How to Find the Normal to a ln Curve?

[Doc G]

I was wondering if anyone could help me with a question I was trying to work through today in our A-level math class.

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

Homework Equations

I don’t know which to use

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

 

[Hootenanny]

How do you get that equation?

 

[Doc G]

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]

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;

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

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

 

[HallsofIvy]

Almost. That is the slope of the tangent line. The slope of the normal line is -3q

 

[Doc G]

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]

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]

My mistake, stupid slips, duly corrected.

 

[Dick]

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.

 

[Doc G]

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 don't really have a good understanding of this area of maths yet..

 

[Dick]

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...