# Thread: tangent line equation of curve..

1. ## tangent line equation of curve..

This is im sure very easy, but it was a busy summer. I am going through some questions early in my calc book, and im not sure of the best/easiest way to tackle this question. (please explain steps)

The point P(2, ln 2) lies on the curve y = ln x. Using the point Q(2.001, ln 2.001) on the curve, a reasonable equation of the tangent line to the curve at P(2, ln 2) is?

Cheers

2. Originally Posted by captain_jonno
This is im sure very easy, but it was a busy summer. I am going through some questions early in my calc book, and im not sure of the best/easiest way to tackle this question. (please explain steps)

The point P(2, ln 2) lies on the curve y = ln x. Using the point Q(2.001, ln 2.001) on the curve, a reasonable equation of the tangent line to the curve at P(2, ln 2) is?

Cheers

I suppose that by "reasonable" they mean you calculate the equation of the line passing through the points P and Q . Use the folllowing, which is Junior H.S. stuff:

** Equation of the line passing through $(x_1,y_1)$ and with slope equal to $m$ : $y-y_1=m(x-x_1)$

** Slope of the line passing through the points $(x_1,y_1)\,,\,(x_2,y_2)\,:\;\frac{y_2-y_1}{x_2-x_1}$ , unless $x_1=x_2$ and then the line passing through these points is a vertical one and its slope is undefined.

Tonio

3. Ok super, thanks.

so y= x/2 -lnx +1 or what do you think

4. Originally Posted by captain_jonno
Ok super, thanks.

so y= x/2 -lnx +1 or what do you think

$m=\frac{\ln(2.001)-\ln 2}{2.001-2}\cong \frac{1}{2}$ , so the wanted line is:

$y - \ln 2=\frac{1}{2}(x-2)\Longrightarrow y = \frac{x}{2}+\ln 2 - 1$