okay so third week of calculus has past by and it really wasnt as bad as what i thought it would be. i have to be honest and say that i have no idea what im going to write in this post. i thought that test was going to kill me and especially with that fire drill wasting everyones time.
so, this week was about review, test, and new things. using geogebra mrs robinson taught us about concave graphs and i actually got the concept pretty good it was alot easier than i thought.
i also learned that drawing things really helps solve a problem. aslo, if a line on a graph is going upward its positive and if its going down its negative which is pretty easy to understand. concave up is from that number to infinity concave down is that number to negative infinity. there is also a list to follow. and you need to know about horizontal and vertical lines.
i am still having trouble with some derivatives though i cant give an exact problem for someone to help me with but its usually larger one with alot of steps that mess me up. i try the breaking it down method with like listing if includes quotient rule and product rule etc. i am also still having problems cancelling things i dont need to. everytime i have to cancel something i end up double guessing myself and then i just do the wrong thing. so hopefully theres a way to help with that.
Friday, September 4, 2009
Tuesday, September 1, 2009
the test
Sooooo, I just needed a break from studying and I'm a little nervous for this test.
just saying
just saying
FOR MY GRADE::
so i was wondering...if i can't really explain anything to anyone [or comment] because i don't understand it myself, can i just print out two of the one's i did last week! [not the same as the first time though]
Sunday, August 30, 2009
Post #2
So, my internet decides to work approximately 15 minutes before the blog is due (I've been trying for approximately 2 1/2 days). Lucky me! If it's late, don't blame me.
Ok, this week in Calculus was way harder for me than last week. I cannot simplify! Whereas last week was just the simpler stuff, this week we got into more complex, well, not concepts, but longer problems I guess you could say?
In the first week we learned how to take the quotient and product rules and how to take the derivatives of trig functions. For example, take the derivative of the following:
(x^2 - 2x + 4)/(x-4)
= (x-4)(2x-2) - (x^2 - 2x +4)(1) (quotient rule)
-----------------------------------
(x-4)^2
= (x^2 - 8x + 4)/(x-4)^2
This week, we also reviewed how to do logs before delving into the derivatives of logs. For instance:
Solve for x:
log39=x
All you do is switch it to so you have:
3^x = 9
and rather than taking the log of both sides like you would normally do, the answer is obviously x=2 because 3 squared equals nine.
We then went into taking the derivatives of natural logs, logs, and inverse trig functions. All I remember was that for the first night of homework (I think it was ln), I was really lost because even though I knew how to plug into the formulas, I just couldn't find out where to start or how to end it. But then I went to school and learned how to start. What really helped me was the list of things to do in the problem. And then I got a lesson in simplifying, and I got a little better. So, an example of inverse trig:
derivative of sin(arc cos(t))
= cos(arc cos(t)) * (-1)/(sqrt(1-t^2))
the cos and arccost cancel giving you:
t * (-1)/(sqrt(1-t^2))
which equals:
-t
--------------
sqrt(1-t^2)
For that particular problem, there wasn't much simplifying which is good for me. My only true problem is that I apparently cannot do basic algebra when simplifying.
And at the moment I'm going insane, so yeah, I think I'll end now.
Ok, this week in Calculus was way harder for me than last week. I cannot simplify! Whereas last week was just the simpler stuff, this week we got into more complex, well, not concepts, but longer problems I guess you could say?
In the first week we learned how to take the quotient and product rules and how to take the derivatives of trig functions. For example, take the derivative of the following:
(x^2 - 2x + 4)/(x-4)
= (x-4)(2x-2) - (x^2 - 2x +4)(1) (quotient rule)
-----------------------------------
(x-4)^2
= (x^2 - 8x + 4)/(x-4)^2
This week, we also reviewed how to do logs before delving into the derivatives of logs. For instance:
Solve for x:
log39=x
All you do is switch it to so you have:
3^x = 9
and rather than taking the log of both sides like you would normally do, the answer is obviously x=2 because 3 squared equals nine.
We then went into taking the derivatives of natural logs, logs, and inverse trig functions. All I remember was that for the first night of homework (I think it was ln), I was really lost because even though I knew how to plug into the formulas, I just couldn't find out where to start or how to end it. But then I went to school and learned how to start. What really helped me was the list of things to do in the problem. And then I got a lesson in simplifying, and I got a little better. So, an example of inverse trig:
derivative of sin(arc cos(t))
= cos(arc cos(t)) * (-1)/(sqrt(1-t^2))
the cos and arccost cancel giving you:
t * (-1)/(sqrt(1-t^2))
which equals:
-t
--------------
sqrt(1-t^2)
For that particular problem, there wasn't much simplifying which is good for me. My only true problem is that I apparently cannot do basic algebra when simplifying.
And at the moment I'm going insane, so yeah, I think I'll end now.
Post #2
This week in calculus we learned about line tangents, we did more advanced derivatives, and we also condensed logs. I didn’t have any problems with solving the logs but I kind of forgot how to condense logs. I can’t belive I forgot how to do logs. O.o I’m pretty sure I grasp the arc equations such as: [arcsin], [arctan], and etc., but I still have trouble with derivatives kind of because I always stop to earlier or mess up on my algerbra but that’s something I have to practice. I also don’t understand how to use (e) fully. Some problems that will really help me out if:
Just condense the problem, #32 last week’s homework.1) 2[lnx-ln(x+1)-ln(x-1)] and
Find an equation of the tangent line #50 last week’s homework.2) y= e-2x+x^2, (2, 1) .
Other than everything we learned; this week was pretty easy if you ask me.
Well I have to do a lot of studying and extra work but with a few examples like this I should be able to learn from what I know and what everyone helps me out with.
I don’t know if I’m supposed to be asking about this because it’s not from last week but for vertical and horizontal asymptotes . I understand vertical asymptotes but I don’t know how to get horizontal.
Like for:
3/x^2-4 you get 3/(x-2)(x+2)
Which gives you vertical asymptotes at x=2, and x=-2but I don’t know how to find the Horizontal ones
Thanks I hope I was able to ask that.
Mabile
Just condense the problem, #32 last week’s homework.1) 2[lnx-ln(x+1)-ln(x-1)] and
Find an equation of the tangent line #50 last week’s homework.2) y= e-2x+x^2, (2, 1) .
Other than everything we learned; this week was pretty easy if you ask me.
Well I have to do a lot of studying and extra work but with a few examples like this I should be able to learn from what I know and what everyone helps me out with.
I don’t know if I’m supposed to be asking about this because it’s not from last week but for vertical and horizontal asymptotes . I understand vertical asymptotes but I don’t know how to get horizontal.
Like for:
3/x^2-4 you get 3/(x-2)(x+2)
Which gives you vertical asymptotes at x=2, and x=-2but I don’t know how to find the Horizontal ones
Thanks I hope I was able to ask that.
Mabile
Post 2
So this week in calculus was intense. It seemed we were going at such a fast pase this week. For the most part, I grasped the concept of things. I understood the very beginning of everything, but got lost during the way. I'm not sure if there's anything I fully understand in calculus from this week, but I'll try to recall something I understood more than other things.
On Monday, we defined a derivitave again, but for different reasons. We learned how to find the equation of a tangent line.
Come to think of it, this is the only new thing we really learned this week, ha. Anyway, I pretty much understood this, so I'll explain it to the best of my abilities
When looking for the equation of a tangent line, once you get your x value, y value, and slope, you are to plug all thoes in to your point slope formula. To get these values, you are to do these steps:
The problem will give you an equation
Ex: f(x)=(x - 1)(x^2 - 2)
The problem will also give you a point
Ex: (0, 2)
1. The first thing to do is to find your slope. You will need to take the derivitave of the equation
Ex: (x - 1)(2x) + (x^2 - 2)(1)
When the derivitave is simplified, it will look like
3x^2 - 2x - 2
2. You then plug in your x value into your x, giving you your slope.
Ex: 3(0)^2 - 2(0) - 2
this will leave you with (-2) being your slope.
3. You thel plug in all your values into point slope form (y - y1 = m(x - x1)
When the x value, the y value, and the slope are plugged into this form, you will get
y - 2 = -2(x)
I also understood most of the log stuff.
I still have a hard time simplifying things. I see things that look alike and I try to combine them, when they can't be combined, or I don't combine things that can be combined. I think I just need more practice on it.
On Monday, we defined a derivitave again, but for different reasons. We learned how to find the equation of a tangent line.
Come to think of it, this is the only new thing we really learned this week, ha. Anyway, I pretty much understood this, so I'll explain it to the best of my abilities
When looking for the equation of a tangent line, once you get your x value, y value, and slope, you are to plug all thoes in to your point slope formula. To get these values, you are to do these steps:
The problem will give you an equation
Ex: f(x)=(x - 1)(x^2 - 2)
The problem will also give you a point
Ex: (0, 2)
1. The first thing to do is to find your slope. You will need to take the derivitave of the equation
Ex: (x - 1)(2x) + (x^2 - 2)(1)
When the derivitave is simplified, it will look like
3x^2 - 2x - 2
2. You then plug in your x value into your x, giving you your slope.
Ex: 3(0)^2 - 2(0) - 2
this will leave you with (-2) being your slope.
3. You thel plug in all your values into point slope form (y - y1 = m(x - x1)
When the x value, the y value, and the slope are plugged into this form, you will get
y - 2 = -2(x)
I also understood most of the log stuff.
I still have a hard time simplifying things. I see things that look alike and I try to combine them, when they can't be combined, or I don't combine things that can be combined. I think I just need more practice on it.
week 2
hellllooo claculus.
so to start off on a good note, i atleast get half of the stuff we did this week =)
first of all, i spent all of last year failing at life on our advanced math
tests because i couldn't grasp the concepts of logs if my life depended
on it, but it's a new year babyyy. when i first looked at our work sheet
i wanted to qluit life, but then it all clicked and i was super excited,
which i still am if you can't tell. though i am a little aggravated that
i spent so much time sucking at such simple concepts.
Now when i see someting like
evaluate the expression log279
i actually know what to do with it and not just put a ridiculous answer.
also, the equation of a tangent line, thankfully,
was extremely easy for me. i understood from the first example
Recall :
• A Tangent Line is a line which locally touches
a curve at one and only one point.
• The slope-intercept formula for a line is y = mx + b,
where m is the slope of the line and b is the y-intercept.
• The point-slope formula for a line is y – y1 = m (x – x1).
This formula uses a point on the line, denoted by (x1, y1),
and the slope of the line, denoted by m, to
calculate the slope-intercept formula for the line.
• The first derivative is an equation for the slope of a tangent
line to a curve at an indicated point.
The equation for the slope of the tangent line to
f(x) = x2 is f '(x), the derivative of f(x).
f(x) = x2
f '(x) = 2x (1)
Therefore, at x = 2, the slope of the tangent line is f '(2).
f '(2) = 2(2)
= 4
Now , you know the slope of the tangent line, which is 4.
All that you need now is a point on the tangent line to be
able to formulate the equation. To find that point, simply plug
the coordinate of the shared point into the original equation, this gives you (2,4)
The only step left is to use the point (2, 4) and slope, 4,
in the point-slope formula for a line. Therefore: Y-4=4(x-2)
NOW FOR WHAT I DON'T UNDERSTAND.
.. they whole arc trig formulas really throw me off and i keep getting
confused on what exactly i do first, or what part of the problem i need
to simplify first, but since b-rob posted that response earlier i
think i understand a little bit better, i just need to stay motivated
and make myself work more problems becuase i know that is
the only way i'm fully going to understand and remember all this.
so to start off on a good note, i atleast get half of the stuff we did this week =)
first of all, i spent all of last year failing at life on our advanced math
tests because i couldn't grasp the concepts of logs if my life depended
on it, but it's a new year babyyy. when i first looked at our work sheet
i wanted to qluit life, but then it all clicked and i was super excited,
which i still am if you can't tell. though i am a little aggravated that
i spent so much time sucking at such simple concepts.
Now when i see someting like
evaluate the expression log279
i actually know what to do with it and not just put a ridiculous answer.
also, the equation of a tangent line, thankfully,
was extremely easy for me. i understood from the first example
Recall :
• A Tangent Line is a line which locally touches
a curve at one and only one point.
• The slope-intercept formula for a line is y = mx + b,
where m is the slope of the line and b is the y-intercept.
• The point-slope formula for a line is y – y1 = m (x – x1).
This formula uses a point on the line, denoted by (x1, y1),
and the slope of the line, denoted by m, to
calculate the slope-intercept formula for the line.
• The first derivative is an equation for the slope of a tangent
line to a curve at an indicated point.
The equation for the slope of the tangent line to
f(x) = x2 is f '(x), the derivative of f(x).
f(x) = x2
f '(x) = 2x (1)
Therefore, at x = 2, the slope of the tangent line is f '(2).
f '(2) = 2(2)
= 4
Now , you know the slope of the tangent line, which is 4.
All that you need now is a point on the tangent line to be
able to formulate the equation. To find that point, simply plug
the coordinate of the shared point into the original equation, this gives you (2,4)
The only step left is to use the point (2, 4) and slope, 4,
in the point-slope formula for a line. Therefore: Y-4=4(x-2)
NOW FOR WHAT I DON'T UNDERSTAND.
.. they whole arc trig formulas really throw me off and i keep getting
confused on what exactly i do first, or what part of the problem i need
to simplify first, but since b-rob posted that response earlier i
think i understand a little bit better, i just need to stay motivated
and make myself work more problems becuase i know that is
the only way i'm fully going to understand and remember all this.
"pzost 2" haha
The second week of Calculus AB was harder than the first week, although it wasn’t too bad. It really was a continuation of last week’s materials, but we just dug deeper into the information and the problems got harder. In the beginning of the week, we did more advanced product and quotient rule problems. Then, towards the end of the week we learned the derivative rules for arc trig functions.
I understood it for the most part, I just still have some problems with simplifying. Mrs. Robinson practiced a lot with us on it this week, so I don’t think I will too much more trouble with it and it should only get easier. I wasn’t really confused with anything this week because it was just more advanced than last week. So I’ll just explain how to take the derivative of the arc trig functions.
The formula for arccos and arcsin are the same, (1/√1-u^2)(u’). The formulas for arctan and arccot are also the same, (1/1+u^2)(u’), the only difference is that arccot has a negative in front. For example, if you had arctan(3x^2 + 4x) you would plug it into the arctan or tan^-1 formula. (1/1+(3x^2 + 4x)^2 (6x+4). Then you would simplify it. (6x+4/1+9x^4+24x^3+16x^2). These formulas are not hard to understand, it’s just very easy to make simple mistakes. The only thing I’m confused about still is the simplifying, and when to or when not to take the derivative of something while you’re simplifying. I know you are supposed to work from the outside in whenever you are taking derivatives, and that did help me a lot this week. Just need to keep practicing J
I understood it for the most part, I just still have some problems with simplifying. Mrs. Robinson practiced a lot with us on it this week, so I don’t think I will too much more trouble with it and it should only get easier. I wasn’t really confused with anything this week because it was just more advanced than last week. So I’ll just explain how to take the derivative of the arc trig functions.
The formula for arccos and arcsin are the same, (1/√1-u^2)(u’). The formulas for arctan and arccot are also the same, (1/1+u^2)(u’), the only difference is that arccot has a negative in front. For example, if you had arctan(3x^2 + 4x) you would plug it into the arctan or tan^-1 formula. (1/1+(3x^2 + 4x)^2 (6x+4). Then you would simplify it. (6x+4/1+9x^4+24x^3+16x^2). These formulas are not hard to understand, it’s just very easy to make simple mistakes. The only thing I’m confused about still is the simplifying, and when to or when not to take the derivative of something while you’re simplifying. I know you are supposed to work from the outside in whenever you are taking derivatives, and that did help me a lot this week. Just need to keep practicing J
Ash's 2nd post
Okay, well. I'm pretty sure the second week in Calculus was a trillion times harder than the first. Algebra. I think I need to re-take that class. Or at least a class in patience. Either one will work.
So, before I get myself into a mess, let me clarify something: I'm having a hard time remembering what we did this week and HOW to do it, so if I screw up, I blame my memory.
I think the safest thing for me to explain is the Trig Inverses. I half-way like them. Except the fractions.
Let's try an example problem! Arctan(3x^2+4x)
1. Find and state the formula for Arctan.
1
--------- * u'
1+u^2
2. Replace u for what is inside of the parenthesis.
1
--------------- * (3x^2+4x)'
1+(3x^2+4x)^2
3. Take the derivative of what you are multiplying it by.
(3x^2+4x) = (6x+4)
4. So now you would get
1
--------------- * (6x+4)
1+(3x^2+4x)^2
5. Since (6x+4) = (6x+4)/1, when you multiply, you will get this
(6x+4)
---------------
1+(3x^2+4x)^2
6. Now, foil out (3x^2+4x)^2
(6x+4)
------------------------
1+9x^4 + 24x^3 + 16x^2
7. Simplify what you can and you're finished!
2(3x+2)
------------------------
9x^4 + 24x^3 + 16x^2 + 1
Let's try another half-way simple problem.
Arccos(1/x^2)
1. Find and state the formula for Arccos. (Sense a pattern here?)
-1
----------- * u'
sqrt(1-u^2)
2. Replace u with what is in the parenthesis.
-1
----------------- * (1/x^2)'
sqrt(1-(1/x^2)^2)
3. Take the derivative of what you're multiplying by.
(1/x^2) = (-2/x^3)
4. Multiply to get
2
-------------------
x^3(sqrt(1-(1/x^4))
5. Simplify (completely simplified already) and you're finished!!
I don't know if that was clear or not. I hope so. If someone finds a mistake, let me know please. :)
As for my question...
I completely forgot logs (figures, used to be my favorite).
Can someone explain how to do this problem, I think it's the hardest:
ln ((x^2-1)/x^3)^3
Logs + Fractions + Derivatives /=/ Happiness.
Thanks!!
So, before I get myself into a mess, let me clarify something: I'm having a hard time remembering what we did this week and HOW to do it, so if I screw up, I blame my memory.
I think the safest thing for me to explain is the Trig Inverses. I half-way like them. Except the fractions.
Let's try an example problem! Arctan(3x^2+4x)
1. Find and state the formula for Arctan.
1
--------- * u'
1+u^2
2. Replace u for what is inside of the parenthesis.
1
--------------- * (3x^2+4x)'
1+(3x^2+4x)^2
3. Take the derivative of what you are multiplying it by.
(3x^2+4x) = (6x+4)
4. So now you would get
1
--------------- * (6x+4)
1+(3x^2+4x)^2
5. Since (6x+4) = (6x+4)/1, when you multiply, you will get this
(6x+4)
---------------
1+(3x^2+4x)^2
6. Now, foil out (3x^2+4x)^2
(6x+4)
------------------------
1+9x^4 + 24x^3 + 16x^2
7. Simplify what you can and you're finished!
2(3x+2)
------------------------
9x^4 + 24x^3 + 16x^2 + 1
Let's try another half-way simple problem.
Arccos(1/x^2)
1. Find and state the formula for Arccos. (Sense a pattern here?)
-1
----------- * u'
sqrt(1-u^2)
2. Replace u with what is in the parenthesis.
-1
----------------- * (1/x^2)'
sqrt(1-(1/x^2)^2)
3. Take the derivative of what you're multiplying by.
(1/x^2) = (-2/x^3)
4. Multiply to get
2
-------------------
x^3(sqrt(1-(1/x^4))
5. Simplify (completely simplified already) and you're finished!!
I don't know if that was clear or not. I hope so. If someone finds a mistake, let me know please. :)
As for my question...
I completely forgot logs (figures, used to be my favorite).
Can someone explain how to do this problem, I think it's the hardest:
ln ((x^2-1)/x^3)^3
Logs + Fractions + Derivatives /=/ Happiness.
Thanks!!
post 2
Well another week down in calculus and once again I understand some things but I don't understand some things. One thing i understand is get the equation of tangent lines. All you do is take a point and plug it into the derivative of the formula. But if only an x is given then you plug the x into the formula to get your y. And finally you have to put the slope you get when you plug it in the derivative and point in point slope form.
An example is x^3+2x^2-x+1 (2,-1). Then the derivative is 3x^2+4x-1. Now you plug the point in to get the slope. 3(2)^2+4(2)-1. This equals 19 which is your slope. Finally all you do is plug your point and slope into point slope form which is y+1=19(x-2). And even thought this can be condensed into a simpler form if you do it would not be in point slope form. So when you get something like that just leave it.
Also I understand how to get the equation of a normal line. You do the same things as you do for tangent lines except you just make the slope perpendicular. Using the same example above the answer would be y+1=-1/19(x-2). All you do to get the inverse of the slope is take the negative reciprocal.
But on the other hand, I don't understand how to do problems such as ln(lnx^2) or ln(t)/t. I don't even know how to start on these. If someone can get me started and explain it to me it would be greatly appreciated.
An example is x^3+2x^2-x+1 (2,-1). Then the derivative is 3x^2+4x-1. Now you plug the point in to get the slope. 3(2)^2+4(2)-1. This equals 19 which is your slope. Finally all you do is plug your point and slope into point slope form which is y+1=19(x-2). And even thought this can be condensed into a simpler form if you do it would not be in point slope form. So when you get something like that just leave it.
Also I understand how to get the equation of a normal line. You do the same things as you do for tangent lines except you just make the slope perpendicular. Using the same example above the answer would be y+1=-1/19(x-2). All you do to get the inverse of the slope is take the negative reciprocal.
But on the other hand, I don't understand how to do problems such as ln(lnx^2) or ln(t)/t. I don't even know how to start on these. If someone can get me started and explain it to me it would be greatly appreciated.
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