qestions involving derivative of lns; finding the equation

[Guest]

1) Find the equation of the tangent to the curve defined by y = ln x - 1 that is parallel to the straight line with equation 3x - 6y - 1 = 0.

. . .y' = 1/x

. . .3x - 6y - 1 = 0
. . .-6y = 1 - 3x
. . .y = (1 - 3x)/(-6)
. . .y = 1/2x - 1/6
. . .y' = 1/2

If the lines are parallel, they have the same slope, so:

. . .1/x = 1/2

. . .x = 2

Now I'm having trouble getting the right answr for the equation:

. . .y = mx + b

. . .y = ln(2) - 1

But then what?

2) If tangents to the curve defined by y - 6x + 3 = 0, find the point where the tangents touch the curve.

. . .y' = 2x + 4(1/x)
. . .y' = 2x + 4/x
. . .y' = (2x^2 + 4)/x
. . .y' = 2(x^2 + 2)/x

. . .y = 6x - 3
. . .y' = 6

. . .1/6 - 2(x^2 + 2)/x
. . .x = 12x^2 + 24
. . .0 = 12x^2 - x + 24

I've found the slopes of each and set them equal to each other. Then I set them to zero. Now how do I solve for x?

Thank you for your help!

 

[pka]

\(\displaystyle m = \frac{1}{2}\quad \& \quad \left( {x_1 ,y_1 } \right) = (2,\ln (2) - 1)\)
Now write the equation.

 

[wjm11]

Now im having trouble getting the right answr for the equation:

y=mx+b

y=ln(2)-1
then..??

You've just solved for the point of tangency: (2, ln(2) - 1)
You know the slope: m = 1/2
You can put this info into the point-slope form of a line:

y - (ln(2) - 1) = (1/2)(x - 2)

If you want it in slope-intercept form, just rearrange it.

Ooops. I see pka was too quick for me.