Question

calculus help

Answer 1

Go ahead :D

Answer 2

can you help me with no 10

Answer 3

anyone?

Answer 4

both parts?

Answer 5

just one

Answer 6

part A?

Answer 7

yes please

Answer 8

Okay :)

Answer 9

Well, what you have is an initial value problem.

So, as you may know, when you take an antiderivative (or integrate a function) you always have a +C at the end. Well, initial value problems help you solve the exact C value. So for the one you're given f'(x)=1/x^2+x you want to first integrate it.

\[f(x)=\int\limits x^{-2}+x dx\]
\[f(x)=-x^{-1}+(1/2)x^2+C\]

Answer 10

From here, you want to plug in your initial value. Namely, f(2)=1.
\[f(2)=-(1/2)+(1/2)(2^2)+C\]
That has to equal 1. or
\[f(2)=-(1/2)+(1/2)(2^2)+C=1\]
adding the 1/2 and multiplying out the 2^2 and the (1/2) you get:
\[(3/2)=2+C \rightarrow C=-(1/2)\]

Answer 11

Any questions? :P

Answer 12

i was totally confused
there are two number 10s

Answer 13

I did part A. Where are you confused?

Answer 14

i am confused what you did at adding and multiplying

Answer 15

where is the equation?

Answer 16

where did you get 3/2 at the left?

Answer 17

Okay. That is just 1+(1/2)

Answer 18

no i'm not confused about what you did
i was confused which problem to do

Answer 19

A is done you could help me with B

Answer 20

what problem do you want me to solve?

Answer 21

pat, I just wanted you to solve A. I think myini is doing B

Answer 22

A for number 10?

Answer 23

yes. The answer was c=-1/2 right?

Answer 24

Yes :)

Answer 25

\[\frac{d^2y}{dt^2}=2-6t\]
\[frac{dy}{dt}=2t-\frac{6t^2}{2}+C_1\]
\[\frac{dy}{dt} (at; t=0)=2(0)-3(0)^2+C_1=4, C_1=4\]
\[y(t)=\frac{2t^2}{2}-\frac{3t^3}{3}+4t+C_2\]
\[y(0)=0^2-0^3+4(0)+C_2=1\]
\[C_2=1\]
y(t)=t^2-t^3+4t+1

Answer 26

\[\int\limits_{?}^{?}(x ^{-2}+x)dx\]
=\[=(x ^{-1})/-1 + (x ^{2})/2\]
=\[-x ^{-1}+(x ^{2})/2\] + c
1=-1/2 + 2 + c
c=1+1/2-2
c=-1/2

Answer 27

my latex messed up on line two
can you make it out?

Answer 28

i cant understand the third one myininaya

Answer 29

i evaluated y' at t=0
becuase we are given y'(0)=4

Answer 30

hey pat, x>0 right? so how is answer less than 0? I did this stuff at begining of the year so i am really confused

Answer 31

yeah, i am sorry. i forgot

Answer 32

yeah, x>0, and we have x=2

Answer 33

how 2?

Answer 34

thanks myininaya for B

Answer 35

:)

Answer 36

pat, you there?

Answer 37

myininaya if you still there can you look ar pats solution and say how is the ansewer 2?

Answer 38

its c=-1/2
f(x)=-1/x+x^2/2-1/2

Answer 39

is that your question?

Answer 40

yes thanks