In Calculus this week, I learned about derivatives, average speed, and instantaneous speed. For each of these math terms, there are different steps to follow. Many formulas are used and it is necessary to remember all of them. We learned around ten derivative formulas which include quotient rule, product rule, sum and difference rules, trig function rules, and rules for numbers with and without variables, and taking derivates of quadratics.
I am very comfortable with taking derivatives by using the product rule. The product rule is used when you are multiplying two terms such as this problem x² (x+4). First, you must know the formula, which is uv^1 + vu^1. In simpler terms, this formula means (copy the first term)(derivative of the second term) + (copy the second term)(derivative of the first term). Following along with the problem so far, you would have (x²)(1) + (x+4)(2x). You get x² because as the formula states, the first part of the problem is copying the first term. Next, you have to take the derivative of (x+4). Knowing the derivative rule for variables without an exponent is equal to 1 and the derivative of any number is equal to 0, you obtain (1+0) for the derivative of (x+4). So far your problem should look like (x²)(1). The next step is to place the addition sign next in the equation. Then you should copy the second term (x+4), and take the derivative of the first. The derivative of x² is obtained by multiplying the exponent by the coefficient and then subtracting one from the exponent. (x²) = (2)(1) x²-1 = 2x. Therefore your problem will then become (x²)(1) +(x+4)(2x). Simple algebra is used from then on, you should distribute the x² to the one and distribute the 2x to (x+4). Then add like terms to obtain your final answer of 3x² +8x.
One thing I am not comfortable with is finding instantaneous speed. Problems such as finding the instantaneous speed at t=2 where y=16t² confuses me. At first I though you just plug it in, but then I realized there was a special formula that confuses me. I know the formula, but I don’t understand how you decipher what is your “h” and what is the f(x). After the formula is plugged in and you do the algebra, what number are you supposed to plug in?
Among many things I learned in Calculus class this week, product rule is the thing I am most comfortable with and finding instantaneous speed is the only thing that confuses me.
Saturday, August 22, 2009
Post #1
On the first day I went into calculus class and we started going over everything that we had to do and be prepared for, I thought i was going to fail. Math is Not my best subject and i struggle with it. But after a few days in the class i realized that i am going to be fine and that i will learn the concepts and will be able to make good grades in this class. Sure, there are still a few things that i'm not totally getting right now, but i will learn it and be able to grasp the concepts after a little more practice.
I'm excited that i actually am starting to get derivitives. When i first saw all those formulas i thought there was no way i was going to be able to remember all of those. But once we started going over the easy ones first.. i felt better. Like the rule that states if you have just a number, then the derivitive is 0. Or the rule that says if you have something like 5x^3, to take the derivitive, you would multiply the 3 and the 5 to get the number in front, and then you would subtract one from the three to give you 2. the final answer would be 15x^2. I'm even starting to get a better grasp on the cos and sin formulas as well. The derivitive of sin is consu x u^1 and the cos formula is -sinu x u^1.
There are a few things that i am still struggling with though. I sort of get the product rule, but when you combine it with hard numbers like 3 square root of t(t^2 + 4). The 3 and the square root throw me off the most. As far as the quotient rule goes, i think i have a pretty good handle on it but on problems like 3 square root of x / x^3 +1.. i get confused on what the do with the top. i know you have to make the radical into a fraction but i don't understand what happens after that. As far as those hard ones go, i believe i have a pretty good handle on these.
I am looking forward to having a great year in calculus this year and i'm excited about having the chance to get college credit by taking the AP exam. I believe this class will help me with understanding my math more and helping me be more successful in math. I know my friends and B-rob are there to help me if i get stuck and need help. I am looking forward to having a great year with all of you :)
I'm excited that i actually am starting to get derivitives. When i first saw all those formulas i thought there was no way i was going to be able to remember all of those. But once we started going over the easy ones first.. i felt better. Like the rule that states if you have just a number, then the derivitive is 0. Or the rule that says if you have something like 5x^3, to take the derivitive, you would multiply the 3 and the 5 to get the number in front, and then you would subtract one from the three to give you 2. the final answer would be 15x^2. I'm even starting to get a better grasp on the cos and sin formulas as well. The derivitive of sin is consu x u^1 and the cos formula is -sinu x u^1.
There are a few things that i am still struggling with though. I sort of get the product rule, but when you combine it with hard numbers like 3 square root of t(t^2 + 4). The 3 and the square root throw me off the most. As far as the quotient rule goes, i think i have a pretty good handle on it but on problems like 3 square root of x / x^3 +1.. i get confused on what the do with the top. i know you have to make the radical into a fraction but i don't understand what happens after that. As far as those hard ones go, i believe i have a pretty good handle on these.
I am looking forward to having a great year in calculus this year and i'm excited about having the chance to get college credit by taking the AP exam. I believe this class will help me with understanding my math more and helping me be more successful in math. I know my friends and B-rob are there to help me if i get stuck and need help. I am looking forward to having a great year with all of you :)
Post #1
Between August 13th and August 21st in Calculus we learned about derivatives. There were the thirty-six formulas that we copied down in order to help us. I was freaked out because it looked hard. I thought it was going to be the hardest week of my life. Well, it was up there! Even after B-rob explained it the next 3 days I continued to stay lost. Then I decided that I was not going to give up and that I would end up learning it, so I put my mind to it and actually got it! GO MEE! ha.
However, the formulas help a lot once you understand the concept. For instance, if you want to take the derivative of ANY number, it will ALWAYS be zero. However, anything with an exponet will have a derivative by multipling the exponet with the number in front of the variable. Once you multiply that, your answer will take the place of the number in front of the variable. But, you're not done yet! Take the exponet and subtract one. Your answer will take the place of the exponet. Then you solve. Remember, though, that if you have a negative exponet you have to put it at the bottom of a fraction in order for it to become positive. [You can't have a negitive exponet!]
We also learned the product and quotient rule. For the product rule, you must have multiplication. You will keep the first part of the problem and multiply it by the derivative of the second part, and then add the second part of the formula, which is to keep the second part of the problem and multiply it by the derivative of the first. You'll solve and that's your answer. I need a little more practice on the harder ones and the ones that have and "x" with an exponet then sin(x) or something like that. Is that the product rule, or do you just take the derivative? For the quotient rule you have a fraction. You take the bottom and then multiply it by the derivative of the top then you'll sbtract the top multiplied by the derivative of the bottom. This will all be over the bottom squared. Then you solve, and that's your answer.
We then had average and instantaneous velocity. Average velocity is when you take the derivative of your problem and plug in the two numbers from the problem. Then you'll take the answer of t1 and t2 in order to subtract t1 from t2. After, you will divide it by the numbers they gave you for t1 and t2, so you'll do...t2 minus t1 -from the problem not the answer you get when you plug them into the problem- then solve for your answer. However, for instantaneous velocity you take the derivative of the problem and put in the one number they give you. What you get is your answer!
Oh, and don't forget, we took a quiz on the trig chart as well as the formulas! KNOW THIS FOR LIFE!!
And what I need help on is when you have a square root or a fraction and you have to find the derivative. I understand the concept, but I think I might need help sometimes because I get confused if I can do what I did or not.
However, the formulas help a lot once you understand the concept. For instance, if you want to take the derivative of ANY number, it will ALWAYS be zero. However, anything with an exponet will have a derivative by multipling the exponet with the number in front of the variable. Once you multiply that, your answer will take the place of the number in front of the variable. But, you're not done yet! Take the exponet and subtract one. Your answer will take the place of the exponet. Then you solve. Remember, though, that if you have a negative exponet you have to put it at the bottom of a fraction in order for it to become positive. [You can't have a negitive exponet!]
We also learned the product and quotient rule. For the product rule, you must have multiplication. You will keep the first part of the problem and multiply it by the derivative of the second part, and then add the second part of the formula, which is to keep the second part of the problem and multiply it by the derivative of the first. You'll solve and that's your answer. I need a little more practice on the harder ones and the ones that have and "x" with an exponet then sin(x) or something like that. Is that the product rule, or do you just take the derivative? For the quotient rule you have a fraction. You take the bottom and then multiply it by the derivative of the top then you'll sbtract the top multiplied by the derivative of the bottom. This will all be over the bottom squared. Then you solve, and that's your answer.
We then had average and instantaneous velocity. Average velocity is when you take the derivative of your problem and plug in the two numbers from the problem. Then you'll take the answer of t1 and t2 in order to subtract t1 from t2. After, you will divide it by the numbers they gave you for t1 and t2, so you'll do...t2 minus t1 -from the problem not the answer you get when you plug them into the problem- then solve for your answer. However, for instantaneous velocity you take the derivative of the problem and put in the one number they give you. What you get is your answer!
Oh, and don't forget, we took a quiz on the trig chart as well as the formulas! KNOW THIS FOR LIFE!!
And what I need help on is when you have a square root or a fraction and you have to find the derivative. I understand the concept, but I think I might need help sometimes because I get confused if I can do what I did or not.
Friday, August 21, 2009
Post #1
So I want to start this post off with saying that this week was not what I expected it to be. I expected the first week to be really hard, especially after seeing ALL of the formulas for the derivatives. However, it turned out not to be as intimidating as I thought. I think I really grasped hold on the concepts behind taking a derivative rather well. I guess it helps that through Mu Alpha Theta I had already learned the basic derivative formula...the whole limit as h approaches 0 thing, as well as taking a derivative of a polynomial the short way. I guess I was just more scared about Calculus because I know that I really need to focus on doing everything right and understanding it completely in order to do well on the AP exam...
Firstly, I think the only thing I didn't grasp at first was the weird wording on the word problems. There seems to be so many different words for the same exact thing, and it just gets really confusing. I think the only thing I had problems with on the packet was the average speed thing...and I think I figured out why. In our notes, we used an example problem with "the first two seconds". I did not realize at first that it doesn't have to be the "first n seconds" it can be any interval, such as from 2 to 4. It was also kind of confusing there because we used the long way to find the derivative, and I got lost in my notes...I guess I should learn to take better notes in math.
One thing that I really like the feel of that I think I could help you guys with is when taking care of things like:
1
---
x(2/3)
I think everyone gets the concept of bringing it back up but I'm more than sure that a lot of you guys are forgetting the derivative rule when doing this. You bring the exponent to the front and then SUBTRACT one. You have to remember the subtract. I don't know how many times you guys are telling me that -(2/3) - 1 = (1/3). It becomes -(5/3), and therefore the answer would be
-2
---
3x(5/3)
I hope that clears up some problems for some of you guys...
Anyway, as far as our first week, I think we are doing strong as a class, and we can move forward next week even stronger...We are doing really well so keep up the good work guys!
Be ready for the quiz on Tuesday :-p
Firstly, I think the only thing I didn't grasp at first was the weird wording on the word problems. There seems to be so many different words for the same exact thing, and it just gets really confusing. I think the only thing I had problems with on the packet was the average speed thing...and I think I figured out why. In our notes, we used an example problem with "the first two seconds". I did not realize at first that it doesn't have to be the "first n seconds" it can be any interval, such as from 2 to 4. It was also kind of confusing there because we used the long way to find the derivative, and I got lost in my notes...I guess I should learn to take better notes in math.
One thing that I really like the feel of that I think I could help you guys with is when taking care of things like:
1
---
x(2/3)
I think everyone gets the concept of bringing it back up but I'm more than sure that a lot of you guys are forgetting the derivative rule when doing this. You bring the exponent to the front and then SUBTRACT one. You have to remember the subtract. I don't know how many times you guys are telling me that -(2/3) - 1 = (1/3). It becomes -(5/3), and therefore the answer would be
-2
---
3x(5/3)
I hope that clears up some problems for some of you guys...
Anyway, as far as our first week, I think we are doing strong as a class, and we can move forward next week even stronger...We are doing really well so keep up the good work guys!
Be ready for the quiz on Tuesday :-p
Thursday, August 20, 2009
Algebra...
ok then...since tab just failed me... I was just wondering if i can manipulate my problem before taking the derivative, or will that just change the derivative completely?
question
Okay, so im having problems understanding instantaneous velocity. Its not like i don't get the concept of it its just that I get stuck on where certain numbers should go in the equation. So I tried to practice some problems to help me with understanding what to do. For example, ( starting at a height of 4 feet, a ball is thrown upwards with an initial velocity of 160 feet/ sec. The ball's height after t seconds is s(t)=-16t2+160+4. Find the average velocity from t=2 to t=4.) Okay so i want to plug in the 2 and the 4 but im having trouble deciding if its the right thing to do, and i have no idea what to do with that 4 at the beginning of the equation. Do i use it or is it just there for nothing??
Im also having trouble with finding the derivative of this problem. Suppose m'(-2)=72 and m(-2)=0. Find a(-2) assuming (m/a)'(-2)=6. Okay, im not sure what to do with this either because there is more than one m. I dont know its just really confusing to me. Which one do i use? Or am i supposed to use both of them?
Im also having trouble with finding the derivative of this problem. Suppose m'(-2)=72 and m(-2)=0. Find a(-2) assuming (m/a)'(-2)=6. Okay, im not sure what to do with this either because there is more than one m. I dont know its just really confusing to me. Which one do i use? Or am i supposed to use both of them?
Wednesday, August 19, 2009
question on packet
when it says find the average i know what to do
when it says find the instantanious i know what to do
buttt when it says JUST find the velocity...which one does that mean??
when it says find the instantanious i know what to do
buttt when it says JUST find the velocity...which one does that mean??
Tuesday, August 18, 2009
BTW...
If you are ever asking questions on the blog...in order for the question to look a little more...understandable, you can use superscript tags or subscript tags.
>SUB<>/SUB< >SUP<>/SUP<
Use less than sign first, then greater at the end... I just can't do it because it will actually try to make it subscript haha. So try it out in a comment if you want...
Anyway...
x2+log12x4x
>SUB<>/SUB< >SUP<>/SUP<
Use less than sign first, then greater at the end... I just can't do it because it will actually try to make it subscript haha. So try it out in a comment if you want...
Anyway...
x2+log12x4x
HELP!!
ok so when you have ...
Given h(x) = 5cot(x) + 2tan(x), what is h'(x)?
i don't understand what h' means or how to get it...i know it's prime but i don't understand what prime means either..
Given h(x) = 5cot(x) + 2tan(x), what is h'(x)?
i don't understand what h' means or how to get it...i know it's prime but i don't understand what prime means either..
Monday, August 17, 2009
Sunday, August 16, 2009
First Post
Okay, if this thing doesn't work....someone's gonna have issues
And if it does hurray, let's go get cookies!
btw, it's Ashley
And if it does hurray, let's go get cookies!
btw, it's Ashley
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