Question

Find the exact curve where the following curve is the steepest:
y= 50/ (1+6e^-2t) for t>or =to 0

Answer 1

exact point or exact curve?

Answer 2

sorry exact point. thanks

Answer 3

wouldnt the steepest be the max slope?

Answer 4

solve this equation to get the value of x and you can find y by substituting value of x to your equation:
dy/dx = 0

Answer 5

Well I wld say the min slope

Answer 6

to find the max of a 1st derivative, derive again right?

Answer 7

ya

Answer 8

What saruz i didn't get what u ssaid

Answer 9

max or min, i was thinking physically lol

Answer 10

you need to equate first derivative to zero.......you will get value of x , i.e 't' in above equaiton.

Answer 11

ya

Answer 12

The first derivative seems like it never equals zero

Answer 13

dy/dx would be the exaft opposite of "steepest"

Answer 14

you need to solve 1st derivative=0

Answer 15

it seems to be undefined at x=0 does that mke sense?

Answer 16

exact opposite means @amistre64?

Answer 17

I think he means that when the original function is at ii max or min then the derivtive is equal to 0?

Answer 18

wait, let me solve whole equation and check!

Answer 19

Thnaks I appreciate that

Answer 20

right, when dy/dx = 0 we are on a completely horizontal slope

Answer 21

you want the steepest slope possible, not the flatest

Answer 22

ya so u find what t equals when the second derivative equals zero

Answer 23

correct, at least in my understanding

Answer 24

which makes sense since inflection points change slope directions

Answer 25

But for some reason when I start doing it it just doesn't work out

Answer 26

i gave up, how can we find steepest and not the flatest @amistre64?

&&
@amistre64 are you from georgia tech institute?

Answer 27

Well the first derivative is the slope.

Answer 28

So we have to find the maximum of the slope since that is where the original function is the steepest

Answer 29

So the second derivative when it equals zero that is where the first deriviative is at its max or min

Answer 30

\[y= 50 (1+6e^{-2t})^{-1}\]
\[y'= -50 (1+6e^{-2t})^{-2}*-12e^{-2t}\]
\[y'= 600 (1+6e^{-2t})^{-2}*e^{-2t}\]
\[y''= \frac{-1200 }{e^{2t}(1+6e^{-2t})^{2}}+\frac{14400}{e^{2t}\ e^{2t}(1+6e^{-2t})^{3}}\]
im gonna have to chk with the wolf to see if i did the mathing right, but when tis =0 or is undefined are good places to check

Answer 31

K thanks

Answer 32

wolf shows no solution!!

Answer 33

You may have not used enough brackets

Answer 34

\[ y''=-\frac{1200 e^{2t} (e^{2t}-6))}{(e^{2t}+6)^3}\]
is what the wolf gives up, that might be a simplification of mine; or something completely different ;)

Answer 35

k i will play around

Answer 36

But when you make it equal to zero it becomes undefined

Answer 37

http://www.wolframalpha.com/input/?i=+-%281200+e%5E%282+t%29+%28e%5E%282+t%29-6%29%29%2F%28e%5E%282+t%29%2B6%29%5E3%3D0

Answer 38

undefined is just as good as zero, it just means that you need to test that point

Answer 39

HUH?

Answer 40

So then it is undefined at (0,0)?

Answer 41

i m guessing that it never touches x-axis

Answer 42

derivatives are finicky. they can give false readings if we assume that dy/dx = 0 is a min or a max. all the critical points to check are: =0, or undefined

Answer 43

But when it is undefined You can't solve it

Answer 44

if the original equation is undefined at the point given by the derivative; then yeah ....

Answer 45

See cus you gotta do ln to both sides (That doesn't make sense whtvr) and ln 0 is undefined

Answer 46

you can overcome that problem by finding second derivative. if second derivative is less than zero , its max and if more than zero , its min.......(doesnt work for second degree eqn)

Answer 47

im not following your reasoning here ... let me reread this

Answer 48

In order to find what t is equal you must use the natural logarithm for both sides

Answer 49

since t is an exponent

Answer 50

and when you do ln0 it is undefined

Answer 51

you mean when this = 0 right?
\[y''=-\frac{1200 e^{2t} (e^{2t}-6))}{(e^{2t}+6)^3}\]

which can only be zero when e^(2t) = 6

Answer 52

1200e^(2t) is never zero; then denominator is never zero; so that leaves:

e^(2t) - 6 = 0

Answer 53

Oh I see

Answer 54

ya so I messed up

Answer 55

Thanks for ur help ;D

Answer 56

;) youre welcome

Answer 57

can you come help me with another question i posted

Answer 58

A plot is attached.

Answer 59

Thanks I didn't need help with this question but with another one I posted but Thanks neways