Question

How do I do this? CALCULUS @satellite73 @zepdrix

Answer 1

Find the slope of any such line.

You have two points (0,0) and (a,b) = (a,ln(a/3)). Calculate the slope at x = a. You should be about done.

Answer 2

What does the (a, lin(a/3)) represent?

Answer 3

ln, sorry. not lin

Answer 4

That's where the line is tangent to the curve. What is the slope, there?

Answer 5

So I have to first solve for x in the equation?

Answer 6

to get the value of a?

Answer 7

Ooo this is a neat problem :O

Answer 8

Can you help?

Answer 9

Derivative might be a good place to start :)
You're given something about slope...
and you know that f'(x) tells you something ... about... slope... right? :)

Answer 10

the slope of the equation given in the question equals the derivative of the line. But I can't figure out the slope of the equation, because the ln is confusing me

Answer 11

So you're having trouble taking the derivative of In(x/3)?

Answer 12

What's your derivative? Gimme.

If you then let f'(x)=m,
you can determine `where` (which x value) gives you this slope value.
That x value is useful, it tells us where the tangent line and curve intersect.

Answer 13

So the derivative of In(x) is 1/x. So we can use that and the chain rule to take the derivative of In(x/3). With that in mind, any ideas what the derivative would be?

Answer 14

Wait, so I find the derivative of the equation they give? Why?

Answer 15

Just do ittttt >.<

Answer 16

You mentioned finding the slop by taking the derivative earlier.

Answer 17

lol zeppy

Answer 18

lol, ok. So the derivative of ln(x/3) would be 1/x ?

Answer 19

Cool.

And you're given that f'(x)=m (slope of the line tangent to the curve at some x).
So then m=1/x, solve for x.

Answer 20

What now, gurl?

Answer 21

sorry, glitching. x = 1/m right?

Answer 22

So there's some point where the tangent line intersects the graph of In(x/3). Let's say that happens at x=a ... m would be the slop of the tangent line.

Answer 23

So the point where they intersect then has coordinates (a,In(a/3)).

Answer 24

Ah, I see

Answer 25

And we also know that the tangent passes through the origin. So the point (0,0).
What would the slop of a line that passes through the points (a,In(a/3)) and (0,0) be?

Answer 26

It would be \(\Large \frac{a}{\ln (a/3)}\) right?

Answer 27

Close. Other way around. Change in y over change in x. So In(a/3)/a.

Answer 28

oh, right. sorry, my brain's slow today xD

Answer 29

Mine too actually. Eh.

Now we also mentioned that since this is a tangent line that the slop would also be given by the derivative of the function. Which we said is 1/x. So at the slope would be 1/a. So this gives us that In(a/3)/a = 1/a .

Answer 30

which means ln(a/3) =1

Answer 31

right?

Answer 32

In(a/3)= 1.

Answer 33

kay, so what next?

Answer 34

Well we need to solve the equation for a. What do we get if we do that?

Answer 35

Would I get 1/x = e?

Answer 36

1/a = e I mean

Answer 37

Hm. Could you show me how you arrived at that result, please? c:

Answer 38

Um...don't you take the e of both sides?

Answer 39

After we cancel out the a's in the denominator, yep. Which would give us a/3.e.

Answer 40

oh, right. Sorry I wrote the equation down wrong...seriously, this is wha tI get for doing math this late

Answer 41

No worries, so what would a equal?

Answer 42

a =3e

I have to go eat dinner. Thanks for all your help. I'll check back tomorrow to see your response

Answer 43

Good. And we know that the slope of the tangent line is 1/a. So slope would be what?

Answer 44

Oh, yes, go eat dinner lol. If I'm not on tomorrow hopefully zeppy or tk will help you out with the rest (only a tiny bit to go) ttyl! :P