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

1. 1. take the derivative of both sides
2. 2. everytime you take the derivative of y note it with dy/dx or y^1
3. 3. solve for dy/dx

Second Derivative

1. first you find the first derivative and solve it for dy/dx by using the steps for the first derivative steps.
2. 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.
3. 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.

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.

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!

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 :)

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..

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?

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!

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!