Question

Find the equation of the curve that passes through the point (1,3) and has a slope of y/x^2 at the point (1,3). i used y-y1=m(x-x1) and ended up with y=3x, however that is linear but the problem said it was a curve. Does that seem right?

Answer 1

So x=1,y=3
what is the slope?

Answer 2

it says the general form for the slope is y/x^2

Answer 3

So plug in and evaluate

Answer 4

have you replaced y with 3 and x with 1 yet?

Answer 5

Yes i think they are looking for a tangent line to the curve at (1,3)
tangent lines are linear and have the form you speak of

Answer 6

wait unless ...

Answer 7

maybe it does actually mean find the equation to the curve
is this calculus 2?

Answer 8

sweetlove333 are you there?

Answer 9

So we are given y'=y/x^2
We can use separation of varaibles to find the equation of the curve y

Answer 10

then we have the point (1,3) so we can find the constant C after integrating

Answer 11

i'm in ap calc ab, can you show the steps?

Answer 12

have you talk about integration?

Answer 13

yeah, do you think i should integrate the y/x^2 equation? and then plug in the values for x,y to find c?

Answer 14

First y'=y/x^2
We need to separate the variables
put all of our y's on one side and all of our x's on the other

Answer 15

\[\frac{dy}{dx}=\frac{y}{x^2}\]

Answer 16

So we need to multiply 1/y on both sides given us
\[\frac{1}{y} \frac{dy}{dx}=\frac{1}{x^2}\]
And then we need to multiply dx on both sides given us
\[\frac{1}{y} dy =\frac{1}{x^2} dx\]
Now this gives you permission to integrate both sides
Can you do that part?
Tell me what you get...

Answer 17

after integrating, I got ln absolute value y = (but how do you antiderive( dx/x^2)?

Answer 18

can i use the trig fcn of arctan?

Answer 19

\[\text{ recall from algebra you can rewrite } \frac{1}{x^2} \text{ as } x^{-2}\]

Answer 20

Do you know the power rule for integration?

Answer 21

so it would -x^-1?

Answer 22

:)

Answer 23

ok and then +c on of of the sides

Answer 24

one of*

Answer 25

\[\ln|y|=\frac{-1}{x}+C\]

Answer 26

right?

Answer 27

okay that's what i got, then do i plug in the points and figure out c?

Answer 28

So we can find C since we are given a point on the curve (1,3)

Answer 29

lol yes sweetlove
great! :)

Answer 30

thanks!, and after finding c, what equation should i plug c into then?

Answer 31

would it be y=lny+ x^-1 +c? and substitute the c i found?

Answer 32

You are looking for the curve (we just found the general form for the curve) so when you find C replace C in the equation and you will have your curve y

Answer 33

would that be the equation of the curve?

Answer 34

where that extra y come from

Answer 35

\[\ln|y|=\frac{-1}{x}+C \]
This is the general form for the curve

Answer 36

okay, thanks a lot! =)

Answer 37

NP
What did you get for C?

Answer 38

for c i got 2.0986

Answer 39

Are you allowed to approximate?

Answer 40

i hope so=P

Answer 41

\[\ln|3|=\frac{-1}{1}+C\]
So this is what you have where you replace x with 1 and y with 3

Answer 42

now this means we have
ln(3)=-1+C

Answer 43

So adding 1 on both sides will isolate C

Answer 44

ln(3)+1=C
C=ln(3)+1=ln(3)+ln(e)=ln(3e) <---you don't have to write this way (I just think it looks prerrier; you can leave it as ln(3)+1)

Answer 45

So you can write
the equation of the curve like this
\[\ln|y|=\frac{-1}{x}+(\ln(3)+1)\]
OR (my favorite)
\[\ln|y|=\frac{-1}{x}+\ln(3e)\]

Answer 46

okay, that was very helpful! ^______^ thanks!

Answer 47

:) great sweetlove
I actually like this problem

Answer 48

You know after I read your problem again it does not say anything about the tangent line
so I'm sure that that is our answer above