Homework Help v1.3

[o({})o]

Originally Posted by Ice(v)an

How do I make each move square root of five distance

Moving 2 spaces in one direction and 1 space in the other. 2^2 + 1^2 = 5... so distance = sqrt(5)

[Ice(v)an]

I'm not really sure what you mean by that.

[MuKen]

Originally Posted by Ice(v)an

I'm not really sure what you mean by that.

Picture the grid like a small chess board, every move must be like a knight's move.

[o({})o]

Originally Posted by Ice(v)an

I'm not really sure what you mean by that.

                                       
                                       
      |\                                 
      | \                                
      |  \                               
      |   \ √(5)                              
   2  |    \                             
      |     \                            
      |      \                           
      --------                                 
          1                             


(1)2 + (2)2 = 5 = (√(5))2 

[Keerua]

I have a questions,

Given that f(x,y)= ((x^-1)+(y^-1)) find F'x and F'y

I just guessed but is it:

F'x= -1/(x)^2 and F'y= -1/(y)^-2

Also

Given z= F(x,y) and x=f(t) while y= g(t,s) express dz/dt and dz/ds

Here i dont know how to make a guess, is the first one, dz/dt simply x'? But how do you express x as y??

Edit: Lol sorry for the bad english, I did this on my ipod touch on the trainstation cause this is urgent (paper due 2 monday) I have sort of missed 99% of classes, relying entirely on the book but I am sort of fucked cause alot of the questions are from lectures

[Keerua]

^
Found some relevant info on the internet so i just need to know if what i have done is correct.

f'x= -1/x^2
f'y= -1/y^2

dz/dt= F'x(x,y)(dx/dt) + F'y(x,y) dy/dt

dz/ds= F'y(x,y) dy/ds

In the last one, do i take x as a constant since x isn't a function of s? Is the last one correct? I am pretty sure the rest is ok

[SHADENINJA]

First question is about "U Substitution"

Depending on the integration it can be done many different ways. I'm usually able to find the easiest scenario, but is there a rule of thumb for it that makes choosing the scenario simpler?

Also, concerning Newton's Law, I understand the concept of it, but how do i know when to stop? How do i know when I've passed the true zero?

[junkdemon]

I could prolly help out with mathematics as well.

[Keerua]

Originally Posted by ShadeNinja

First question is about "U Substitution"

Depending on the integration it can be done many different ways. I'm usually able to find the easiest scenario, but is there a rule of thumb for it that makes choosing the scenario simpler?

Also, concerning Newton's Law, I understand the concept of it, but how do i know when to stop? How do i know when I've passed the true zero?

From personal experience with u substitution you simply have to do alot of them, and after a while you will be able to see almost instantly what to pick.

[Nohia]

Originally Posted by ShadeNinja

First question is about "U Substitution"

Depending on the integration it can be done many different ways. I'm usually able to find the easiest scenario, but is there a rule of thumb for it that makes choosing the scenario simpler?

Also, concerning Newton's Law, I understand the concept of it, but how do i know when to stop? How do i know when I've passed the true zero?

Integration by substitution really does just take a lot of practice and examples in order to master. Unless you have some abstract integral that has the derivatives of weird functions inside (like inverse trig functions or hyperbolics which you'll usually use a integral/derivative table for) things will typically divide out smoothly, otherwise you use integration by parts or trig intergration, or just algebra manipulation. One inverse trig function that you should memorize as it pops up in many applications is the arctan. dy/dx arctan = 1/(1 + x^2)

Your question about Newton's Law, I'm not following. Newton has several main laws of motion (inertial, acceleration, reciprocation) that I am familiar with, but don't know what this "true zero" is. It's probably something I just haven't learned yet, but if you could elaborate I might be able to dig something up.

[Keerua]

^ I think he is talking about the newton quotient, the limit formula, not sure though

[Dendra]

I know theres a homework help thread but nobody has used it in a long time so I was afraid that people wouldn't have seen my post. If this thread has to be closed then its okay, I can wait.

But I'm in grade 10 applied math and so far we're studying y=mx+b, slope, y-intercepts etc. This week i have mid-term exams because teachers don't want to "stress" us out at the end of the year, I'm very grateful for that. But can anybody tell me how I would find an equation of a line by a) looking at a graph or b) using points: (1,3) & (5,8). Which is y2-y1/x2-x1.

If anybody knows this can you please help. I ask my teacher but his method is hard and i can't understand. So if you do know then help, if you don't then please don't put anything down, because it will confuse me even more. If nobody knows, thats ok, i'll just ask somebody else.

[Keerua]

Originally Posted by Dendra

I know theres a homework help thread but nobody has used it in a long time so I was afraid that people wouldn't have seen my post. If this thread has to be closed then its okay, I can wait.

But I'm in grade 10 applied math and so far we're studying y=mx+b, slope, y-intercepts etc. This week i have mid-term exams because teachers don't want to "stress" us out at the end of the year, I'm very grateful for that. But can anybody tell me how I would find an equation of a line by a) looking at a graph or b) using points: (1,3) & (5,8). Which is y2-y1/x2-x1.

If anybody knows this can you please help. I ask my teacher but his method is hard and i can't understand. So if you do know then help, if you don't then please don't put anything down, because it will confuse me even more. If nobody knows, thats ok, i'll just ask somebody else.

a= (Y2-Y1)/(X2-X1)

Y= a(x-x1)+y1 equation for a straight line

that is what i vaguely remember it as, thread might get locked soon lol for you posted in the wrong place.

a= the slope, which i hope you know how to calculate (even gave u the formula)

x1: 1
x2: 5
Y1: 3
y2: 8

[Dendra]

Yeah i know its going to be closed, but i really need help. I have the slowest brain in the world. Its so stressful. But i don't mind, i just want to see if people know so they can help. And where would it have gone? I looked at I didn't know where to put it..

Y= a(x-x1)+y1 equation for a straight line this was the equation you gave. Is this y=mx+b? Slope is rise over run. I know that. I know how to calclate slope, but how do you find the y-intercept?

[Keerua]

the homework thread

yes that is y= mx+b once you solve it out

Y= ax- ax1 +y1
-ax1+y1= b
ax= mx

where -ax1+y1 is the intercept, on a graph the intercept is the point where the graph intercepts the y.axis. In the equation it is the constant (the part without X)

[Dking]

how the hell is y=mx+b 10th grade math? I learned that when i was in 7th.

Have you even gotten to parabolas or quadratics? I doubt your in 10th grade..but whatever.

Anyways,

y=mx+b basically means this:
Every line has atleast 2 points.
Every point has atleast 2 cords (x,y) You basically plug in those cords in the formula.

Ex. (3,4)(5,6)
you chose any pair of the 2, it doesnt matter which.
in this case, ill chose (3,4)
Here, x = 3 and y = 4
so the formula would be

4=m3+b (after you plug it in)

M = the slope, while B = the y intercept (where the line crosses the y axis)

to get the slope, there is a formula.

m = y2 - y1 / x2 - x1.
This basically means that you have to plug in the y's and the x's with the given cords.

Here are the cords we used again: (3,4)(5,6)

y2 = 6 - y1 = 4 / x2 = 5 - x1 = 3

So to get the slope, after you plugged it in, it would be:

6-4/5-3 = 1

So M = 1

4 = 3(1) + b

b is obviously 1 looking from the equation.

You can also plug in the other cords to check your answer. (5,6)

6 = 5(1) + 1

So now know b = 1 and m = 1, the final equation is

y=(1)x+1

--------------------------------------------

Now, finding out the equation from just looking at the graph isnt much different.

If no points are given, then chose 2 points that lay on the line in the graph.

Do the slope formula (m = y2 - y1 / x2 - x1) and find the y intercept by looking where the line crosses the y axis. Its that easy.

--------------------------------------------

You also gave specified cords (1,3) (5,8)

y = mx + b

lets find M:

8 - 3 / 5 - 1 = M
M = 5/4

now, plug in any pair, ill chose (1,3)

y = (5/4)x + b
3 = (5/4)1 + b
3 = 5/4 + b
(subtract 5/4 from both sides)
b = 1.75

Now, knowing that M = 5/4, and that B = 7/4 (or 1.75, same thing) plug it in.

y = mx + b
y = 5/4x + 7/4

that simple.

[Dendra]

^ Dking, I live in Mississauga. Me not knowing this proves that i'm in APPLIED gr.10. Why do you always have to start beef? But thank you for helping me. And same with the other guy, but why are you using a as slope? Isn't it m?

Okay so basically, all i do is use the points given to me to find the whole equation..i guess? (3,4)(5,6) These are your points. So when I do y2-y1/x2-x1, i use that number and thats m. But how come your putting it in fraction form? Then after i find slope, i choose any 2 points, so, 3,4. And i 4 would be y, and 3 would be x. So 4=(3/2)3+y Where do you subtract 3/2 from?

edit: acutally dking i'm going to copy/paste your example, print it and study off of it, (ha thanks!). But i still need an answer to the where you sub from. Other then that my question has been answer.

[Bara]

Please use the homework thread.

btw: homework thread lasted used April 5th. Not that long ago.

[o({})o]

Next time you disrespect those people who actively participate in the homework thread by discrediting it in your self-important blabber about it being an inactive thread, I'll just remove your thread before you get any answers. What's more, you openly acknowledged that you posted this in the wrong place... I'm really getting tired of you trolling about on this forum, doing as you please on a whim. That type of attitude is to be kept in The Lounge and/or GFX Spam area.

That stated, I'll merge this with the HW thread and then reply to it accordingly.

Your trouble seems to be fairly basic. You haven’t paid attention in class, or genuinely don’t understand the method by which your teacher told you to solve the problem. Given your personality and my profession, I’d tend toward siding with the teacher on this one. Still, I will extend to you the benefit of the doubt.

Slope-intercept Form for the equation of a line:
y = m * x + b

Point-slope Form for the equation of a line:
y – y₁ = m * (x - x₁)

Slope:
m = rise / run
m = (y₂ - y₁) / (x₂ - x₁)

Lets examine the meaning of all of these variables. All explanations apply to any standard equation form for a line...

x = The independent variable that we can replace with whatever value we want.
y = The dependent variable that relies on the value of x. We must plug a value into x before we can get a y-value back out of the equation. Once we plug in all the available x-values into the equation, we will have found all the corresponding y-values and will be able to plot points in the form (x, y) which will eventually make up a line.
m = The slope of the line associated with the equation. This is in the form of rise over run. Rise over run indicates that we are going up a certain number of units on our graph for every unit we move to the right (positive x side of the coordinate plane).
b = The y-intercept of the line associated with the equation. The y-intercept is where the line intersects (crosses) the y-axis. At this point, x = 0 and y = b. Therefore, it has coordinates (0, b) on the coordinate plane.

x₁ = The first value of x that we are given.
y₁ = The y-value in the equation that we would find if we plugged x₁ into the equation. y₁ depends on x₁.
x₂ = The second value of x that we are given.
y₂ = The y-value in the equation that we would find if we plugged x₂ into the equation. y₂ depends on x₂.

(x₁, y₁) and (x₂, y₂) are points on the line associated with our equation.

Lets now look at a graph and work an example:



In this graph, we see several points that the line clearly passes through.

From the graph, what is the y-intercept?

(0, 2) is the y-intercept as a point. Written as b, we would say b = 2. This is simply the y-value of the coordinate.

What is the slope of the line?

For every one unit we move to the right, we are going up one unit. This means that we have a rise of 1 for every 1 unit step to the right. This means we have a slope m = 1 / 1 = 1.

How could we have calculated the slope with an applied method?

To move from any x₁ to any other x₂, we must move x₂ - x₁ units. Similarly, to move in a vertical manner, we must move y₂ - y₁ units in order to go from any y₁ to y₂.

As always, we can use any x and y values that are on the graph. I picked them as follows:
(0, 2) and (1, 3) are clearly on the line representing our mystery equation.
x₁ = 0
y₁ = 2
x₂ = 1
y₂ = 3

Plugging these values into our m = (y₂ - y₁) / (x₂ - x₁) equation:

m = (3 - 2) / (1 - 0) = (1) / (1) = 1

This method has uncovered the same slope m = 1 through calculation as the previous method did through observation.

We have now discovered m and b, so we can write our equation. Which form should the equation of the line be in, and what is it?

Slope-intercept Form for the equation of a line:
y = m * x + b
y = 1 * x + 2
y = x + 2

So, our simplified, final equation is y = x + 2.

Lets now solve the same problem given only the same two points on the line that we noticed earlier. Assume you no nothing else about the line itself.

What form will this new question take on a test or worksheet?

Find the equation of the line passing through the points (0, 2) and (1, 3).

What info does this question provide for us?

x₁ = 0
y₁ = 2
x₂ = 1
y₂ = 3

Can we find the slope? If so, what is it?

This time, observation is out of the question. We must use the numbers this time:
m = (y₂ - y₁) / (x₂ - x₁)
m = (3 - 2) / (1 - 0) = (1) / (1) = 1

Now that we have the slope, do we need to find b?

No. One of our equation forms has no b in it and includes only information we have uncovered thus far.

Which form of the equation for a line should we use here?

Point-slope Form for the equation of a line:
y – y₁ = m * (x - x₁)

Use the form specified above to find the true equation of this line as requested by the question.

Point-slope Form for the equation of a line:
y – y₁ = m * (x - x₁)

x₁ = 0
y₁ = 2

m = 1

y - 2 = 1 * (x - 0)
y - 2 = 1 * (x) <----- We don't need to write the "- 0", because adding or subtracting 0 does not change the equation.
y - 2 = x <----------- We don't need to write the "1 *" because multiplying by 1 does not change the equation.
y - 2 + 2 = x + 2 <-- Add 2 to both sides of the equation. We are looking to cancel out the "- 2" on the left with the "+ 2" that we've added.
y = x + 2 <----------- The "- 2" and the "+ 2" canceled out as we had hoped, and we are left with a final equation.

Notice that we got the same equation using entirely different methods and entirely different starting questions.

Hopefully you can now apply this method to the example points you provided.

Originally Posted by Dendra

Okay so basically, all i do is use the points given to me to find the whole equation..i guess? (3,4)(5,6) These are your points. So when I do y2-y1/x2-x1, i use that number and thats m. But how come your putting it in fraction form? Then after i find slope, i choose any 2 points, so, 3,4. And i 4 would be y, and 3 would be x. So 4=(3/2)3+y Where do you subtract 3/2 from?

edit: acutally dking i'm going to copy/paste your example, print it and study off of it, (ha thanks!). But i still need an answer to the where you sub from. Other then that my question has been answer.

You need to seriously reconsider what x₁, y₁, x₂ and y₂ are:

x₁ = 3
y₁ = 4
x₂ = 5
y₂ = 6

m = (y₂ - y₁) / (x₂ - x₁)
m = (6 - 4) / (5 - 3) = (2) / (2) = 1

Now we have a slope of m = 1. We now can use point-slope form to finish this problem off:

Point-slope Form for the equation of a line:
y – y₁ = m * (x - x₁)

x₁ = 3
y₁ = 4

m = 1

Note that (3, 4) is ONE point, not two as you have written in your solution. Plugging in values directly:

y – y₁ = m * (x - x₁)
y - 4 = 1 * (x - 3)
y - 4 = x - 3
y - 4 + 4 = x - 3 + 4
y = x + 1

Thus, we have a final equation for our line of y = x + 1.