MHB How can the equation ln(1+d)=d be used to approximate p= Ae^-0.004h/30?

[Bawx]

Hi guys,
I'm having a lot of trouble with this question and any help would be greatly appreciated, thanks!

1. Sketch the curve y = ln x and find the tangent line to this curve at the point where the curve crosses the x-axis. Deduce that, for small delta,

ln(1+d) '=' d

2. Use the approximation from the previous part to deduce that

p = p(h) '=' Ae^-0.004h/30

*the '=' are supposed to be the squiggly approximately signs but I'm not sure how to type those sorry

-I think I've found the equation for the tangent line, as ln(1)=0 and y'=1/x so by using y-y1=m(x-x1) the tangent line is y=x-1 but I'm not sure how this relates to delta at all?

- I'm not really sure how to approach the second part at all

Thanks :)

 

[topsquark]

Hi guys,
I'm having a lot of trouble with this question and any help would be greatly appreciated, thanks!

1. Sketch the curve y = ln x and find the tangent line to this curve at the point where the curve crosses the x-axis. Deduce that, for small delta,

ln(1+d) '=' d

Good so far. The rest relies on that line. The line approximates the value of ln(x) in the area near to x = 1. According to your line, any value of ln(1 + d) is going to be close to the value on the tangent line, in this case when x = 1 + d, the y value will be y = (1 + d) - 1 = d.

2. Use the approximation from the previous part to deduce that

p = p(h) '=' Ae^-0.004h/30

What's p?

-Dan

 

[MarkFL]

This same problem was recently posted by another member, and you may find the discussion helpful:

http://mathhelpboards.com/calculus-10/y-ln-x-10679.html

 

[Bawx]

Oh ok, so since ln(x) '=' x-1 you can just sub (1+d) to both sides?

Ah oops, I forgot to add the explanation above the questions:

Air pressure decays approximately exponentially at about 0.4 per cent for each rise of 30 metres above sea level. If we let p = p(h) denote air pressure (measured in some appropriate units) at h metres above sea level, then we can model this phenomenon using the equation

p = p(h) = Ae^kh

for some appropriate constants A and k.

 

[Bawx]

This same problem was recently posted by another member, and you may find the discussion helpful:

http://mathhelpboards.com/calculus-10/y-ln-x-10679.html

Oh oops, I didn't realize that. Thanks for pointing it out! That one seems to get stuck on finding the tangent line tough, I hope it's ok if I keep using this thread. Sorry about that.

 

[MarkFL]

Oh oops, I didn't realize that. Thanks for pointing it out! That one seems to get stuck on finding the tangent line tough, I hope it's ok if I keep using this thread. Sorry about that.

There's nothing wrong with posting a problem that has been posted before (it happens from time to time, we only discourage the same person from posting a problem more than once), I just wanted you to be able to see the discussion there as well. :D

 

[chisigma]

If You can demonstrate that is...

$\displaystyle \lim_{d \rightarrow 0} \frac{\ln (1 + d)}{d} = 1\ (1)$

... then You can say that for 'small d' is $\displaystyle \ln (1 + d) \sim d$. The (1) is easily demonstrable using the concept of derivative but let's suppose You don't have yet such a concept and the only You know is the definition of exponential...

$\displaystyle e^{x} = \lim_{n \rightarrow \infty} (1 + \frac{x}{n})^{n}\ (2)$

... and the fact that logarithm is the inverse of exponential. What You can do is to compute the limit...

$\displaystyle \lim_{d \rightarrow 0} e^{\frac{\ln (1+d)}{d}} = \lim_{d \rightarrow 0} (1+ d)^{\frac{1}{d}} = \lim_{n \rightarrow \infty} (1+ \frac{1}{n})^{n} = e\ (3)$

... and (1) is demonstrated...

Kind regards

$\chi$ $\sigma$