• Support PF! Buy your school textbooks, materials and every day products Here!

Finding normal to ln curve

  • Thread starter Doc G
  • Start date
18
0
I was wondering if anyone could help me with a question I was trying to work through today in our A-level math class.

1. Homework Statement

a. Sketch the graph of y= ln(3x)
b. The normal to the curve at Point Q, with x coordinate q, passes through the origin. Find the equation of the normal and the value of q


2. Homework Equations

I don’t know which to use

3. The Attempt at a Solution

Part a was not too difficult and I was able to do that

However I have no idea how to tackle part b. I have only been able to express the gradient of the normal as -q, but cannot solve it. I think [but am not sure] that I can get to the equation

ln3q + q^2 = 0

but I cannot solve this.

I would be very grateful for some help.

Thanks
 

Answers and Replies

Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,598
6
How do you get that equation?
 
18
0
Since when the normal and the curve y=ln(3x) intersect, the x coordinate is q, the the y coordinate is thus ln3q.

And since the normal passes through (0,0), the gradient of the normal can also be expressed as (ln3q)/q

I believe also that the gradient of the normal can be expressed as -q, thus

(ln3q) / q= -q

Rearranging gives

ln3q + q^2 = 0

However I don't know any method for solving such an equation
 
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,598
6
Since when the normal and the curve y=ln(3x) intersect, the x coordinate is q, the the y coordinate is thus ln3q.
Good
And since the normal passes through (0,0), the gradient of the normal can also be expressed as (ln3q)/q
Would it not be better to express the gradient as;

[tex]m=\frac{dy}{dx}\ln\left(3x\right) = \frac{1}{x}[/tex]

[tex]\left.\frac{dy}{dx}\right|_{\left(q,\ln(q)\right)} = \frac{1}{q}[/tex]
 
Last edited:
HallsofIvy
Science Advisor
Homework Helper
41,736
894
Almost. That is the slope of the tangent line. The slope of the normal line is -3q
 
18
0
Thanks for the replies.

Okay we haven't really covered much differentiation of ln graphs yet, but using what hootenanny/HallsofIvy posted, should the equation therefore be

ln(3q)/q = -3q

giving:

ln(3q) + 3(q^2) = 0

is this right? and how could i solve that?

thanks again
 
Dick
Science Advisor
Homework Helper
26,258
618
Good

Would it not be better to express the gradient as;

[tex]m=\frac{dy}{dx}\ln\left(3x\right) = \frac{1}{3x}[/tex]
It might be if it were correct. But it's NOT. 1/ x not 1/(3x). Think chain rule. I think Doc G's original equation is actually correct.
 
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,598
6
My mistake, stupid slips, duly corrected.
 
Dick
Science Advisor
Homework Helper
26,258
618
His problem doesn't seem to be finding the normal, etc. It's finding a solution to ln(3*x)+x^2=0 (I think this what he actually means by q). And I do think you can only solve it numerically. By inspection I would say you have a root just below 1/3.
 
18
0
Thanks, should I now attempt to find a close solution by iteration [we've only just covered the basics of which in class]?

And would the correct formula for which be:

[tex]x_{n+1} = x_{n} - \frac{\ln3x_{n} + x^2_{n}}{1/x_{n} + 2x}[/tex] ?


i tried putting in the equation ln3q + q^2 = 0 into Newton's Method. Is this okay? - i dont really have a good understanding of this area of maths yet..
 
Last edited:
Dick
Science Advisor
Homework Helper
26,258
618
Unless this is a numerical analysis course I wouldn't open a second front on root finding algorithms. Either use some canned software or just do bisection to get a few decimal points if you want. This is easy enough to do with a calculator and a little patience...
 

Related Threads for: Finding normal to ln curve

Replies
1
Views
3K
Replies
4
Views
1K
  • Last Post
Replies
6
Views
979
Replies
3
Views
2K
  • Last Post
Replies
1
Views
787
  • Last Post
Replies
4
Views
10K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
4
Views
777
Top