Question

Consider the curve y=ln(3x-1).let p be the point on the curve where x=2.

a. 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 coordinate of R. (is the gradient of P 3/(3x-1)?

Answer 1

a) First off:

For y = Inx dy/dx = 1/x

For y = In(3x-1) dy/dx = 1/(3x-1) * 3 = 3/(3x-1).

So yes the gradient at P is = 3/(3x-1)

But also remember at P, x = 2. Therefore gradient = 3 / (3*2 -1) = 3 /(6-1)

Gradient, m at P is actually = 3 /5 = 0.6

b) The normal to curve:

m1 * m2 = -1. Condition for normal (perpendicular)

m2 = -1 / (3/5) = -5/3

Recall at point P, x =2, y = In(3x-1) = In(3*2-1) = ln5

Hence Point P = ( 2, In5)

Equation of gradient and one point:

y - y1 = m(x - x1). Here x1 = 2, y1 = In5

y - ln5 = (-5/3)(x - 2)

3y - 3ln5 = -5x + 10

3y = -5x + 10 + 3ln5 , 3y = -5x + 10 + ln5^3

3y = -5x + 10 + ln125

Hence normal to the curve is 3y = -5x + (10 + ln125)

When normal cuts the x axis, at that point y = 0.

3y = -5x + (10 + ln125)

3*0 = -5x + (10 + ln125)

0 = -5x + (10 + ln125)

5x = (10 + ln125) Divide both sides by 10

x = (10 + ln125) / 5 = 2.9657.

Coordinates of R = ( 2.97, 0).