# Finding normal to ln curve

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

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Hootenanny
Staff Emeritus
Gold Member
How do you get that equation?

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
Gold Member
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;

$$m=\frac{dy}{dx}\ln\left(3x\right) = \frac{1}{x}$$

$$\left.\frac{dy}{dx}\right|_{\left(q,\ln(q)\right)} = \frac{1}{q}$$

Last edited:
HallsofIvy
Homework Helper
Almost. That is the slope of the tangent line. The slope of the normal line is -3q

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
Homework Helper
Good

Would it not be better to express the gradient as;

$$m=\frac{dy}{dx}\ln\left(3x\right) = \frac{1}{3x}$$
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
Gold Member
My mistake, stupid slips, duly corrected.

Dick
Homework Helper
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.

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:

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

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