multiply by "one", in this case since x^2 is the highest value, you would multiply by (1/x^2)/(1/x^2)
you would then get [(6/x^2)+(8/x)-5]/[2-(3/x)+(1/x^2)] this would simplify to (0+0-5)/(2-0+0) simplifying to -5/2 which is the limit as x-> infinity
we then worked on a handout call IVP and lim x->infinty f(x) + review
Last night in homework we found the slope of the tangent line of y=x^2 – 4x using the delta process. Today we learned how to use the delta process to find the equation of the line. So first if we can find the tangent’s slope using a graph and table first that will show how the actual equation to use as a check later in the process. So the graph looks like a regular parabola with x-intercepts at x=0 and x=4. Then we found different values for the slope of the tangent line at different x values.
x mtan
1 -2
2 0
3 2
4 4
So from the graph and from these values we found the lines equation to be mtan= 2x – 4. Then, we can use our new process to find this equation using the delta process. So we decide the fixed point would be (x, x^2 – 4x) and the moving point would be ((x+h),
(x+h)^2 – 4(x+h)). So to set up the equation it would be msec= [(x+h)^2—4(x+h)] – (x^2—4x) / (x+h) – x So after tons of algebra you would get msec=h/h (2x + h – 4). So then we can say that mtan= lim h/h (2x + h – 4) = 2x – 4 h → 0
The equation that we developed is called a derived formula or a derivative function since we took steps to derive it. In class we did a practice problem so here it is:
Find the derivative function for y=x^3 – 9x
*answer is 3x^2 – 9
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)
9/17/08
Today we reviewed the shortcuts:
if f(x)=x^N then f'(x)=Nx^n-1
if f(x)=Kx^N then f'(x)=KNx^N-1
if f(x)= ax^2+bx^2+c then f'(x)=2ax+b
Mr. Hansen passed out a worksheet that focused on the derivative of functions. We completed the quick review and a few of the discussion questions.
Mr. Hansen passed out tests back and we spent the remainder of the class on our test questions.
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)
Friday in class we talked about tangency. First we went over the question that Mr. Hansen posed to us on Thursday’s homework, how can we tell if a particle’s speed is increasing or decreasing? First, we decided that speed is equal to |v(t)| or the magnitude of the velocity. Then we could tell that if the v(t) and a(t) have the same sign (+ or – ) the speed is increasing, and if opposite signs it is decreasing. The book states that as “a particle is being pushed in the same way it is moving , speed is increasing.” Then we need to review two facts:
1. On the graph of f, the value of f’ tells me the slope of the tangent line.
2. If s(t) is a position function, s’(t) tells me the instantaneous velocity.
Then, from looking at pictures and deciding which one illustrated “tangency” we must really define what “tangency” means. So we decided that “tangency” means:
1. The lines/curves must intersect
2. The lines/curves must have the same slope where/when they intersect
Then after defining this we used this definition and our knowledge of derivatives to solve practice problems on our sheet handed out to us by Mr. Hansen.
Well today we began the day with a review of the homework. Then Mr. Hansen told us not to use a certain word phrasing; the derivative is the tangent line. Which is incorrect because the derivative is the equation of the tangent line. We had a quiz on the reading. The reading was about using tangent lines as estimates to more complicated functions. We used an example of square root of x. We needed to find square root of 9.16 so we found the tangent to a close easy to find out point. We used 9 because its square root is just 3. This meant that the solution would be close to 3 just slightly above. Tomorrow we have a half period quiz so im off to study.
On September 22, we first practiced using the delta process on the function of x^(1/2), to find this value we multiplied the the value [(x+h)^(1/2)-x^(1/2)]/h by the conjugate radical.
Afterwards, we practiced finding the point of intersection for tangent lines to a curve, to do this, we found the equations for both tangent lines and then used these equations to find the point of intersection.
Finally, we practiced finding the coefficients for curves tangent to each other. For a curve to be tangent to another, they have to share a point, and at that point, their derivatives have to be equal.
Mr. Hansen said that sometimes, if we are stuck on a math problem, we should draw a picture.
Quick comment about the phrase "The derivative is the tangent line".
The output values of the derivative of function f are equal to the slopes of tangent lines to f; we still do NOT have the equation of the tangent line until we use this slope together with the point of tangency to write an equation for the line tangent to f
On September 30th, 2008 we covered a lot of topics. First we went over homework problems from the night prior. We went over page 325 #3 and page 252 #15. We reviewed the differentials and more. After reviewing the homework, the class continued onto the lab for the day. We were handed a sheet and we started to do the sheet in pairs. On the first side we had a parabola, and we found the tangent line at certain points. We then plotted the ordered pairs (x, mtan) on the axes on the sheet. On the following sheet we had a graph and we had to graph the derivative of it. The line of y prime was under the x-axis only when the slope of y was negative. When the line of y prime was above the x-axis the slope of y was positive. We then tried estimating this on the next page and continued to test this theorem.
On October 1st, we reviewed how the values of the derivitive of a function are equal to the slope of the function. Also, we stated these three fact; first, when the graph of the function is becoming larger and has a positive slope, the derivitive of the function is above the x-axis. Second, when the graph of the function is decreasing and has a negative slope, the graph of the derivative is below the x-axis. Lastly, when the slope of the function is 0, the graph of the derivitive hits the x-axis. We also talked about what a terrace point is. A terrace point is a point on the graph where the funciton lingers near the x-axis before and after it hits the axis. An example of this occurs at the origin of the graph of y=x^3.
On Monday October 3rd, We worket on two worksheets. One of the worksheets helped us review the grapghs of functions and their derivatives in 2 ways. We had to sketch the graph of the derivative from looking at graph of the original function. We also had to look at a graph which had an original function, it's first derivative, and it's second derivative, and then we had to identify which is which. Also, on this worksheet, we had to use the equations for osition, velocity and acceleration. The original equation is position, the first derivative is velocity, and the second derivative is acceleration. On the other worksheet, we worked on finding anti-derivatives. To find an Anti-Derivative, you add 1 to the exponent of the derivative and divide the coefficient by the new exponent.
Today we reviewed:
the values of f' equal the slope on f
today we graphed f from the f' graph
Every time the f' graph hits the x-axis there is either a relative maximum, relative minimum or a terrace point
Also, if the graph of f' is below the x-axis, then the slopes on f is decreasing, and if the points on f' are above the x-axis, then the slopes on f is increasing
Mr. Hansen handed out a worksheet and we did three practice problems
Well, today we established that our goal for the next part of this unit is to figure out the derivatives for sin and cos. So, to begin Mr. Hansen passed us out a piece of paper, which we used as our notes today. We reviewed trig ideas and indeterminate limits first. Then we used a graph of what looks like sin x and drew its derivative. From drawing this graph we could estimate that the derivative of sin x seems to be cos x. But we cannot determine this for sure yet. Then we used our calculator to estimate the limit as x approaches 0 of sin x divided by x. This came out to equal one. Then to end we used a unit circle and the questions on Mr. Hansen’s sheet to prove that this limit has to equal one. So beginning here we start to discover the derivatives of trigonometric functions.
Today Oct. 21, we went over the reasons why some functions are not differentiable. There are three ways, when the graph has a corner/clasp, when there is a vertical asymptote or when the graph/function is not continuous. We did a couple examples to prove this. Then we worked on 3 and 4 from the worksheet and saw how parabolas, quadratics, cubics, etc. when zoomed in upon enough you will see that it becomes a line. But when you do this for a graph/function without a differentiable and there is a corner then zooming further and further in does nothing. We then talked about statements, their converse and the contra-positive for the statement. When one of the first two is true then the contra-postive must also be true. Then we went of a homework problem to close class.
Today we had our quiz over differentiability and then we started talking about concave and inflection points. Concave up is when the tangent lines are below the curve and concave down is when the tangent lines are above the curve. We then found out how to find the inflection point, which is by finding the derivative of the original function and when that function reaches a relative minimum or maximum, and then to find that exact point, you have to set the second derivative equal to zero. The x-coordinate you get from that is the inflection point.
this is really interesting believe me a limit of a function is really very curious subject to understand. Limit Calculator | Problem solver
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