Question

find the exact length of the parametric curve x= e^t cost y= e^t sint 0 11 years ago

Answer 1

\[x = e^t cost\]

Answer 2

\[y= e^t sint\]

Answer 3

hello

Answer 4

kinda confused
you mention 4 variables...

Answer 5

but thats whats on my homework

Answer 6

or is t meant to be theta

Answer 7

i think that t means theta

Answer 8

that would be great then :p

Answer 9

so have you found dx/dt and dy/dt

Answer 10

do we get rid of the t's in this problem?

Answer 11

we don't need to

Answer 12

why not

Answer 13

\[L=\int\limits_0^{\frac{\pi}{5}}\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2} dt\]

Answer 14

the derivative of e^t is just e^t right?

Answer 15

yah!

Answer 16

okay so dx is e

Answer 17

\[x=e^t \cos(t) \\ \\ \text{ To find } \frac{dx}{dt} \text{ use product rule }\]

Answer 18

dx = -te^t sin(t)

Answer 19

\[\frac{dx}{dt}=\frac{d(e^t)}{dt} \cdot \cos(t)+\frac{d (\cos(t))}{dt} \cdot e^t\]

Answer 20

remember product rule is
(uv)'=u times v'+v times u'

Answer 21

so dx= te^t cos(t) + - e^t sin(t)

Answer 22

i thought you said derivative of e^t was e^t
where does the extra t come from

Answer 23

i typed too fast lol sorry

Answer 24

\[\frac{dx}{dt}=e^t \cos(t)-\sin(t)e^t \\ \frac{dx}{dt}=e^t(\cos(t)-\sin(t))\]

Answer 25

now we need to find dy/dt

Answer 26

dy = e^t sin(t) + e^t cos(t)

Answer 27

\[x=e^t \cos(t) \\ \frac{dx}{dt}=e^t(\cos(t)-\sin(t)) \\ y=e^tsin(t) \\ \frac{dy}{dt}=e^t \sin(t)+ e^t \cos(t) =e^t(\sin(t)+\cos(t))\]
great now we got square these ugly thingys

Answer 28

\[(\frac{dx}{dt})^2=e^{2t}(\cos(t)-\sin(t))^2=e^{2t}[\cos^2(t)-2\cos(t)\sin(t)+\sin^2(t)] \\ (\frac{dx}{dt})^2=e^{2t}[1-\sin(2t)]\]
you try to simplify dy/dt now

Answer 29

okay

Answer 30

by the way you might find using (a+b)^2=a^2+2ab+b^2 useful :)

Answer 31

okay

Answer 32

for the trig identities right?

Answer 33

I don't know what you mean

Answer 34

like for what you do for the (cos(t) - sin(t))^2

Answer 35

I used (a-b)^2=a^2-2ab+b^2

Answer 36

yep do the same to dy

Answer 37

i got e^2t[1+cos(2t)] for dy

Answer 38

let me check

Answer 39

\[(\frac{dy}{dt})^2=e^{2t}(\sin^2(t)+2\sin(t)\cos(t)+\cos^2(t)) \\ (\frac{dy}{dt})^2=e^{2t}(1+\sin(2t))\]
the only thing I disagree about is you making 2sin(t)cos(t) equal to something other than sin(2t)

Answer 40

\[(\frac{dx}{dt})^2+(\frac{dy}{dt})^2=e^{2t}(1+\sin(2t))+e^{2t}(1-\sin(2t))\]
combine like terms

Answer 41

distribute first then combine like terms

Answer 42

you will see somethign really cool happen :)

Answer 43

for the dy why is the 2sin(t)cos(t) equal to sin(2t)

Answer 44

it is always equal to that

Answer 45

it is a trig identity

Answer 46

sin(2t)=2sin(t)cos(t)
cos(2t)=cos^2(t)-sin^2(t)

Answer 47

oh i need to remember that one

Answer 48

@El_Arrow are you still simplifying the thingy I asked you too?

Answer 49

\[(\frac{dx}{dt})^2+(\frac{dy}{dt})^2=e^{2t}(1+\sin(2t))+e^{2t}(1-\sin(2t)) \]
this thingy
combing like terms?

Answer 50

yes trying to

Answer 51

\[e^{2t} +e^{2t} \sin(2t)+e^{2t}-e^{2t}\sin(2t)\]



===
you know u+u=? and v-v=?

Answer 52

it would be 2e^2t

Answer 53

\[L=\int\limits\limits_0^{\frac{\pi}{5}}\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2} dt \]

\[L=\int\limits\limits_0^{\frac{\pi}{5}}\sqrt{2e^{2t}} dt \]

Answer 54

well its