Question

Calculus

Answer 1

hey

Answer 2

which question are you having trouble with?

Answer 3

#21

Answer 4

@welshfella @freckles

Answer 5

yes I see you have posted 21
but there are 4 questions there
are you saying you need help on all of them?

Answer 6

yep

Answer 7

are you having trouble finding when the fraction y/(2x) is undefined ?

Answer 8

For A I think it is x because it is the denominator via the hint we are given. when x=0

Answer 9

right y/(2x) is undefined when x=0

Answer 10

hint for b lines are parallel when their slopes are the same (different y-intercept of course)

Answer 11

what is the slope of the line that is on the x-axis ?

Answer 12

ok so lines are parallel when their slopes are equal i know that.... which line though? Graph the function given?

Answer 13

the one that is the x-axis

Answer 14

hint: what is the slope of any horizontal line....

Answer 15

0

Answer 16

right so the slope of the line that is known as the x-axis is 0

Answer 17

aka the slope of y=0 is 0

Answer 18

anyways find when y/(2x) is 0

Answer 19

well when y=0

Answer 20

and x can be anything.

Answer 21

ok
the 3rd question is just pluggin in two points

Answer 22

So just plug into the given function?

Answer 23

you do know that dy/dx is the slope of y

Answer 24

they are asking for the slopes at (1,2) and (-1,-2)

Answer 25

they are asking you to evaluated y' at (1,2) and (-1,-2)

Answer 26

well they would be 0 then.

Answer 27

how ?

Answer 28

if we plug in we get 1 and the derivative of that is 0

Answer 29

-2/(2*-1) is 1

Answer 30

right the slope at (-1,-2) is 1 not 0

y/(2x) is only ever 0 when y is 0 provided x is not 0

Answer 31

bu y' is the derivative right so the derivative of the function is 0?? or are we not taking the derivative?

Answer 32

you are already given the derivative aka slope of y...

Answer 33

it is dy/dx

Answer 34

y'=dy/dx

Answer 35

ohhhhhhh so we have the slope of 1 at poth of the points given so the answer is just 1?

Answer 36

yes

Answer 37

gotcha. Ok so part d

Answer 38

but if for some reason you wanted to find the slope of dy/dx
that would be d^2y/dx^2
you would find the derivative before plugin in the point
you would not plug in the point and then take derivative

Answer 39

can i hear what you think about d

Answer 40

if you really understand what we did in b it shouldn't be too hard

Answer 41

they want the line parallel to the point that has a slope of .5

Answer 42

they want the line that consists of line segments parallel to the line y=0.5x

remember lines that are parallel have to same ....

Answer 43

remember lines that are parallel have the same ....
*

Answer 44

slope so they have to have a slope of .5

Answer 45

right so what equation are you going to be solving ?

Answer 46

the function that was given y/2x or the y=mx+b ??

Answer 47

you know that dy/dx is the thing that is representing are slope field?
the line segments in dy/dx is what we are interested in

Answer 48

ok so use the y=mx+b?

Answer 49

how would you use that?

Answer 50

using the slope of .5 and then finding points that fit it?

Answer 51

dy/dx is the slope of y
.5 is the slope of y=0.5x

parallel means lines with same slope (different y-intercept)

Answer 52

same means equal

Answer 53

you are wanting to solve dy/dx=.5

Answer 54

right so the slopes of both have to be .5

Answer 55

we could use (1,1)

Answer 56

do you know to find the line such that dy/dx=.5 ?

Answer 57

not exactly just plug and chug...

Answer 58

you are given dy/dx=y/(2x)

Answer 59

right and if we set it equal to .5 we have two variables to solve for.... but if we plug i got 1,1 as a point...

Answer 60

that is a point
there are many points in which the line segments will have slope .5
according to your question you should get a line that consists of infinitely many line segments where their slopes are .5

Answer 61

dy/dx=.5
y/(2x)=.5
y=2x(.5)
y=x

Answer 62

ohhh i see now... so no matter what the points are since they are the same they will have the same slope ...

Answer 63

so just put y=x?

Answer 64

yes y=x is the line they are looking for

Answer 65

(1,1) is just one point on that line
there are many points on that line where the line segment will have slope .5

Answer 66

ok i have another one.

Answer 67

ok i will brb

Answer 68

ok

Answer 69

are you having trouble with both questions?

Answer 70

yeah all of them again

Answer 71

have you ever seen euler's method?
it is just a formula

Answer 72

i have seen it.... just confused on how it works. Can you post the formula and tell me how to plug in and such?

Answer 73

\[y_n=y_{n-1}+f(x_{n-1},y_{n-1})(x_{n}-x_{n-1}) \\ \text{ assuming we are giving the point } (x_{n-1},y_{n-1}) \\ \\ \]



\[y'=f(x,y) =\frac{y}{x}+1\]

Answer 74

the step side is just that change in x

Answer 75

what they are calling h in the second question

Answer 76

ok

Answer 77

so that yn equation being a... and then how would we approximate for b?

Answer 78

I would probably input (x_(n-1),y_(n-1)) into the function called f(x,y) which in this case is y/x+1

but anyways we are given h=.01 this is the thing you can replace that change of x with I mentioned above
start with your initial solution for y which you are given y(1)=0
that is (x0,y0)=(1,0)
then use this to try to find y(1.03) this means find the approximately solution y at x=1.03

Answer 79

ok hold up so our h value goes where?

Answer 80

that change of x up there I mentioned

Answer 81

so we are using the equation we were given.... right so where would the change in x fall?

Answer 82

in the formula I mentioned
....
\[y_n=y_{n-1}+f(x_{n-1},y_{n-1})(x_{n}-x_{n-1}) \\ \text{ assuming we are giving the point } (x_{n-1},y_{n-1}) \\ \]

do you see the factor that contains the change of x?

Answer 83

\[y_n=y_{n-1}+f(x_{n-1},y_{n-1})(h) \\ \text{ assuming we are giving the point } (x_{n-1},y_{n-1}) \\ \\\]

Answer 84

ok so we are using a x value to get the y=1.03? What would be the best way to go about that?

Answer 85

x0=0 -> y0=1
x1=0+.01 -> y1=?
x2=0+2(.01) ->y2=?
.... keep doing this to
you have for some number k
xk=1.03

Answer 86

no you are looking for y(1.03)
not mean y is 1.03

Answer 87

that is going to take awhile
that is 103 steps :p

Answer 88

so that list should have read so you only need 3 steps to get there
x0=1 -> y0=0
x1=1+.01 -> y1=?
x2=1+2(.01) ->y2=?
x3=1+3(.01) ->y3=?

Answer 89

you are using the equation I mentioned earlier where the change in x is .01
aka h is .01

Answer 90

Ok I'm back now sorry it was realllllllly laggy

Answer 91

So we have to go through each step until we get y(1.03)? And we are looking for the x value that makes y(1.03)

Answer 92

@freckles

Answer 93

we are looking for the y value when x is 1.03

Answer 94

use the equation above you are given (x0,y0)
use that to find y1 when x1 is x0+1*h

Answer 95

ok so use the equation they gave us not the one we came up with?
so just plug in x as 1.03?

Answer 96

no.... you have three steps to get to 1.03
the next x is x1 which is 1.01 since the step size is .01
the first x was x0 which was 1