#1

(again, I've deleted the other maths thread I made to stop spam. this way people will actually see the new question I am asking)

I'm revising and need help with this question;

The curve with equation y = ln 3x crosses the x axis at point P(p,0)

a) sketch the graph of y = ln 3x showing the exact value of p

I did that algabraically instead of sketching and p is 1/3

Then I get to the next question

The normal to the curve at the point Q, with x coordinate q, passes through the origin.

b) Show that x=q is a solution to the equation x^2 + ln 3x

How do you do (b)?

I'm revising and need help with this question;

The curve with equation y = ln 3x crosses the x axis at point P(p,0)

a) sketch the graph of y = ln 3x showing the exact value of p

I did that algabraically instead of sketching and p is 1/3

Then I get to the next question

The normal to the curve at the point Q, with x coordinate q, passes through the origin.

b) Show that x=q is a solution to the equation x^2 + ln 3x

How do you do (b)?

#2

ummm, im not sure how to go about this, but would differentiating and then putting q into the expression you get, then take the negative reciprocal of it, that would give you the gradient of the normal in terms of q. c is 0 as it goes through the origin.

im just throwing an idea at you lol. im not really sure, that seems quite hard.

im just throwing an idea at you lol. im not really sure, that seems quite hard.

#3

I think xX_Jimi_Xx has something, but I don't remember derivatives of natural logs.

You just have to figure out the x-coordinate on the graph where the normal goes through the origin, then plug that number into the second equation x^2 + ln 3x (= 0?)

You just have to figure out the x-coordinate on the graph where the normal goes through the origin, then plug that number into the second equation x^2 + ln 3x (= 0?)

#4

eww...

#5

the derivative of lnx is 1/x

ln3x is also ln3 + lnx.

therefore the derivative of ln3x is also 1/x i think. it is possible im wrong though lol

ln3x is also ln3 + lnx.

therefore the derivative of ln3x is also 1/x i think. it is possible im wrong though lol

#6

Yeah that's right. dy/dx = 1/x

#7

#8

#9

Hey I have a quick question Logic_Smogic. I was reading in your profile that you are a Physics major and I was wondering what math classes have/will you take in college? Right now I'm a junior in high school and in Calc. 1 and next year I will take Calc. 2 but I was curious what my options are for college as far as math goes.

#10

Hey I have a quick question Logic_Smogic. I was reading in your profile that you are a Physics major and I was wondering what math classes have/will you take in college? Right now I'm a junior in high school and in Calc. 1 and next year I will take Calc. 2 but I was curious what my options are for college as far as math goes.

For me, the story goes like this:

AP Calculus (AB) - High School

College (UW-Madison):

Calculus II

Calculus III (dropped, and started new 'Honors' sequence instead, listed below)

Calculus I (proof-based)

Calculus II (proof-based)

Multi-Variable Calculus and Linear Algebra (proof-based)

Multi-Variable Calculus and Differential Equations (proof-based)

Probability

..and I will likely take courses in Real Analysis and Advanced Algebra (and/or complex).

I've learned a lot of math in my physics courses, though. Topics such as the Hilbert space, non-closed integration methods, and Fourier analysis have been exclusively covered in my physics courses, for example.

#11

Thanks I was just wondering what kind of classes are out there.

#12

On that picture q is positive but you've labelled the green line with a -q gradient when the gradient should be positive.

#13

Thanks I was just wondering what kind of classes are out there.

No problem. A solid math sequence would go (in my opinion):

Calc 1

Calc 2

Calc 3

Linear Algebra

Ordinary Differential Equations

Topics in Partial Differential Equations or Fourier Analysis

Probability

Real Analysis

Modern Algebra

Complex Analysis

#14

On that picture q is positive but you've labelled the green line with a -q gradient when the gradient should be positive.

Good point, I'll change it.

EDIT: Done.

*Last edited by Logic Smogic at Jan 17, 2007,*

#15

Ok heres one.

T = 40 x W + 20

write down the formula for T in terms of W

T = 40 x W + 20

write down the formula for T in terms of W

#16

that's already in terms of W, if you wanted W in terms of T then:

W = (T/40) - 20 i believe

W = (T/40) - 20 i believe

#17

^ You mean W = (T-20)/40.

#18

Make V the subject of the formula m(v-u)=I

#19

Make V the subject of the formula m(v-u)=I

m(v-u) = I

mv - mu = I

mv = I + mu

v = (I + mu)/m

i think.