Alright, so this week:
Study for Test
Study for Test
Test
Test
Limits
I mostly understand the limits.
There are rules when dealing with these things:
(Note: These apply only when the limit is to infinity)
Rule #1 - When the degree (exponent) of the bottom is GREATER than the degree of the top, the limit is Zero.
Rule #2 - When the degree (exponent) of the bottom is SMALLER than the degree of the top, the limit is infinity. (positive or negative)
Rule #3 - When the degrees are equal, the limit is the coeffecients.
Here are some tips that we got in class:
1) When plugging in a limit and you get something over zero, do not just automatically put ZOMG! TLDNE! (THE LIMIT DOES NOT EXIST!). There are two other choices:
1. Factor and cancel (Not sure what this means...can someone clarify please?)
2. Use the calculator (table function) (take the --> # and add and minus .1, .01, and .001)
2) Limits have to match from left to right (OMG! Can someone help with this? I have no idea how to do left and right limits!)
3) When there is a radical, the limit will always be (highest top coef)/(sqrt(highest bottom coef))
Alex then put some stuff on the board that I'm kinda iffy about:
1) lim sinax
x->0 bx
The limit is a/b
2) lim tanax
x->0 bx
The limit is a/b
Okay, so, The Things I Don't Understand!!
1. The concept of factoring and cancelling when there is something over zero...what does it do? Is that your limit or what?
2. How to find the limit using left and right hand. I have noooo idea how to even begin!
Saturday, November 7, 2009
Friday, November 6, 2009
Post #12
Ok so week 12 of Calculus is over with. This week was stressful let me just say that! We’re going over everything once again and for some reason I’m STILL unable to grasp a lot of the problems. We had the packets that were due Friday [today] on all of the stuff we needed help on. It helped me a lot on my test. Our test was broken up into two parts, the multiple choice one day and the free response the next. Goshh…… I’m glad I did okay on the multiple choice because I know I did pretty badly on the free response. I don’t know why but I just blanked when I got the test in my hand.
After taking the multiple choice part of the test I realized I might have a chance, so I went home and studied everything that I didn’t understand once again. WOW all that work for nothing. After the test was over with we were all relieved. However, today Mrs. Robinson told us that we did HORRIBLE on the limits part, so we got a packet on limits today. I LOVE LIMITS PERSONALLY because that’s one of the things I feel confident on...Ha!
So I guess I can explain limits right.
Umm let’s say you have to find the limit of x = 4 with some equation, all you have to do is plug the equation into your y= part of your calculator and remember if you have a fraction that the numerator and the denominator are in parenthesis because the calculator needs to know what’s with what. After that press second graph in order to get to table and remember that your table set function should be set to ask on independent. If its x=4 for the limit you’re trying to find, plug in 4-.1…4-.01…4-.001…4+.001…4+.01…4+.1… and the answers you get is what you look at to find the answer. You look at all the – and all the + and whatever they are going to is what the limit is!
If you have x=4ֿ then you just plug in the minus’s and see what it’s going to, and for x=4(+) then you would just plug in the plus’s and see what it’s going to.
Remember all the rules for the limit going to 0 and to infinity!! WE NOW HAVE THAT IN OUR NOTES FROM TODAY!
Just a review:
• a differential is when something is solved for dy or dx like:
y=cos(x)---solve for dy
dy/dx = -sin(x)
dy = [(dx)*(-sin(x))]
• When we use differentials to approximate there are steps involved:
1. Identify Equation
2. f(x) + f'(x)*dx
3. determine dx
4. determine x
5. Plug in and solve!
For example: (number one on the Approximate part of our wednesday night
homework) to approximate the square root of 99.4 we use the
steps...
~the equation is -- f(x) = the square root of (x)
~the square root of [x] + [(1)/(2 * the square root of [x]) multiplied by (dx)]
~ dx = .4
~ x = 99
~ plug in: [the square root of 99] + [(1) / (2 * the square root of 99) multiplied
by (.4)]
~when solved:....9.96997, the error is 0.0002
After taking the multiple choice part of the test I realized I might have a chance, so I went home and studied everything that I didn’t understand once again. WOW all that work for nothing. After the test was over with we were all relieved. However, today Mrs. Robinson told us that we did HORRIBLE on the limits part, so we got a packet on limits today. I LOVE LIMITS PERSONALLY because that’s one of the things I feel confident on...Ha!
So I guess I can explain limits right.
Umm let’s say you have to find the limit of x = 4 with some equation, all you have to do is plug the equation into your y= part of your calculator and remember if you have a fraction that the numerator and the denominator are in parenthesis because the calculator needs to know what’s with what. After that press second graph in order to get to table and remember that your table set function should be set to ask on independent. If its x=4 for the limit you’re trying to find, plug in 4-.1…4-.01…4-.001…4+.001…4+.01…4+.1… and the answers you get is what you look at to find the answer. You look at all the – and all the + and whatever they are going to is what the limit is!
If you have x=4ֿ then you just plug in the minus’s and see what it’s going to, and for x=4(+) then you would just plug in the plus’s and see what it’s going to.
Remember all the rules for the limit going to 0 and to infinity!! WE NOW HAVE THAT IN OUR NOTES FROM TODAY!
Just a review:
• a differential is when something is solved for dy or dx like:
y=cos(x)---solve for dy
dy/dx = -sin(x)
dy = [(dx)*(-sin(x))]
• When we use differentials to approximate there are steps involved:
1. Identify Equation
2. f(x) + f'(x)*dx
3. determine dx
4. determine x
5. Plug in and solve!
For example: (number one on the Approximate part of our wednesday night
homework) to approximate the square root of 99.4 we use the
steps...
~the equation is -- f(x) = the square root of (x)
~the square root of [x] + [(1)/(2 * the square root of [x]) multiplied by (dx)]
~ dx = .4
~ x = 99
~ plug in: [the square root of 99] + [(1) / (2 * the square root of 99) multiplied
by (.4)]
~when solved:....9.96997, the error is 0.0002
Wednesday, November 4, 2009
Post 11?
Ok, so as I was coming do my comments today, I saw that I totally forgot to do my blog for last week all together, so I'll do it now.
So last week was completley crazy. I forgot to do all my homework and this blog, but last week we did go over a lot. Although we did not learn related rates and angle of elevation last week, we continued on going over it.
Related rates are very similar to optimization; they make optimization look like a piece of cake. Anyway, the steps for related rates are:
1. identify all variables and equations
2. identify what you are looking for
3. make a sketch and lable
4. write an equation involving variables
5. take the derivative with restect to time (dy/dt instead of dy/dx) and solve
This basically helps you solve a problem when something in the problem is changing (a rate; derivatives are rates).
We also went over angle of elevation. I am still pretty bad at these. I can never decide what number in the problem goes with which variable. I particularly have trouble with problems that involve air planes.
EX. When a problem says a plane is flying at 200 mi/hr and starts rising at 20 degrees
I think that's angle of elevation. I'm not completley sure though?
Last week we learned linerization and differentials. Mrs. Robinson described linerization as taking, say, a parabola and streaching it out so it becomes a line. When there is a linerization problem, the key word the problem will say is approximate. The equation for this is:
f(x)=f(c)=f '(c) (x-c)
Differentials are when something is solved for dy or dx.
So last week was completley crazy. I forgot to do all my homework and this blog, but last week we did go over a lot. Although we did not learn related rates and angle of elevation last week, we continued on going over it.
Related rates are very similar to optimization; they make optimization look like a piece of cake. Anyway, the steps for related rates are:
1. identify all variables and equations
2. identify what you are looking for
3. make a sketch and lable
4. write an equation involving variables
5. take the derivative with restect to time (dy/dt instead of dy/dx) and solve
This basically helps you solve a problem when something in the problem is changing (a rate; derivatives are rates).
We also went over angle of elevation. I am still pretty bad at these. I can never decide what number in the problem goes with which variable. I particularly have trouble with problems that involve air planes.
EX. When a problem says a plane is flying at 200 mi/hr and starts rising at 20 degrees
I think that's angle of elevation. I'm not completley sure though?
Last week we learned linerization and differentials. Mrs. Robinson described linerization as taking, say, a parabola and streaching it out so it becomes a line. When there is a linerization problem, the key word the problem will say is approximate. The equation for this is:
f(x)=f(c)=f '(c) (x-c)
Differentials are when something is solved for dy or dx.
Sunday, November 1, 2009
Ash's 11th Post
First off..it's 11:11...make a wish...
Okay, now that that's over with, let's get down to business.
Number one: Is anyone else's wiki page saying there's an error? Maybe because it's almost midnight on a Sunday night and they're fixing it...idk
Anyway..this week: I honestly don't remember much except study guides and..wow..Linearization.
I half-way understood it whenever she was going over it in class, but now, I have no idea what I'm doing. So, to solve that problem, I'm going to go and read everyone's blog about 13 times each (lucky number) and hopefully something hits me (physically or mentally, it doesn't matter which =]).
How about I explain something I understand a little more? Hmm...Related Rates? Sure, that doesn't sound TOO bad..
Alright, Related Rates it is! :)
Steps:
1. Identify all variables and equations
2. Identify what you are looking for
3. Make a sketch and label
4. Write an equation(s) involving your variables (only have 1 unknown)
5. Take the derivative with respect to TIME! (ohh...I should have remembered that..)
6. Substitute in the Derivative and solve
Okay, For those of whom are still scratching their heads in confusion, let's take a different approach to this ;)
Steps:
1. Think of it as step one in Chemistry (for those who have it): Plan
2. Step two in Chemistry: Analyze
3. "Sketch and Label"
4. Use yer fermulas (either given or basic (haha) Geometry formulas)
5. Take d(enter variable here)/dt
6. Substitute and solve (come on, you've been doing it since elementary right? That can't be so bad...haha.. -.-)
Example Problem:
The variables x and y are differentiable functions of t. They are related by the equation y=2x^3-x+4. When x=2, dx/dt = -1. Find dy/dt when x=2.
If you don't get it, break it down (also first step):
1. X and Y are differentiable; y=2x^3-x+4; x=2 so dx/dt=-1.
2. Find: dy/dt
3. Can't really sketch this ^^
4. Your formula was given: y=2x^3-x+4
5. dy/dx = 6x^2dx/dy-dy/dy
dy/dx = 6 (2)^2(-1)-(-1)
dy/dx=-23
Ta Da!! That wasn't so bad now was it? :)
Okay, now that that's over with, let's get down to business.
Number one: Is anyone else's wiki page saying there's an error? Maybe because it's almost midnight on a Sunday night and they're fixing it...idk
Anyway..this week: I honestly don't remember much except study guides and..wow..Linearization.
I half-way understood it whenever she was going over it in class, but now, I have no idea what I'm doing. So, to solve that problem, I'm going to go and read everyone's blog about 13 times each (lucky number) and hopefully something hits me (physically or mentally, it doesn't matter which =]).
How about I explain something I understand a little more? Hmm...Related Rates? Sure, that doesn't sound TOO bad..
Alright, Related Rates it is! :)
Steps:
1. Identify all variables and equations
2. Identify what you are looking for
3. Make a sketch and label
4. Write an equation(s) involving your variables (only have 1 unknown)
5. Take the derivative with respect to TIME! (ohh...I should have remembered that..)
6. Substitute in the Derivative and solve
Okay, For those of whom are still scratching their heads in confusion, let's take a different approach to this ;)
Steps:
1. Think of it as step one in Chemistry (for those who have it): Plan
2. Step two in Chemistry: Analyze
3. "Sketch and Label"
4. Use yer fermulas (either given or basic (haha) Geometry formulas)
5. Take d(enter variable here)/dt
6. Substitute and solve (come on, you've been doing it since elementary right? That can't be so bad...haha.. -.-)
Example Problem:
The variables x and y are differentiable functions of t. They are related by the equation y=2x^3-x+4. When x=2, dx/dt = -1. Find dy/dt when x=2.
If you don't get it, break it down (also first step):
1. X and Y are differentiable; y=2x^3-x+4; x=2 so dx/dt=-1.
2. Find: dy/dt
3. Can't really sketch this ^^
4. Your formula was given: y=2x^3-x+4
5. dy/dx = 6x^2dx/dy-dy/dy
dy/dx = 6 (2)^2(-1)-(-1)
dy/dx=-23
Ta Da!! That wasn't so bad now was it? :)
post # 11
Alrighty then, I learned related rates this week because I wasn't at school the week before, and I also learned linearization.
Related Rates:
1: identify all variables in equations
2: identify what you are looking for
3: sketch and lebel
4: write an equation involving your variables. (you can only have one unkown so a secondary equation may be given)
5: take the derivative with respect to time.
6: substitute derivative and solve.
Example: the variables x and y are functions of t and related by the equation y=2x^3-x+4 when x=2, dy/dt=-1. Find dy/dt when x=2
alright, so you put down the equation, y=2x^3-x+4.
Then you take the derivative of that, so you get dy/dt=6x^2(dx/dt)-(dx/dt)
then you plug in to find that dy/dt=6(2)^2(-1)-(-1)
and that is further simplified to, dy/dt=-23.
Linearization:
f(x)=f(c)+f'(c)(x-c)
example: Approximate the tangent line to y=x^2 at x=1
you find all the different values: dy/dx=2x dy/dx=2 y=(1)^2=1
then you plug into the formula to get: f(x)=1+2(x-1)
example 2: use differentials to approximate: sq root(16.5)
steps:
1: identify an equation--- f(x)=sq root(x)
2:f(x)+f;(x)dx--- sqrt(x)+ (1/(2sqrt(x)))(dx)
3:determine dx-- .5
4:determine x--- 16
5:plug in--- sqrt(16)+(1/2sqrt(16))(.5)= 4.0625
error= .0005
so I did learn how to do that stuff, and did pretty well on that quiz that we had earlier in the week. I am actually quite happy because I have been starting to pass my quizzes, so I am glad we started this new stuff because I am actually understanding it quite well.
One thing I do not understand is angle of elevation, I don't really have any notes on it, just an example problem, so I have not been able to really learn it just by looking at that. If anyone has any advice, it would be greatly appreciated.
Related Rates:
1: identify all variables in equations
2: identify what you are looking for
3: sketch and lebel
4: write an equation involving your variables. (you can only have one unkown so a secondary equation may be given)
5: take the derivative with respect to time.
6: substitute derivative and solve.
Example: the variables x and y are functions of t and related by the equation y=2x^3-x+4 when x=2, dy/dt=-1. Find dy/dt when x=2
alright, so you put down the equation, y=2x^3-x+4.
Then you take the derivative of that, so you get dy/dt=6x^2(dx/dt)-(dx/dt)
then you plug in to find that dy/dt=6(2)^2(-1)-(-1)
and that is further simplified to, dy/dt=-23.
Linearization:
f(x)=f(c)+f'(c)(x-c)
example: Approximate the tangent line to y=x^2 at x=1
you find all the different values: dy/dx=2x dy/dx=2 y=(1)^2=1
then you plug into the formula to get: f(x)=1+2(x-1)
example 2: use differentials to approximate: sq root(16.5)
steps:
1: identify an equation--- f(x)=sq root(x)
2:f(x)+f;(x)dx--- sqrt(x)+ (1/(2sqrt(x)))(dx)
3:determine dx-- .5
4:determine x--- 16
5:plug in--- sqrt(16)+(1/2sqrt(16))(.5)= 4.0625
error= .0005
so I did learn how to do that stuff, and did pretty well on that quiz that we had earlier in the week. I am actually quite happy because I have been starting to pass my quizzes, so I am glad we started this new stuff because I am actually understanding it quite well.
One thing I do not understand is angle of elevation, I don't really have any notes on it, just an example problem, so I have not been able to really learn it just by looking at that. If anyone has any advice, it would be greatly appreciated.
post #11
This week we pretty much reviewed for a test.
One thing that i feel pretty sure of myself with is related rates
1) first determine what is given in the problem. All given variables and even what your looking for.
2)set up your equation. Most of the time this equation will be simple such as patheagorean theroem or area or something like that.
3) you then take the derivative and put dy/dx,dx/dr, etc. the correct term for each variable.
4) plug in your given and then you have set up a simple equation.
5) once your equatin is set up and all given is plugged in you then solve for you unknown.
Because this week was hecktic and busy i guess i just missed linearization or something but i have no clue what to do with it. I need help
One thing that i feel pretty sure of myself with is related rates
1) first determine what is given in the problem. All given variables and even what your looking for.
2)set up your equation. Most of the time this equation will be simple such as patheagorean theroem or area or something like that.
3) you then take the derivative and put dy/dx,dx/dr, etc. the correct term for each variable.
4) plug in your given and then you have set up a simple equation.
5) once your equatin is set up and all given is plugged in you then solve for you unknown.
Because this week was hecktic and busy i guess i just missed linearization or something but i have no clue what to do with it. I need help
post 11
After another week in calculus we have learned about linearization, but we also spent some time reviewing for a test on everthing we have learned up to this point. The reason why we are doing this is because we are now done completely with derivatives. But the way to pick out these problems is when it asks you to approximate. Also you use differentials to approximate the linearization. Differential is something that has been solved for dy or dx.
The steps for solving linearization problems are:
1. Pick out the equation
2. f(x)+f`(x)dx
3. Figure out your dx
4. Figure out your x
5. Plug in everything you get
Also I am very comfortable with tangent lines. All you do is:
1. Take the derivative of the equation
2. Plug in the x value which gives you your slope
3. Use the slope you get and the point given and plug into slope intercept form (y-y1)=slope(x-x1)
*If a point is not given and only an x value is given plug the x value into the original which will give you a y value creating a point.
I did also learn a trick that helped me some with optimization. I learned that if you are looking for the optimization of a rectangle the answer will turn out to be a square.
I am not comfortable with problems like number 2 on the packet she gave us. It has to do with velocity. Also if someone could refresh my memory on instantaneous speed steps it would be appreciated because I can not seem to be able to find my notes on this.
The steps for solving linearization problems are:
1. Pick out the equation
2. f(x)+f`(x)dx
3. Figure out your dx
4. Figure out your x
5. Plug in everything you get
Also I am very comfortable with tangent lines. All you do is:
1. Take the derivative of the equation
2. Plug in the x value which gives you your slope
3. Use the slope you get and the point given and plug into slope intercept form (y-y1)=slope(x-x1)
*If a point is not given and only an x value is given plug the x value into the original which will give you a y value creating a point.
I did also learn a trick that helped me some with optimization. I learned that if you are looking for the optimization of a rectangle the answer will turn out to be a square.
I am not comfortable with problems like number 2 on the packet she gave us. It has to do with velocity. Also if someone could refresh my memory on instantaneous speed steps it would be appreciated because I can not seem to be able to find my notes on this.
post 11
this week has been crazy tiring w/ homecoming and such but i still ended up gettin alot accomplished. this week we learned linearization. linearization has to do w/ the word approximate. anytime you see the word "approximate" in a problem. it means to use linearization.
the formula for linearization: f(x) = f(c) + f(c)(x-c)
we went over related rates and had a quiz on it.
you're given a problem like, you have a spherical object moving , rate of change is ___ and dy/dx = ____ so find dx/dt. that type of problem requires you to draw out the picture yourself.
but you can also be given a picture with the problem, which can be very helpful.
steps, once again, are:
1. first identify all variables in equation
2. figure out what you want to find(what you are solving the equation for)
3. dont forget to sketch and label! *extra points
4. write an equation involving all your variables(plug in)
5. take derivative
6. solve
im much beetter at optimization after doing it over and over again. i also know horiztontal tangents
Determine all values of x, if any, at which the graph of the function has a horizontal tangent.
y = x^3 + 12x^2 + 5
All you have to do is take the derivative and set it equal to zero, then solve for x.
3x^2
3x^2 + 24x = 0
3x( x + 8) = 0
X = 0, x = -8
im not too good at angles of elevation really. anyonoe wanna explain a problem?
the formula for linearization: f(x) = f(c) + f(c)(x-c)
we went over related rates and had a quiz on it.
you're given a problem like, you have a spherical object moving , rate of change is ___ and dy/dx = ____ so find dx/dt. that type of problem requires you to draw out the picture yourself.
but you can also be given a picture with the problem, which can be very helpful.
steps, once again, are:
1. first identify all variables in equation
2. figure out what you want to find(what you are solving the equation for)
3. dont forget to sketch and label! *extra points
4. write an equation involving all your variables(plug in)
5. take derivative
6. solve
im much beetter at optimization after doing it over and over again. i also know horiztontal tangents
Determine all values of x, if any, at which the graph of the function has a horizontal tangent.
y = x^3 + 12x^2 + 5
All you have to do is take the derivative and set it equal to zero, then solve for x.
3x^2
3x^2 + 24x = 0
3x( x + 8) = 0
X = 0, x = -8
im not too good at angles of elevation really. anyonoe wanna explain a problem?
Post #11
Lets just jump right in to an example problem since I'm great at these:
The radius, r, of a circle is increasing at a rate of 2 centimeters per minute. Find the rate of change of area, A, when the radius is 4.
A= pi(r)2 dr/dt=2 dA/dt=?
2(pi)r(dy/dt)
2(pi)(4)2
2(pi)8
= 16pi
These are real simple. First, I identify what the problem is trying to find. Then I copy what is given. These problems usually involve the area of a circle or volume of a cone or sphere. So, we should probably start memorizing those. After, take the derivative of the equation and just plug in.
Another example problem, just because I love doing these:
A point is moving along the graph of the function y=sin3x such that dx/dt=2 centimeters per second. Find dy/dt when x=pi/7.
y=sin3x dx/dt=2 dy/dt=?
dy/dt= cos3x(3)dy/dx
cos3(pi/7)(3)2
dy/dt= cos 3pi/7(6)
dy/dt- 6cos(3pi/7)
Also, I understand how to find d^2y/dx^2, but for some reason I have trouble getting the right answer. I think I might just make stupid little mistakes.
Soooo for what I'm still having major trouble with:
Angle of Elevation
linearization: like use differentials to approximate the square root of 16.5
Help please :)
The radius, r, of a circle is increasing at a rate of 2 centimeters per minute. Find the rate of change of area, A, when the radius is 4.
A= pi(r)2 dr/dt=2 dA/dt=?
2(pi)r(dy/dt)
2(pi)(4)2
2(pi)8
= 16pi
These are real simple. First, I identify what the problem is trying to find. Then I copy what is given. These problems usually involve the area of a circle or volume of a cone or sphere. So, we should probably start memorizing those. After, take the derivative of the equation and just plug in.
Another example problem, just because I love doing these:
A point is moving along the graph of the function y=sin3x such that dx/dt=2 centimeters per second. Find dy/dt when x=pi/7.
y=sin3x dx/dt=2 dy/dt=?
dy/dt= cos3x(3)dy/dx
cos3(pi/7)(3)2
dy/dt= cos 3pi/7(6)
dy/dt- 6cos(3pi/7)
Also, I understand how to find d^2y/dx^2, but for some reason I have trouble getting the right answer. I think I might just make stupid little mistakes.
Soooo for what I'm still having major trouble with:
Angle of Elevation
linearization: like use differentials to approximate the square root of 16.5
Help please :)
11
This week in calc was basically a review with a couple new things added on here and there.
We continued to work on implicit derivatives, related rates and angle of elevation which i am not completely comfortable with. We were also introduced to linearization at the end of the week.
The key word in linearization is approximate
EQUATION: f(x) = f(c) + f'(c) (x-c).
related rates:
1. Identify all of the variables and equations.
2. Identify what you want to find.
3. Sketch and label.
4. Write an equations involving your variables.
5. Take the derivative with in terms of time.
6. Substitute the derivative in and solve.
implicit derivatives:
First Derivative
Second Derivative
We continued to work on implicit derivatives, related rates and angle of elevation which i am not completely comfortable with. We were also introduced to linearization at the end of the week.
The key word in linearization is approximate
EQUATION: f(x) = f(c) + f'(c) (x-c).
related rates:
1. Identify all of the variables and equations.
2. Identify what you want to find.
3. Sketch and label.
4. Write an equations involving your variables.
5. Take the derivative with in terms of time.
6. Substitute the derivative in and solve.
implicit derivatives:
First Derivative
- 1. take the derivative of both sides
- 2. everytime you take the derivative of y note it with dy/dx or y^1
- 3. solve for dy/dx
Second Derivative
- first you find the first derivative and solve it for dy/dx by using the steps for the first derivative steps.
- you then take the second derivative of the solved equation. Plugging in d^2y/d^2x everytime you take the derivative of y again. and where you have dy/dx you plug in your solved equation for that.
- once you have everything plugged in and ready to go you then solve for d^2y/d^2x
EXAMPLE:
y=SIN(x)(solve for dy)
dy/dx = COS(x)
dy = [(dx)*(COS(x))]
what i don't understanddddddddddd. I think i would be able to grasp almost everything except i havent takin the time to really sit down and try to understand it, especially since everything was so busy last week, i was just exausted. but anyways, i don't really know what linearization is and the purpose of it
posting...#11
This week that we call the 11th week of blogging; we didn’t really learn anything new we just mostly reviewed old stuff such an angle of elevation and related rates which I still have a lot of trouble with just like everything else. But the new thing we did learn I actually understand I was surprised at first. We learned linearization at the end of the week and I know how to do it. I think it’s called linearization. We also got this big packet that I really don’t know how to do that’s do Monday so im nervous about that.
Related Rates: 1. identify all variables and equations.2. identify what you are looking for.3. make a sketch and label.4. write an equation involving your variables. you can only have an unknown so a secondary equation may be given.5. take the derivative with respect to time.6. substitute in derivative and solve.an example problem would be
the variable x and y are differentiable functions of t and are related by the equation
y=2x^3-x+4 when x=2 dx/dt=1. find dy/dt when x=2.dy/dt=?x=2 dx/dt=-1 y=2x^3-x+4dy/dt=6x^2dx/dt-dx/dtdy/dt=6(2)^2(-1)+(1)dy/dt=-23
Now for everything I don’t know which is a lot my biggest problem is still trying to find the equation for the problem and im bad at related rates angle of inclination and optimization so any help that you can give and im also bad at that packet so if anyone wants to help me out be my guess.
Related Rates: 1. identify all variables and equations.2. identify what you are looking for.3. make a sketch and label.4. write an equation involving your variables. you can only have an unknown so a secondary equation may be given.5. take the derivative with respect to time.6. substitute in derivative and solve.an example problem would be
the variable x and y are differentiable functions of t and are related by the equation
y=2x^3-x+4 when x=2 dx/dt=1. find dy/dt when x=2.dy/dt=?x=2 dx/dt=-1 y=2x^3-x+4dy/dt=6x^2dx/dt-dx/dtdy/dt=6(2)^2(-1)+(1)dy/dt=-23
Now for everything I don’t know which is a lot my biggest problem is still trying to find the equation for the problem and im bad at related rates angle of inclination and optimization so any help that you can give and im also bad at that packet so if anyone wants to help me out be my guess.
eleven
week 11 in calculus. we learned linearization.
linearization = approixmate. anytime you see the word "approximate" in a problem. it means to use linearization.
the formula for linearization: f(x) = f(c) + f(c)(x-c)
also, a differential is when something is solved for dy or dx.
alrighttttt, so. related rates are super easy now. ill explain those :)
you're given a problem usually something like, you have a spherical balloon going up, rate of change is ___ and dy/dx = ____ so find dx/dt. something like that.
^^^that type of problem requires you to draw out the picture yourself.
but you can also be given a picture with the problem, which can be very helpful.
steps, once again, are:
1. first identify all variables in equation
2. figure out what you want to find(what you are solving the equation for)
3. dont forget to sketch and label! *extra points
4. write an equation involving all your variables(plug in)
5. take derivative
6. solve
simple, right? kind of like optimization. but easier!
another recent thing we learned that is very simple. or i caught onto easily at least is implicit derivatives. you know it's an implicit derivative if you are given an x and y value in your equation. you simply take the derivative, noting all derivatives of y-values with dy/dx or y'. then solve for dy/dx or y'. it's quite easy.
what i don't understand: linearization and how to find angles of elevation.
linearization = approixmate. anytime you see the word "approximate" in a problem. it means to use linearization.
the formula for linearization: f(x) = f(c) + f(c)(x-c)
also, a differential is when something is solved for dy or dx.
alrighttttt, so. related rates are super easy now. ill explain those :)
you're given a problem usually something like, you have a spherical balloon going up, rate of change is ___ and dy/dx = ____ so find dx/dt. something like that.
^^^that type of problem requires you to draw out the picture yourself.
but you can also be given a picture with the problem, which can be very helpful.
steps, once again, are:
1. first identify all variables in equation
2. figure out what you want to find(what you are solving the equation for)
3. dont forget to sketch and label! *extra points
4. write an equation involving all your variables(plug in)
5. take derivative
6. solve
simple, right? kind of like optimization. but easier!
another recent thing we learned that is very simple. or i caught onto easily at least is implicit derivatives. you know it's an implicit derivative if you are given an x and y value in your equation. you simply take the derivative, noting all derivatives of y-values with dy/dx or y'. then solve for dy/dx or y'. it's quite easy.
what i don't understand: linearization and how to find angles of elevation.
Post #11
This week in calculus, we learned linearization, took a quiz, and had review packets for our test Wednesday.
The key word in linearization is approximate and that f(x) = f(c) + f'(c) (x-c).
For linearization you can be asked to approximate the tangent line to y= x^3 at x=2
Take the derivative : 3x^2
Plug in your x: 3(2)^2 = 12
Find y by plugging in for x: y= (2)^3 = 8
Plug into your equation: f(x) = 8+12(x-8)
You may be asked to use differentials to approximate something. The steps for doing these problems are:
1. Identify an equation
2. f(x)+f'(x)dx
3. Determine dx and x
4. Plug into your equation
EXAMPLE:
1. the square root of 65.4
Equation: the square root of x
f(x) + f'(x) dx = the square root of x + 1/ 2(the square root of x) dx
dx= .4
x= 65
Plug in: the square root of 65+ 1/ 2(the square root of 65) (.4)
= 65.155
We also learned how to solve for dy.
Example:
y= 5x^2-6
dy/dx= 10x
dy= 10x(dx)
I am still having problems with angle of elevation and I also have a question on number 22 on the packet I have. The question says: A man 6 feet tall walks at a rate of 4 feet per second away from a light that is 15 feet about the ground. When he is 10 feet from the base of the light, at what rate is the length of his shadow changing?
I know when you are looking for the rate of the tip of his shadow, you have to set up a proportion and solve for a, then take the derivative and plug in for dx/dt, but since this problem is asking for the rate of the length I think you have to work it differently. So if anyone knows how to work it please fill me in.
Post Number Eleven
Wowww I just did my whole blog but when I clicked publish post there was a conflicting error and it deleted the whole thing. Here I go againnnnnnnnnnn
This week in Calculus we learned linearization. The first thing you need to know about this is the word approximate. Any time you see this word know that it is a linearization problem. The formula for linearization is f(x) = f(c) + f(c ) (x – c). Differential means when something is solved for dy or dx.
Example:
Approximate the tangent line to y = x^2 at x=1.
Dy/dx = 2x ** derivative
y = (1)^2 = 1 ** plug in x into original
dy/dx = 2 ** plug in x into derivative
f(x) = 1 + 2 (x – 1)
Fairly simple right?
Another example is
Use differentials to approximate the square root of 16.5
1. Identify an equation: f(x) = square root of x
2. f(x) + f’(x)dx: square root of x + (1)/2(square root of x)(dx)
3. Determine dx: (decimal and what follows) .5
4. Determine x: (before the decimal) 16
5. Plug in: square root of 16 + (1)/2(square root of 16)(.5)
=4.0625
Error = .0005
I thought I understood linearization when we learned it in class and when I went home and did the homework, but now I am starting to second guess myself.
One thing I am completely comfortable with is horizontal tangents:
Determine all values of x, if any, at which the graph of the function has a horizontal tangent.
y = x^3 + 12x^2 + 5
All you have to do is take the derivative and set it equal to zero, then solve for x.
3x^2
3x^2 + 24x = 0
3x( x + 8) = 0
X = 0, x = -8
One thing I am not completely comfortable with is related rates. I understand how to work the problems where everything is point blank given but I have the most trouble with the recognizing what is what inside the word problem. If anyone knows how to help with this it’d be greatly appreciated.
Also, angles of elevation have blown my mind. I need serious help with this if anyone wishes to.
The most trouble I find I am having is knowing what is given to me and what to classify it as. I know all the steps to all the types of problem and can work them once I recognize what is what, it’s just getting to that point that confuses me. I don’t know if anyone can help me with this, I know it takes a lot of practice it just is so frustrating!
This week in Calculus we learned linearization. The first thing you need to know about this is the word approximate. Any time you see this word know that it is a linearization problem. The formula for linearization is f(x) = f(c) + f(c ) (x – c). Differential means when something is solved for dy or dx.
Example:
Approximate the tangent line to y = x^2 at x=1.
Dy/dx = 2x ** derivative
y = (1)^2 = 1 ** plug in x into original
dy/dx = 2 ** plug in x into derivative
f(x) = 1 + 2 (x – 1)
Fairly simple right?
Another example is
Use differentials to approximate the square root of 16.5
1. Identify an equation: f(x) = square root of x
2. f(x) + f’(x)dx: square root of x + (1)/2(square root of x)(dx)
3. Determine dx: (decimal and what follows) .5
4. Determine x: (before the decimal) 16
5. Plug in: square root of 16 + (1)/2(square root of 16)(.5)
=4.0625
Error = .0005
I thought I understood linearization when we learned it in class and when I went home and did the homework, but now I am starting to second guess myself.
One thing I am completely comfortable with is horizontal tangents:
Determine all values of x, if any, at which the graph of the function has a horizontal tangent.
y = x^3 + 12x^2 + 5
All you have to do is take the derivative and set it equal to zero, then solve for x.
3x^2
3x^2 + 24x = 0
3x( x + 8) = 0
X = 0, x = -8
One thing I am not completely comfortable with is related rates. I understand how to work the problems where everything is point blank given but I have the most trouble with the recognizing what is what inside the word problem. If anyone knows how to help with this it’d be greatly appreciated.
Also, angles of elevation have blown my mind. I need serious help with this if anyone wishes to.
The most trouble I find I am having is knowing what is given to me and what to classify it as. I know all the steps to all the types of problem and can work them once I recognize what is what, it’s just getting to that point that confuses me. I don’t know if anyone can help me with this, I know it takes a lot of practice it just is so frustrating!
11th post
This week in calculus we learned linearization and mrs. robinson made us write down our top five things we did not understand so far in calculus. She then took the major problems we were having and made a packet out of them so we could practice for the test on wednesday.
I finally have a grip on the horizontal tangent problems. First you take the derivative of the function and set it equal to zero and solve for your x values. Whatever you get for x is your values that have horizontal tangents.
What i understand the most out of all we have learned so far is related rates.. i feel very comfortable working these problems. let's look at an example of one.
They tell you that the radius of a right circular cylinder is [the square root of(3t+5)] and that the height is t^7. They want you to find the rate of change of the volume of the cylinder.
1. The first step is figuring out the formula for the volume of a right circular cylinder which is V= (pie) r^2h.
2. Next plug in what you are given V= (pie) (the square root of 3t+5)^2 (t^7)
Also note that the square root will cancel because it is being raised to the second power.
So now we are left with V= (pie) (3t+5) (t^7)
3. Now distribute the t^7 with the 3t+5 to give you:
V= (pie) (3t^8 +5t^7)
4. Now take the derivative of the function:
dV/dt= (pie) (24t^7 + 35 t^6)
5. Now you can take out a t^6
6. The answer will be dV/dt= (pie) (t^6) (35 + 24t)
What i still do not understand is angle of elevation. I still do not understand what i am looking for and how to find it. I know it is like related rates at one point but i do not understand how to get to that point. Another point i am confused on is linearization. I do not quite grasp the concept of it and where and how it is used. If anyone can help with these problems i would greatly appreciate it. Good luck to everyone on the test this week :)
I finally have a grip on the horizontal tangent problems. First you take the derivative of the function and set it equal to zero and solve for your x values. Whatever you get for x is your values that have horizontal tangents.
What i understand the most out of all we have learned so far is related rates.. i feel very comfortable working these problems. let's look at an example of one.
They tell you that the radius of a right circular cylinder is [the square root of(3t+5)] and that the height is t^7. They want you to find the rate of change of the volume of the cylinder.
1. The first step is figuring out the formula for the volume of a right circular cylinder which is V= (pie) r^2h.
2. Next plug in what you are given V= (pie) (the square root of 3t+5)^2 (t^7)
Also note that the square root will cancel because it is being raised to the second power.
So now we are left with V= (pie) (3t+5) (t^7)
3. Now distribute the t^7 with the 3t+5 to give you:
V= (pie) (3t^8 +5t^7)
4. Now take the derivative of the function:
dV/dt= (pie) (24t^7 + 35 t^6)
5. Now you can take out a t^6
6. The answer will be dV/dt= (pie) (t^6) (35 + 24t)
What i still do not understand is angle of elevation. I still do not understand what i am looking for and how to find it. I know it is like related rates at one point but i do not understand how to get to that point. Another point i am confused on is linearization. I do not quite grasp the concept of it and where and how it is used. If anyone can help with these problems i would greatly appreciate it. Good luck to everyone on the test this week :)
11th post
okay so this is the eleventh week of calculus and this week was about related rates more than anything. i think im okay a little with related rates i just need to practice.
Related Rates:
1. identify all variables and equations.
2. identify what you are looking for.
3. make a sketch and label.
4. write an equation involving your variables. you can only have an unknown so a secondary equation may be given.
5. take the derivative with respect to time.
6. substitute in derivative and solve.
example: the variable x and y are differentiable functions of t and are related by the equation y=2x^3-x+4 when x=2 dx/dt=1. find dy/dt when x=2.
dy/dt=?
x=2 dx/dt=-1 y=2x^3-x+4
dy/dt=6x^2dx/dt-dx/dt
dy/dt=6(2)^2(-1)+(1)
dy/dt=-23
Implicit Derivatives:
1. involves xs and ys
2.STEPS:
a. take the derivative like normal of both sides
b. everytime you take the derivative of y note it with dy/dx or y1
c. solve for dy/dx
3. if you want the slope you must plug in a x and a y value.
Intermediate Value Theorem:\
1. if f is continuous on [a,b] and k is any number between f(a)and f(b), then there is at least 1 number c when f(c)=k.
* basically you cannot skip any y value
HOW TO FIND THE EQUATION OF A TANGENT LINE:
1. take f1(x)
2. plug x in to find your slope m
3. plug x into f(x)to get y
4. using m and (x,y) plug it into the equation (y-y1)=m(x-x1).
okay im still having trouble with this optimazation stuff so.....
Optimazation:
1. identify primary and secondary equations
*primary- the one your maximizing or minimizing
*secondary- the other one
2. solve secondary for 1 variable and plug into primary .
*if primary only has 1 variable this step is not necessary.
3. take derivative of primary;set=0;solve for x.
4. plug into secondary equation to find other values. check end points if necessary. see examples for which ones you need to check endpoints for.
okay as i said before, im still having trouble figuring out optimazation and implicit derivatives so if someone can please help me out..
Related Rates:
1. identify all variables and equations.
2. identify what you are looking for.
3. make a sketch and label.
4. write an equation involving your variables. you can only have an unknown so a secondary equation may be given.
5. take the derivative with respect to time.
6. substitute in derivative and solve.
example: the variable x and y are differentiable functions of t and are related by the equation y=2x^3-x+4 when x=2 dx/dt=1. find dy/dt when x=2.
dy/dt=?
x=2 dx/dt=-1 y=2x^3-x+4
dy/dt=6x^2dx/dt-dx/dt
dy/dt=6(2)^2(-1)+(1)
dy/dt=-23
Implicit Derivatives:
1. involves xs and ys
2.STEPS:
a. take the derivative like normal of both sides
b. everytime you take the derivative of y note it with dy/dx or y1
c. solve for dy/dx
3. if you want the slope you must plug in a x and a y value.
Intermediate Value Theorem:\
1. if f is continuous on [a,b] and k is any number between f(a)and f(b), then there is at least 1 number c when f(c)=k.
* basically you cannot skip any y value
HOW TO FIND THE EQUATION OF A TANGENT LINE:
1. take f1(x)
2. plug x in to find your slope m
3. plug x into f(x)to get y
4. using m and (x,y) plug it into the equation (y-y1)=m(x-x1).
okay im still having trouble with this optimazation stuff so.....
Optimazation:
1. identify primary and secondary equations
*primary- the one your maximizing or minimizing
*secondary- the other one
2. solve secondary for 1 variable and plug into primary .
*if primary only has 1 variable this step is not necessary.
3. take derivative of primary;set=0;solve for x.
4. plug into secondary equation to find other values. check end points if necessary. see examples for which ones you need to check endpoints for.
okay as i said before, im still having trouble figuring out optimazation and implicit derivatives so if someone can please help me out..
Post # 11
This week in Calculus we reviewed and I began to understand many new things. Things that i understand include related rates, they came to my understanding with time. I still have trouble with the problems when it isn't clear how to figure out what everything is like your dy/dx and that kind of stuff.
So, related rates:
1. Pick out and identify all variables and equations
2. Find out what you're looking for
3. Make a sketch and label what you know so far and what you're looking for
4. Find all equations
5. Take derivative with time in mind..
DON'T TAKE DERIVATIVE UNTIL YOU CHECK HOW ANSWER CHOICES ARE
6. Plug in derivative and solve
Example Problem:
dx/dt y + dy/dt = 0
now plug inn..
dy/dt = -10y/8
= -5y/4
not so hard..they are really easy if you practice them alottt.
So, the thing i don't understand is when to put dx/dt behind taking the derivative of x..
do you do it every time, only sometimes, or for certain equations?
So, implicit derivatives for the third week. Are still as easy as ever, i think the hardest part of an implicit derivative is remembering to put the dy/dx behind taking the derivative of y.
BUT THAT IS VERY IMPORTANT, YOU DON'T PUT DY/DX BEHIND THE Y, YOU PUT IT AFTER TAKING TEH DERIVATIVE OF Y.
Now, lets talk about linearization. I completely forgot how to do this..but i don't think i understood it when b-rob taught it..
unless im getting my names mixed up, i think these were word problems too..right?
So, related rates:
1. Pick out and identify all variables and equations
2. Find out what you're looking for
3. Make a sketch and label what you know so far and what you're looking for
4. Find all equations
5. Take derivative with time in mind..
DON'T TAKE DERIVATIVE UNTIL YOU CHECK HOW ANSWER CHOICES ARE
6. Plug in derivative and solve
Example Problem:
dx/dt y + dy/dt = 0
now plug inn..
dy/dt = -10y/8
= -5y/4
not so hard..they are really easy if you practice them alottt.
So, the thing i don't understand is when to put dx/dt behind taking the derivative of x..
do you do it every time, only sometimes, or for certain equations?
So, implicit derivatives for the third week. Are still as easy as ever, i think the hardest part of an implicit derivative is remembering to put the dy/dx behind taking the derivative of y.
BUT THAT IS VERY IMPORTANT, YOU DON'T PUT DY/DX BEHIND THE Y, YOU PUT IT AFTER TAKING TEH DERIVATIVE OF Y.
Now, lets talk about linearization. I completely forgot how to do this..but i don't think i understood it when b-rob taught it..
unless im getting my names mixed up, i think these were word problems too..right?
Post #11
Another week of Calculus gone. During that week we continued to review! For instance we took a quiz and then wrote down five things that we need help with. Mrs. Robinson made us a packet with the top seven types of problems that we needed help on. It's due monday due to some people [NOT ALL] not doing their work in class or asking her to help them. But anyways, I feel that working the problems have helped me a little, however i'm still strugling with Angle of Elevation and the problems that say something like the building is 220 and the initial velocity is -12 ft per second, then an equation is given and it asks you what is the velocity at a certain time..TO ME, that's veryyy confusing!
Wednesday we learned Linearization and the any time we see the word APPROXIMATE we will be using linerarization. For example:
Approximate the tangent line to y=(x)^2 at x=5
that means that dy/dx is 2x
so when we plug in 5 for x we get 2(5) which is 10. 10 is our slope!. Now we will take the original and plug in 5 for x giving us y=(5)^2, so that's y=25.
Since the equation for Linerarization is f(x) = f(c) + f'(c)*(x-c) we will plug in what we have into the equation. we know that f(c)=25...f'(c)=2...and c=5 so when we plug in we get>>>> f(x)=25+2(x-5)
We learned that a differential is when something is solved for dy or dx like:
y=cos(x)---solve for dy
dy/dx = -sin(x)
dy = [(dx)*(-sin(x))]
When we use differentials to approximate there are steps involved:
1. Identify Equation
2. f(x) + f'(x)*dx
3. determine dx
4. determine x
5. Plug in and solve!
For example: (number one on the Approximate part of our wednesday night
homework) to approximate the square root of 99.4 we use the
steps...
~the equation is -- f(x) = the square root of (x)
~the square root of [x] + [(1)/(2 * the square root of [x]) multiplied by (dx)]
~ dx = .4
~ x = 99
~ plug in: [the square root of 99] + [(1) / (2 * the square root of 99) multiplied
by (.4)]
~when solved:....9.96997, the error is 0.0002
I'm still having trouble with the problems on the packet, but other than that I THINK i'm okay!
Wednesday we learned Linearization and the any time we see the word APPROXIMATE we will be using linerarization. For example:
Approximate the tangent line to y=(x)^2 at x=5
that means that dy/dx is 2x
so when we plug in 5 for x we get 2(5) which is 10. 10 is our slope!. Now we will take the original and plug in 5 for x giving us y=(5)^2, so that's y=25.
Since the equation for Linerarization is f(x) = f(c) + f'(c)*(x-c) we will plug in what we have into the equation. we know that f(c)=25...f'(c)=2...and c=5 so when we plug in we get>>>> f(x)=25+2(x-5)
We learned that a differential is when something is solved for dy or dx like:
y=cos(x)---solve for dy
dy/dx = -sin(x)
dy = [(dx)*(-sin(x))]
When we use differentials to approximate there are steps involved:
1. Identify Equation
2. f(x) + f'(x)*dx
3. determine dx
4. determine x
5. Plug in and solve!
For example: (number one on the Approximate part of our wednesday night
homework) to approximate the square root of 99.4 we use the
steps...
~the equation is -- f(x) = the square root of (x)
~the square root of [x] + [(1)/(2 * the square root of [x]) multiplied by (dx)]
~ dx = .4
~ x = 99
~ plug in: [the square root of 99] + [(1) / (2 * the square root of 99) multiplied
by (.4)]
~when solved:....9.96997, the error is 0.0002
I'm still having trouble with the problems on the packet, but other than that I THINK i'm okay!
Post #11
Calculus Week #11
In this week of Calculus, we focused on reviewing for our short quiz we had, and then we all turned in ideas that we were uncomfortable with so that we could get packets on those topics in preparation for our next major test, which is a test on all of derivatives before we move on to another topic in Calculus.
As for my example, I'm going to explain how to do a related rate problem with a conical tank.
A conical tank (with vertex down) is 10 feet across the top and 12 feet deep. If water is flowing into the tank at a rate of 10 cubic feet per minute, find the rate of change of the depth of the water when the water is 8 feet deep.
Now, to start, we know that it is a related rates problem. We also know that dV/dt = 10ft/min and we want to find dh/dt (the depth) when the h=8ft. Also, we know for a conical tank, the volume is given by V = 1/3 pi r^2 h. Now, if we would take the implicit derivative with respect to time at this particular point we would have two unknowns. So, to work this out, we will solve for h in terms of r.
Setting up the proportion:
r/h = 10/12
We can solve this for r to find 5h/6.
Now we can plug this into our original and then take the implicit derivative.
dV 25 pi h^2 dh
-- = --------- --
dt 36 dt
We can pug in for dV/dt and for h=8 then solve for dh/dt to find our answer of:
dh/dt = 9/(40pi)
It's really not that hard of a problem. Just knowing the formula and the fact that you have to use a secondary equation (kind of like optimization) so that you only have one unknown after taking the derivative is all that is needed.
Anyway, good luck on this!
In this week of Calculus, we focused on reviewing for our short quiz we had, and then we all turned in ideas that we were uncomfortable with so that we could get packets on those topics in preparation for our next major test, which is a test on all of derivatives before we move on to another topic in Calculus.
As for my example, I'm going to explain how to do a related rate problem with a conical tank.
A conical tank (with vertex down) is 10 feet across the top and 12 feet deep. If water is flowing into the tank at a rate of 10 cubic feet per minute, find the rate of change of the depth of the water when the water is 8 feet deep.
Now, to start, we know that it is a related rates problem. We also know that dV/dt = 10ft/min and we want to find dh/dt (the depth) when the h=8ft. Also, we know for a conical tank, the volume is given by V = 1/3 pi r^2 h. Now, if we would take the implicit derivative with respect to time at this particular point we would have two unknowns. So, to work this out, we will solve for h in terms of r.
Setting up the proportion:
r/h = 10/12
We can solve this for r to find 5h/6.
Now we can plug this into our original and then take the implicit derivative.
dV 25 pi h^2 dh
-- = --------- --
dt 36 dt
We can pug in for dV/dt and for h=8 then solve for dh/dt to find our answer of:
dh/dt = 9/(40pi)
It's really not that hard of a problem. Just knowing the formula and the fact that you have to use a secondary equation (kind of like optimization) so that you only have one unknown after taking the derivative is all that is needed.
Anyway, good luck on this!
Subscribe to:
Posts (Atom)