Question

please help, explanation would be greatly appreciated (picture included)

Answer 1

y=2*e^cos(x)
\[\frac{ dy }{ dt } = 5 unit/s\]

Answer 2

do a d/dt to both sides of that equation

Answer 3

to which equation?

Answer 4

to dy/d=2(lne)(e^cosx)(-sinx)

Answer 5

i mean that that is the derivative of y=2e^cosx

Answer 6

im sorry but i dont know what that means

Answer 7

y is a function of x is a function of t, y--->x--->t

Answer 8

y ( x(t))

Answer 9

would t be the 5 units per second?

Answer 10

no nvm i know that wrong

Answer 11

to get from y to the t, you do dy/dx * dx/dt

Answer 12

notice that is dy/dt = (dy/dx)*(dx/dt) -- both sides same

Answer 13

alright what do we do next?

Answer 14

the thing is compounded like that
\[y(x[t]) = 2e^{\cos[(x(t)]}\]

Answer 15

and we need to find the derivative of that equation

Answer 16

they give you some info

dy/dx = 5 unit/s always
x = pi/2

they want you to find dx/dt at the instant when the angle x(t) is pi/2

Answer 17

i dont understadn how i would do that

Answer 18

*dy/dt = 5 mistype

Answer 19

i still dont get it im very sorry

Answer 20

dy/dt = dy/dx * dx/dt


take derivative w.r.t. x first, then w.r.t t

Answer 21

whst is w.r.t

Answer 22

with respect to,

Answer 23

dy/dt = dy/dx * dx/dt

Answer 24

derivative of y w.r.t.x times derivative of x w.r.t t

Answer 25

how am i supposed to do that if i dont know what dy/dx is or dx/dt

Answer 26

all i know is that dy/dt is 5 units/sec

Answer 27

like remember doing chain rule derivatives... you have to multiply by the derivative of the inner function

ex

[ d/dx] * cos(x^2+2) = -sin(x^2+2) * ( 2x)

Answer 28

anyway here...

Answer 29

thanks for trying but i have somewhere i be in an hour and i have to leave now, if i wasn't in a hurry i could probably understand it better thank you

Answer 30

\[\large y = 2e^{\cos(x)}\]
derivative with respect to x, d/dx both sides

Answer 31

\[\large \frac{ dy }{ dx } = -2*\sin(x) *e^{\cos(x)}\]

Answer 32

a step further realizing xis a function of t, the chain rule gives you dy/dt
\[\huge \frac{ d }{d t }=\frac{ dy }{ dx }*\frac{ dx }{ dt }\]

Answer 33

but thats what i said was the derivative at the beginning!!

Answer 34

so we just solve for dx/dt next?

Answer 35

no actual x(t) function is said, so just add that dx/dt on the end



\[\frac{ dy }{ dt } = \large \frac{ dy }{ dx } * \frac{ dx }{ dt } = -2*\sin(x) *e^{\cos(x)} * \frac{ dx }{ dt }\]

Answer 36

use that to figure the rate x is changing with the time, dx/dt, when x is that given angle

5 = -2*sin(x)*e^(cos(x)) * [dx/dt]

Answer 37

\[5 = -2*\sin(\pi/2)*e^{\cos(\pi/2)}* \frac{ dx }{ dt }\]

Answer 38

5 = -2 * 1 * 1 * dx/dt


dx/dt = 5/-2 unit/sec

Answer 39

just need to practice the chain rule thing for nested functions

to get dy/dt

you have to do d/dx first and multiply by dx/dt

Answer 40

like A [ B [ C [ D(x) ] ] ]

same, to get to dA/dx there, you need

dA/dB * dB/dC * dC/dD * dD/dx = dA*/Dx

Answer 41

gl