# Find the limiting velocity

• Dec 5th 2009, 02:36 PM
Ife
Find the limiting velocity
This is a question i got, which i have NO CLUE how to solve, or what they are asking. Can I please get some help??

The velocity of a body of mass m falling from rest under the action of gravity is given by the equation $\mbox{v}= \sqrt {\frac{mg}{k}}\mbox{tanh}\ ( \sqrt{\frac{gk}{m} \mbox{t}})$, where k is a constant that depends on the body's aerodynamic properties and the density of the air, g is the gravitational constant, and t is the number of seconds into the fall. Find the limiting velocity, $\lim_{\mbox{t}\rightarrow\infty} \mbox{v}$, of a 340lb skydiver (mg = 340) when k=0.006.
• Dec 5th 2009, 06:21 PM
redsoxfan325
Quote:

Originally Posted by Ife
This is a question i got, which i have NO CLUE how to solve, or what they are asking. Can I please get some help??

The velocity of a body of mass m falling from rest under the action of gravity is given by the equation $\mbox{v}= \sqrt {\frac{mg}{k}}\mbox{tanh}\ ( \sqrt{\frac{gk}{m} \mbox{t}})$, where k is a constant that depends on the body's aerodynamic properties and the density of the air, g is the gravitational constant, and t is the number of seconds into the fall. Find the limiting velocity, $\lim_{\mbox{t}\rightarrow\infty} \mbox{v}$, of a 340lb skydiver (mg = 340) when k=0.006.

Since $\sqrt{\frac{mg}{k}}$ is a constant, we can pull it out of the limit, leaving us with:

$\sqrt{\frac{mg}{k}}\cdot\lim_{t\to\infty}\tanh\lef t(\sqrt{\frac{gk}{m}t}\right)$

As $t\to\infty$, so does $\sqrt{\frac{gk}{m}t}$

so what you need to evaluate is $\sqrt{\frac{mg}{k}}\cdot\lim_{t\to\infty}\tanh(t)= \sqrt{\frac{mg}{k}}$

because $\lim_{t\to\infty}\tanh(t)=1$.

Now plug in the values.
• Dec 10th 2009, 06:33 PM
Ife
Quote:

Originally Posted by redsoxfan325
Since $\sqrt{\frac{mg}{k}}$ is a constant, we can pull it out of the limit, leaving us with:

$\sqrt{\frac{mg}{k}}\cdot\lim_{t\to\infty}\tanh\lef t(\sqrt{\frac{gk}{m}t}\right)$

As $t\to\infty$, so does $\sqrt{\frac{gk}{m}t}$

so what you need to evaluate is $\sqrt{\frac{mg}{k}}\cdot\lim_{t\to\infty}\tanh(t)= \sqrt{\frac{mg}{k}}$

because $\lim_{t\to\infty}\tanh(t)=1$.

Now plug in the values.

Thanks, but there was one part that is confusing me a bit, i guess i am muddling the concepts of limits. How is it that if when t approaches $\infty$ so does $\sqrt{\frac{gk}{m}t}$ and this results to $\lim_{t\to\infty}\tanh(t)$? I am just a bit confused here I guess... :s sorry for the late reply as well, i had some other assignments to do before completing this one, so i didn't get to focus much on this until now.
• Dec 10th 2009, 06:41 PM
redsoxfan325
Quote:

Originally Posted by Ife
Thanks, but there was one part that is confusing me a bit, i guess i am muddling the concepts of limits. How is it that if when t approaches $\infty$ so does $\sqrt{\frac{gk}{m}t}$ and this results to $\lim_{t\to\infty}\tanh(t)$? I am just a bit confused here I guess... :s sorry for the late reply as well, i had some other assignments to do before completing this one, so i didn't get to focus much on this until now.

It was slightly bad notation on my part. I was using a limit substitution without really explicitly stating it. Officially, it would be:

Let $s=\sqrt{\frac{gk}{m}t}$. As $t\to\infty$, $s\to\infty$. Thus, $\lim_{t\to\infty}\tanh\left(\sqrt{\frac{gk}{m}t}\r ight) = \lim_{s\to\infty}\tanh(s)=1
$
.
• Dec 10th 2009, 07:07 PM
Ife
Quote:

Originally Posted by redsoxfan325
It was slightly bad notation on my part. I was using a limit substitution without really explicitly stating it. Officially, it would be:

Let $s=\sqrt{\frac{gk}{m}t}$. As $t\to\infty$, $s\to\infty$. Thus, $\lim_{t\to\infty}\tanh\left(\sqrt{\frac{gk}{m}t}\r ight) = \lim_{s\to\infty}\tanh(s)=1
$
.

Thanks, but i understood that part. what i am not getting is how the limit as the expression tends to $\infty$, the result is 1? I thought that for any integer, the $\lim_{x\to\infty} x^n = \infty$ so how is the answer 1? I am still confused...(Headbang). i guess i am not sure how to treat limits of hyperbolic functions. (that trig function is pretty new to me, somehow i never got to learn it)
• Dec 10th 2009, 07:31 PM
redsoxfan325
Quote:

Originally Posted by Ife
Thanks, but i understood that part. what i am not getting is how the limit as the expression tends to $\infty$, the result is 1? I thought that for any integer, the $\lim_{x\to\infty} x^n = \infty$ so how is the answer 1? I am still confused...(Headbang). i guess i am not sure how to treat limits of hyperbolic functions. (that trig function is pretty new to me, somehow i never got to learn it)

$\tanh(x)=\frac{\sinh(x)}{\cosh(x)}=\frac{\frac{1}{ 2}(e^x-e^{-x})}{\frac{1}{2}(e^x+e^{-x})}=\frac{e^x-e^{-x}}{e^x+e^{-x}}$

As $x\to\infty$, $e^{-x}\to0$, so you end up with $\lim_{x\to\infty}\frac{e^x}{e^x}=1$.
• Dec 10th 2009, 07:51 PM
Ife
Quote:

Originally Posted by redsoxfan325
$\tanh(x)=\frac{\sinh(x)}{\cosh(x)}=\frac{\frac{1}{ 2}(e^x-e^{-x})}{\frac{1}{2}(e^x+e^{-x})}=\frac{e^x-e^{-x}}{e^x+e^{-x}}$

As $x\to\infty$, $e^{-x}\to0$, so you end up with $\lim_{x\to\infty}\frac{e^x}{e^x}=1$.

Thanks a lot for that. I never saw that calculation, we did limits of regular trig functions and other functions, but we came across hyperbolic functions by the way a long time after the fact, so we never got any detail on it... its still quite quite foreign to me. (i guess you didn't think you would've need to read and spell tht part huh?) But thanks a lot, much appreciated.
(Clapping)
• Dec 13th 2009, 07:16 AM
Ife
Quote:

Originally Posted by redsoxfan325
It was slightly bad notation on my part. I was using a limit substitution without really explicitly stating it. Officially, it would be:

Let $s=\sqrt{\frac{gk}{m}t}$. As $t\to\infty$, $s\to\infty$. Thus, $\lim_{t\to\infty}\tanh\left(\sqrt{\frac{gk}{m}t}\r ight) = \lim_{s\to\infty}\tanh(s)=1
$
.

One more thing. How is it that the sqrt fraction goes to 1 as t tends to infinity? this is before you get that it becomes tanh(t). can you give me the breakdown of that calculation?
• Dec 13th 2009, 09:05 AM
birpindy
• Dec 13th 2009, 09:09 AM
Ife
Quote:

Originally Posted by birpindy

Hey you will have to post this as a new thread. Instead of replying to mine. Start a new thread in the forum, and copy your question there, k?