Confusion with a couple p2 questions

[abhiguda]

g'day,

just been doing past papers since the exam is this week!

i've had a little hiccup with some of the following problems any help would be much appreciated;

++1. Consider the curve y=ln(3x-1). Let P be the point on the curve where x = 2

(a) write down the gradient of the curve at P.

(b) the normal to the curve at P cuts the x-axis at R. Find the coordinates of R.

I can do A without any problem really

y'(2) = 1/(3x-1) * Dx

= 1/(3x-1) * 3

= 3/(3x-1)

= 3/(3(2)-1)

= 3/17

i know that the normal is the negative reciprocal of the slope of the tangent

and i understand that i must integrate the tangent and solve for X to find the coordinates of R

but i am confused as how to go about doing this

is it: -(3x-1)/3 or -17/3...im not sure, i would appreciate help with this

++2. Solve the equation, e^x=4sin x for 0 < x <2pi

i got this far

lnx=ln4sinx

therefore,

x=4sinx

i just dont have a clue as to where to go from here, it seems so simple but i just cant figure it out.

thanks in advance

[dessskris]

(1a)

m = 3/(3(2)-1)

= 3/(6-1)

= 3/5

You made a careless mistake here..

(1b)

m of normal = -1 / m of tangent

= -5/3

Find the equation of the normal

Since it passes through point P, let's find the y-coordinate of P first.

y = ln(3x-1)

= ln(5)

The normal passes through (2, ln5). Let's find its equation.

y=mx+c

y = -5x/3 +c

ln5 = -10/3 +c

c= ln5 + 10/3

Therefore y = -5x/3 + ln5 + 10/3

That line cuts the x-axis at R. So the y-coordinate of R is 0.

When y=0,

y = -5x/3 + ln5 + 10/3

0 = -5x/3 + ln5 + 10/3

5x/3 = ln5 + 10/3

x = 3(ln5 + 10/3)/5

x = 3(ln 5)/5 + 2

Since you can use calculator in paper 2,

x = 2.97

Or actually since you can use GDC, you can just plot the normal and y=0 and find the intersection.

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(2)

Plot y=e^x and y=4sinx in your GDC. Find the intersection.