• April 24th 2010, 06:46 AM
Sunyata
A large stone is falling through a layer of mud and its depth x m below ground level at time t minutes is given by x = 12 - 12e^(-0.5t). How long will it take the stone to reach within 1 mm of its final position?
• April 24th 2010, 08:01 AM
sa-ri-ga-ma
Quote:

Originally Posted by Sunyata
A large stone is falling through a layer of mud and its depth x m below ground level at time t minutes is given by x = 12 - 12e^(-0.5t). How long will it take the stone to reach within 1 mm of its final position?

x = 12 - 12e^(-0.5t).

The given problem can be written as
12 - x = 12e^(-0.5t)
!2 - x = 10^-3 m.
So e^(-0.5t) = (10^-3)/12
Or e^(0.5t) = 12000
Take ln on both side and solve for t.
• April 24th 2010, 08:21 AM
Soroban
Hello, Sunyata!

Quote:

A large stone is falling through a layer of mud.
Its depth $x$ m below ground level at time $t$ minutes is given by: . $x \:=\: 12 - 12e^{-\frac{1}{2}t}$

How long will it take the stone to reach within 1 mm of its final position?

The stone reaches its final position as $t \to\infty.$

. . $\lim_{t\to\infty}x \;=\;\lim_{t\to\infty}\left(12 - \frac{12}{e^{\frac{1}{2}t}}\right) \;=\;12 - 0 \;=\;12$

The final positon is 12 meters below ground level.

When will the stone be 11.99 m below ground level?

We have: . $12 - 12e^{-\frac{1}{2}t} \;=\;11.99 \quad\Rightarrow\quad -12e^{-\frac{1}{2}t} \;=\;-0.01$

. . . . . . $e^{-\frac{1}{2}t} \;=\;\frac{0.01}{12} \quad\Rightarrow\quad -\frac{1}{2}t \;=\;\ln\left(\frac{0.01}{12}\right)$

. . . . . . $t \;=\;-2\ln\left(\frac{0.01}{12}\right) \;=\;14.18015367$