Question

Find the point on the curve y=ln(x^2) where the slope of the tangent line is 2/3
find the x intercept of the line tangent to the curve at that point

Answer 1

\[\Large\bf\sf y(x)\quad=\quad \ln(x^2)\]And we want to find the x value that gives us,\[\Large\bf\sf y'(x)\quad=\quad \frac{2}{3}\]

Answer 2

Do you know how to take the derivative of the function?

Answer 3

yes it is 2/x i believe

Answer 4

omy do i just plug in :o

Answer 5

Mmm ok cool.
\[\Large\bf\sf y'(x)\quad=\quad \frac{2}{3}\]Setting our derivative equal to 2/3,\[\Large\bf\sf \frac{2}{x}\quad=\quad \frac{2}{3}\]

Answer 6

Yah the first part isn't too bad :)
Find the x value where this is true.

Answer 7

so x =3

Answer 8

Mmmmk good.

Answer 9

so that is my point i assume
?

Answer 10

Hmm I'm not sure what the fastest method for finding the x-intercept is...
This is the process I took though,

We find an equation for the tangent line at x=3.
\[\Large\bf\sf y_{\tan}\quad=\quad \frac{2}{3}x+b\]
After we find the y-intercept we can plug in y=0 to find the x-intercept.
If there is a faster way you can stop me Kanui :P hehe

Answer 11

well i've learned finding the x you just set the y to zero or if you want to find y you make x 0 but I forgot if thats intercept. :(

Answer 12

Yah that's the right process :)
But what y are we setting equal to zero?
We don't have an equation for the tangent line yet.

After we set it up we can set y=0 to find the x-intercept of that line.

Answer 13

We could do this in point-slope form if you prefer, but slope-intercept seems fine :p
We need a coordinate pair for the line.

So we'll plug x=3 into our function to find a corresponding y value.

Answer 14

\[\Large\bf\sf y(3)\quad=\quad \ln(3^2)\quad=\quad 2\ln(3)\]So our coordinate pair is,\[\Large\bf\sf (3,\;2 \ln3)\]

Answer 15

Umm actually let's do point-slope form. It'll make more sense I think.\[\Large\bf\sf y-y_o\quad=\quad m(x-x_o)\]Plugging in our slope, and our coordinate point,\[\Large\bf\sf y-2\ln3\quad=\quad \frac{2}{3}(x-3)\]This is the equation of our line tangent to the curve at x=3.
(which has slope 2/3 as they requested in the first part).

Now to find our x-intercept we let y=0.

Answer 16

im a little confused

Answer 17

Ya it's a lot going on :(

Answer 18

wait do i only look at point slope form part?

Answer 19

We're looking for the x-intercept of our `tangent line`.
Before we could find the x-intercept we had to establish the equation for the tangent line.

Whether you write the tangent line in `point-slope` form or `slope-intercept` form doesn't matter.

Which form are you more familiar with?
They're both just different ways of writing linear equations.

Answer 20

i think i understand the way it is written in point slope the best

Answer 21

So then hopefully this looks a little bit familiar:\[\Large\bf\sf y-y_o\quad=\quad m(x-x_o)\]
We have to plug in a coordinate pair \(\Large\bf\sf (x_o,y_o)\) and a slope \(\Large\bf\sf m\).

Answer 22

The slope was `given to us` at the start of the problem.
We just had to figure out `where that slope was located`.

Answer 23

so it could be y-yo= 2/3(x-xo) or am i totally off

Answer 24

the Y im not exactly sure where to get

Answer 25

Plugging in the slope they gave us? yes good.

Answer 26

We finished the first part, determining that the `location` of the 2/3 slope is at x=3 right?

Answer 27

sorrry im like eyes half open so i didnt even actually plug in 3 :O

Answer 28

but yes we just determined that

Answer 29

So we need the y coordinate that corresponds to x=3.
So we plug the x=3 into the `original function`.

brb gotta get my food out the oven :OOO

Answer 30

i got 2.19. thank you for your help I think thats the farthest I can go I really need to sleep and shower ! thank you again I appreciate it so much my grade depends on this assignment. I wish i loved math ahha goodnight mr/mrs zepdrix

Answer 31

lol night :D

Answer 32

(3, 2.19)

Plug it into the x0, y0 in the point-slope form.
then y=0 gives you your x value.

go to bed. :P