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?

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- Apr 24th 2010, 06:46 AMSunyataMotion - Please, any help???
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?

- Apr 24th 2010, 08:01 AMsa-ri-ga-ma
- Apr 24th 2010, 08:21 AMSoroban
Hello, Sunyata!

Quote:

A large stone is falling through a layer of mud.

Its depth $\displaystyle x$**m**below ground level at time $\displaystyle t$ minutes is given by: .$\displaystyle x \:=\: 12 - 12e^{-\frac{1}{2}t}$

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

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

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

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

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

Answer: about 14.2 minutes.