Question

Please help with calc problem!!

Answer 1

I'm not sure if I did this correctly....

Answer 2

\[-e ^{4t}\] is the answer

Answer 3

dy/dt is 2e^(2t), dx/dt is -2e^(-2t)

Answer 4

so divide dy/dt by dx/dt

Answer 5

also, we need answer i.t.o. x

Answer 6

so how to you go from t to x?

Answer 7

@Softballgirl372015 I think where you went wrong was law of exponents

Answer 8

doesn't look you did top exponent minus bottom exponent

Answer 9

@freckles thanks! so if I fix that, then I have the right answer right?

Answer 10

and @caozeyuan , what is i.t.o.x?

Answer 11

i.t.o means in terms of

Answer 12

you will have what was mentioned by @caozeyuan above and yes

Answer 13

Ohh okay!! Thanks guys! :) And for future problems, will I always take the derivative of the two equations and then divide them?

Answer 14

\[\frac{dy}{dx}=\frac{dy}{dt} \cdot \frac{dt}{dx}\]

Answer 15

so you can find dy/dt and dx/dt

Answer 16

but then flip dx/dt and multiply

Answer 17

you could of also wrote y in terms of x in found dy/dx

Answer 18

but I always think it as dy/dt divide dx/dt

Answer 19

it is weird for me to think dt/dx,

Answer 20

\[y=\frac{1}{x}+1 \\ y'=\frac{-1}{x^2} \\ x=e^{-2t} \\ y'=\frac{-1}{e^{(-2t)2}} \]

Answer 21

either way

Answer 22

Gotcha! Thank you guys soooo much!! You are a lifesaver!! :)

Answer 23

\[x=e ^{-2t},\frac{ dx }{ dt }=-2e ^{-2t}=-2x\]
\[\frac{ dy }{ dt }=2 e ^{2t}=\frac{ 2 }{ e ^{-2t} }=\frac{ 2 }{ x }\]
\[\frac{ dy }{ dx }=\frac{ \frac{ dy }{ dt } }{ \frac{ dx }{ dt } }=\frac{ 2x }{ \frac{ 2 }{ x } }=?\]

Answer 24

Okay. I really understand this stuff now. Thanks a billion!! :)

Answer 25

correction\[\frac{ dy }{ dx }=\frac{ \frac{ 2 }{ x } }{ -2x }=?\]

Answer 26

yw