Question

help calculus @vocaloid

Answer 1

@Vocaloid

Answer 2

have 5 questions could you help please?

Answer 3

@Ultrilliam

Answer 4

The trick is to separate out the x and y variables
get the y and dy terms together, and the x and dx terms together like so
\[\large y^{\prime} = \frac{1+y^6}{xy^5}\]
\[\large \frac{dy}{dx} = \frac{1+y^6}{xy^5}\]
\[\large xy^5 dy = (1+y^6)dx \ {\color{blue}{\text{Cross multiply}}}\]
\[\large y^5 dy = \frac{(1+y^6)dx}{x} \ {\color{blue}{\text{Divide both sides by x}}}\]
\[\large \frac{y^5}{1+y^6}dy = \frac{dx}{x} \ {\color{blue}{\text{Divide both sides by} 1+y^6}}\]
\[\large \frac{y^5}{1+y^6}dy = \frac{1}{x}dx\]

Answer 5

Then you apply the integral operation to both sides
\[\large \frac{y^5}{1+y^6}dy = \frac{1}{x}dx\]
\[\large \int\frac{y^5}{1+y^6}dy = \int\frac{1}{x}dx\]
To find the integral of \(\large \frac{y^5}{1+y^6}\), you'll use u-substitution with u = 1+y^6, so du/dy = 6y^5 leading to du/6 = y^5*dy. So \(\large \displaystyle \int\frac{y^5}{1+y^6}dy = \int\frac{du/6}{u}\). I'l let you finish up.

The integral of 1/x is the natural log of x. Don't forget about the +C. However, the +C will be replaced with some fixed number after you use the fact that y(1) = 1 which is the same as saying x = 1 and y = 1 pair up together.

Answer 6

got it lol 2 more questions?

Answer 7

@jimthompson5910

Answer 8

@jimthompson5910 you there

Answer 9

yes just trying to phrase it the right way, one sec

Answer 10

okay cool

Answer 11

This page goes over Newton's method if you need a refresher
http://tutorial.math.lamar.edu/Classes/CalcI/NewtonsMethod.aspx

The general form is
\[\Large x_{n+1} = x_{n} - \frac{f(x_{n})}{f'(x_{n})}\]

f(x) = x^3 - 3x + 3
f ' (x) = 3x^2 - 3 .... use power rule

\(\large x_0 = -3\) is the initial guess, which leads to the next iteration value to be...
\[\Large x_{n+1} = x_{n} - \frac{f(x_{n})}{f'(x_{n})}\]
\[\Large x_{0+1} = x_{0} - \frac{f(x_{0})}{f'(x_{0})}\]
\[\Large x_{1} = -3 - \frac{f(-3)}{f'(-3)}\]
\[\Large x_{1} = -3 - \frac{-15}{24}\]
\[\Large x_{1} = -2.375\]

This value is then used to compute \(\large x_2\)
\[\Large x_{n+1} = x_{n} - \frac{f(x_{n})}{f'(x_{n})}\]
\[\Large x_{1+1} = x_{1} - \frac{f(x_{1})}{f'(x_{1})}\]
\[\Large x_{2} = -2.375 - \frac{f(-2.375)}{f'(-2.375)}\]
\[\Large x_{2} \approx -2.375 - \frac{-3.271484}{13.921875}\]
\[\Large x_{2} \approx -2.140011\]
Tell me what you get for \(\large x_3\)

Answer 12

one quick question how did you get the answers for the parts after for example -2.375-f(2.375)/f'(2.375?

Answer 13

wait original function duh

Answer 14

do you see how I calculated \(\large x_1\) to be -2.735?

Answer 15

-2.375 I meant

Answer 16

for f(-2.14) I got -.380344 maybe it's right for the top part ?

Answer 17

I'm getting f(-2.14) = -0.380344 as well
Though your teacher says to use as many digits as your calculator can hold for the intermediate calculations. I'm just noticing that.

Answer 18

so for f' do I use the derivative of original function?

Answer 19

you would use f ' (x) = 3x^2 - 3

Answer 20

2.1046

Answer 21

I'm getting \(\Large x_3 \approx -2.104583\)

Answer 22

for the prior problem I mean

Answer 23

she rounded to 4 decimal places

Answer 24

it should be negative though

Answer 25

For the other problem you just posted, a general solution is something in the form y = f(x)+C where C is some constant. So you'll have a family of functions. Or put another way, you have some function that is shifted up and down like you see here
https://slideplayer.com/slide/4441010/14/images/3/2-6%3A+Families+of+Functions.jpg

A particular solution is exactly one function that passes through the initial point desired. So if you know the initial conditions, you can narrow the infinite member family down to one single answer.

Answer 26

oh poop lol