Log Question (1 Viewer)

[4DOGS]

Could some one please help me with the is

QUESTION
Point A whose y-co-ordinate is 1, lies on the curve y=In(x+1). the tangent to the curve at A cuts the y-axis at P. A line through A perpendicular to the y-axis meets the y -axis at M. P=(0,0), M=(0,0)

i) Show that the x-coordinate of A is (e-1)

ii) show that the equation of the tangent AP is x-ey+1=0

iii) Find the length of PM in terms of e

 

[onebytwo]

Could some one please help me with the is

QUESTION
Point A whose y-co-ordinate is 1, lies on the curve y=In(x+1). the tangent to the curve at A cuts the y-axis at P. A line through A perpendicular to the y-axis meets the y -axis at M. P=(0,0), M=(0,0)

i) Show that the x-coordinate of A is (e-1)

ii) show that the equation of the tangent AP is x-ey+1=0

iii) Find the length of PM in terms of e

for part (i) just let y = 1
so 1 = ln (x+1), so e^1 = x +1, so x = e - 1

 

[onebytwo]

for (ii)
dy/dx = 1/(x+1), so at A(e-1,1), m=1/e
so using point gradient formula, y-1=(1/e)(x-e+1) so ey = x + 1

 

[sando]

geez 4dogs... they are simple

 

[withoutaface]

i) sub in y=1, exp() both sides to get exp(1) = x+1, x=e-1
ii) Find the gradient by differentiating to get m = 1/(x+1) = 1/e at A. Then sub in 1/e (x-e +1) = y-1 and then basic algebra to get x-ey+1=0.
iii) Find P by subbing in x = 0, .'. y = 1/e. The line through A perpendicular to the y axis obviously meets the axis at 1, so the answer is 1-1/e = (e-1) / e

 

[4DOGS]

Thanks for that seems so simple now.:wave: