Finding the terminal velocity using limits in calculus.












1












$begingroup$


So I'm tasked with finding the terminal velocity of a base jumper, which is defined as $operatorname{lim v(t)}limits_{t to infty}$ = $limlimits_{t to infty}$ $sqrt{frac{mg}{k}}tanh(sqrt{frac{mg}{k}} * t)$



I know that the answer to this problem ends up beings $sqrt{frac{mg}{k}}$, but I'm not sure how to reach the answer. Is it a matter of integrals and derivative's, or something more? Any kind of help on how to approach this would be appreciated.



For reference, velocity is equal to $v(t) =sqrt{frac{mg}{k}}tanh(sqrt{frac{mg}{k}} * t)$



Distance is equal to $d(t) = frac{m}{k}*ln(cosh(sqrt{frac{mg}{k}} * t))$



Although I don't believe you need them, but $m=75$ $g=9.8$ and $k=0.2$






calculus







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  • $begingroup$
    The result doesn't change but, for completeness, $v(t) =sqrt{frac{mg}{k}}tanh(sqrt{frac{mk}{g}} * t)$ .
    $endgroup$
    – Tony Piccolo
    Oct 15 '15 at 10:30
















1












$begingroup$


So I'm tasked with finding the terminal velocity of a base jumper, which is defined as $operatorname{lim v(t)}limits_{t to infty}$ = $limlimits_{t to infty}$ $sqrt{frac{mg}{k}}tanh(sqrt{frac{mg}{k}} * t)$



I know that the answer to this problem ends up beings $sqrt{frac{mg}{k}}$, but I'm not sure how to reach the answer. Is it a matter of integrals and derivative's, or something more? Any kind of help on how to approach this would be appreciated.



For reference, velocity is equal to $v(t) =sqrt{frac{mg}{k}}tanh(sqrt{frac{mg}{k}} * t)$



Distance is equal to $d(t) = frac{m}{k}*ln(cosh(sqrt{frac{mg}{k}} * t))$



Although I don't believe you need them, but $m=75$ $g=9.8$ and $k=0.2$






calculus







share|cite|improve this question









$endgroup$












  • $begingroup$
    The result doesn't change but, for completeness, $v(t) =sqrt{frac{mg}{k}}tanh(sqrt{frac{mk}{g}} * t)$ .
    $endgroup$
    – Tony Piccolo
    Oct 15 '15 at 10:30














1












1








1





$begingroup$


So I'm tasked with finding the terminal velocity of a base jumper, which is defined as $operatorname{lim v(t)}limits_{t to infty}$ = $limlimits_{t to infty}$ $sqrt{frac{mg}{k}}tanh(sqrt{frac{mg}{k}} * t)$



I know that the answer to this problem ends up beings $sqrt{frac{mg}{k}}$, but I'm not sure how to reach the answer. Is it a matter of integrals and derivative's, or something more? Any kind of help on how to approach this would be appreciated.



For reference, velocity is equal to $v(t) =sqrt{frac{mg}{k}}tanh(sqrt{frac{mg}{k}} * t)$



Distance is equal to $d(t) = frac{m}{k}*ln(cosh(sqrt{frac{mg}{k}} * t))$



Although I don't believe you need them, but $m=75$ $g=9.8$ and $k=0.2$






calculus







share|cite|improve this question









$endgroup$




So I'm tasked with finding the terminal velocity of a base jumper, which is defined as $operatorname{lim v(t)}limits_{t to infty}$ = $limlimits_{t to infty}$ $sqrt{frac{mg}{k}}tanh(sqrt{frac{mg}{k}} * t)$



I know that the answer to this problem ends up beings $sqrt{frac{mg}{k}}$, but I'm not sure how to reach the answer. Is it a matter of integrals and derivative's, or something more? Any kind of help on how to approach this would be appreciated.



For reference, velocity is equal to $v(t) =sqrt{frac{mg}{k}}tanh(sqrt{frac{mg}{k}} * t)$



Distance is equal to $d(t) = frac{m}{k}*ln(cosh(sqrt{frac{mg}{k}} * t))$



Although I don't believe you need them, but $m=75$ $g=9.8$ and $k=0.2$






calculus




calculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Oct 15 '15 at 3:25









etreeetree

1571616




1571616












  • $begingroup$
    The result doesn't change but, for completeness, $v(t) =sqrt{frac{mg}{k}}tanh(sqrt{frac{mk}{g}} * t)$ .
    $endgroup$
    – Tony Piccolo
    Oct 15 '15 at 10:30


















  • $begingroup$
    The result doesn't change but, for completeness, $v(t) =sqrt{frac{mg}{k}}tanh(sqrt{frac{mk}{g}} * t)$ .
    $endgroup$
    – Tony Piccolo
    Oct 15 '15 at 10:30
















$begingroup$
The result doesn't change but, for completeness, $v(t) =sqrt{frac{mg}{k}}tanh(sqrt{frac{mk}{g}} * t)$ .
$endgroup$
– Tony Piccolo
Oct 15 '15 at 10:30




$begingroup$
The result doesn't change but, for completeness, $v(t) =sqrt{frac{mg}{k}}tanh(sqrt{frac{mk}{g}} * t)$ .
$endgroup$
– Tony Piccolo
Oct 15 '15 at 10:30










1 Answer
1






active

oldest

votes


















0












$begingroup$

HINT:



We have



$$tanh x=frac{e^x+e^{-x}}{e^x-e^{-x}}$$



so that



$$lim_{xto infty}tanh x=1$$



Can you finish now?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Please let me know how I can improve my answer. I really want to give you the best answer I can. If the answer was not useful to warrant an up vote, I am happy to delete.
    $endgroup$
    – Mark Viola
    Nov 11 '15 at 18:50











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

HINT:



We have



$$tanh x=frac{e^x+e^{-x}}{e^x-e^{-x}}$$



so that



$$lim_{xto infty}tanh x=1$$



Can you finish now?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Please let me know how I can improve my answer. I really want to give you the best answer I can. If the answer was not useful to warrant an up vote, I am happy to delete.
    $endgroup$
    – Mark Viola
    Nov 11 '15 at 18:50
















0












$begingroup$

HINT:



We have



$$tanh x=frac{e^x+e^{-x}}{e^x-e^{-x}}$$



so that



$$lim_{xto infty}tanh x=1$$



Can you finish now?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Please let me know how I can improve my answer. I really want to give you the best answer I can. If the answer was not useful to warrant an up vote, I am happy to delete.
    $endgroup$
    – Mark Viola
    Nov 11 '15 at 18:50














0












0








0





$begingroup$

HINT:



We have



$$tanh x=frac{e^x+e^{-x}}{e^x-e^{-x}}$$



so that



$$lim_{xto infty}tanh x=1$$



Can you finish now?






share|cite|improve this answer









$endgroup$



HINT:



We have



$$tanh x=frac{e^x+e^{-x}}{e^x-e^{-x}}$$



so that



$$lim_{xto infty}tanh x=1$$



Can you finish now?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Oct 15 '15 at 3:34









Mark ViolaMark Viola

133k1278176




133k1278176












  • $begingroup$
    Please let me know how I can improve my answer. I really want to give you the best answer I can. If the answer was not useful to warrant an up vote, I am happy to delete.
    $endgroup$
    – Mark Viola
    Nov 11 '15 at 18:50


















  • $begingroup$
    Please let me know how I can improve my answer. I really want to give you the best answer I can. If the answer was not useful to warrant an up vote, I am happy to delete.
    $endgroup$
    – Mark Viola
    Nov 11 '15 at 18:50
















$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can. If the answer was not useful to warrant an up vote, I am happy to delete.
$endgroup$
– Mark Viola
Nov 11 '15 at 18:50




$begingroup$
Please let me know how I can improve my answer. I really want to give you the best answer I can. If the answer was not useful to warrant an up vote, I am happy to delete.
$endgroup$
– Mark Viola
Nov 11 '15 at 18:50


















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