Solution Check

Sully

Just want an extra set of eyes checking my solution for this;

Let v(t) = 2ln(t)

- Find the slop of the function at t=1
- Find the equation of the tangent at t=1

Part 1

v(t) = 2ln(t)
v'(t) = 2(1/t)
= 2/t
=2/1 = 2

2 > 0 therefore slope is increasing at t=1

Part 2

Formula to calculate tangent:
y-y1 = m(x-x1)

Therefore I need (x1, y1) and m.

To find my (x1, y1)..

v(t) = 2ln(t)
= 2ln(1)
= 0

Points are (1,0)

Slope determined already in part 1 as 2.

y-0 = 2[x-(1)]
y-0 = 2x-2
y = 2x-2+0

I cant spot my error here, hoping for an extra pair of eyes to point out where I made the mistake. Thanks!

Michael Collins

Sully wrote: »

Just want an extra set of eyes checking my solution for this;

Let v(t) = 2ln(t)

- Find the slop of the function at t=1
- Find the equation of the tangent at t=1

Part 1

v(t) = 2ln(t)
v'(t) = 2(1/t)
= 2/t
=2/1 = 2

2 > 0 therefore slope is increasing at t=1

Part 2

Formula to calculate tangent:
y-y1 = m(x-x1)

Therefore I need (x1, y1) and m.

To find my (x1, y1)..

v(t) = 2ln(t)
= 2ln(1)
= 0

Points are (1,0)

Slope determined already in part 1 as 2.

y-0 = 2[x-(1)]
y-0 = 2x-2
y = 2x-2+0

I cant spot my error here, hoping for an extra pair of eyes to point out where I made the mistake. Thanks!

I can't see any calulation errors, although I wouldn't say the "slope" is increasing at t = 1. The function is increasing alright, but the slope is, in fact, decreasing.

Sully

Cheers, much appreciated!