Question

Come up with the specific values of coordinates (x,y) of the highest point on the curve parameterized by [x(t), y(t)] = [te^t, 1-t^2] from t being negative infinty to infinity

Answer 1

so we should be able to find out whne the slope of the line tangent to the curve is zero

Answer 2

we can either take the derivative as is; or convert this to an x(y) deal and find when the slope is undefined

Answer 3

so dy/dx?

Answer 4

yes, but lets name is dx/dt and dy/dt just to be proper :)

Answer 5

\[-2t \div t ^{2}e ^{t}\]

Answer 6

and set that to 0?

Answer 7

the set to zero part is good, but lets see fif we can work on that derivatives better

Answer 8

r = < te^t, 1-t^2>

the derivative of a componentwise function is the derivative of the components ....

Answer 9

r' = <t' e^t+t e'^t , (1-t^2)'>

r' = <e^t+t e^t , -2t>

r' = <e^t(1+t) , -2t>

when r' = <0,0> we should be good

Answer 10

our critical point to chk, is (-1,0) do you see why?

Answer 11

because at point -1 thats the max x value? because e^0 = 1

Answer 12

:) remember that we found t values that should suffice; we need to plug these into our original set up to find the actual point that is defined

Answer 13

that might be in error, but it sounds good lol

Answer 14

lol im still a bit confused on what to do!!! so do i use dy/dx??

Answer 15

or just dy/dt and set that equal to 0?

Answer 16

http://www.wolframalpha.com/input/?i=y%3D+te%5Et+and+x%3D1-t%5E2

you set dx/dt = 0 and set dy/dt = zero

this setup relies upon "t" for its values. x(t) and y(t) are function the depend on "t" to define them

Answer 17

the point -1,0 on the graph is the low point

Answer 18

ah and thats how u get -1,0? so what would be the high point?

Answer 19

its a critical point that we need to look at and determine its condition

Answer 20

the graph in the link shows that we are at a minimum and there there is no maximum other than infinity

Answer 21

ok and what do u mean by determine its condition, sorry this is taking me a while to understand

Answer 22

so no highg point exists?

Answer 23

when we place a condition upon something, then we want to know when it is doing what we want it to do.

when the graph is doing a minimum then it is at a low point, a dip in the graph.

when the graph is at a maximum, there will be a high point in the graph.

the derivative simply tells us when the graph quits moving up and down; it doesnt tell us the condition of high and low.

Answer 24

so high point does not exist?

Answer 25

we need the second derivative if we are to NOT guess at this :)

Answer 26

ok so then u set that equal to 0? i got -2,0

Answer 27

xe^x + 2e^x

Answer 28

r' = <e^t(1+t) , -2t>

derive it again for "concavity"

r'' = <e^t(1+t)+e^t , -2>

r'' = <e^t(2+t) , -2> ; this is never at (0,0)

Answer 29

2,-2?

Answer 30

err -2,-2 sorry i meant

Answer 31

asdf;laskjfaf lol so lost

Answer 32

when t=-2, there might be concavity; its hard for me to say cause I cant be to certain :)

but, the value for r' seem to be values of t for each component

Answer 33

when t=-1, x is at a critical point; when t=0 then y is at a critical point

Answer 34

when t=-2, then x should be at a turning point in concavity

Answer 35

hmm

Answer 36

so would be the highest point, like the actual answer?

Answer 37

amistre are you sure we want r'=<0,0> ?
my understanding is a bit different....

Answer 38

i was until i started trying to fit it to the graph :) but now i have to wonder ....

Answer 39

I think we need\[\frac{dy}{dx}={{dy\over dt}\over{dx\over dt}}=0\]

Answer 40

so that's\[\frac{-2t}{e^t(1+t)}=0\implies t=0\]as our only critical point

Answer 41

ah so would it be 0,1 then if u plug the value of t back into its original functions?

Answer 42

if got my x andy backwards in me graph :)

Answer 43

and that would be my high point? 0,1?

Answer 44

yes, I believe the point in question is <0,1>
whether we can show it is a max and not a min I'm not sure about

Answer 45

http://www.wolframalpha.com/input/?i=x%3D+te%5Et+and+y%3D1-t%5E2%2C+t+from+-2+to+2

i had this thing laying on its side lol

Answer 46

well yeah u can u choose a point before and a point after and see that it is less then 0,1

Answer 47

yeah, but I wanted to do it with concavity
oh well, it is clearly the max, so there ya go :D

Answer 48

yeah, second derivative is\[\frac{-2}{e^t(1+t)}<0\]so there we can show it with concavity

Answer 49

thank u guys!

Answer 50

good job :)