Question

help derivative question I dont understand

Answer 1

suppose x and y are both functions of t that x^2+2y^2=99 and that dy/dt=3 when x=-7 and y=5 then find dx/dt when x=-7 and y=5

Answer 2

a little intuition, and imagination never hurt anybody ^.^
\[\Large \frac{dx}{dt}= \frac{ \ \frac{dy}{dt} \ }{\frac{dy}{dx}}\]

Answer 3

so where do I o from there?

Answer 4

You're already given dy/dt, so might I suggest finding dy/dx? ^.^

Answer 5

but why does dx=dy/dt?

Answer 6

I didn't say that :D
All I said was this:\[\Large \frac{dx}{dt}= \frac{ \ \frac{dy}{dt} \ }{\frac{dy}{dx}}\]

Answer 7

but that implies that dx=dy/dt and dt=dy/dx

Answer 8

LOL You know that's not true :P
Case in point...
\[\Large \frac12 = \frac24\]
This doesn't mean 1 = 2 or 2 = 4 though XD

Answer 9

no but if you're given x=3 and y=4 and youre asked what does x/y equal that implies that x is 3 aand y is 4

Answer 10

That didn't make sense...

Answer 11

if you have x/y and you are given x=2+y^3 then you know that having x on top means your real equation is 2+y^3/y

Answer 12

kind of like in physics where you are given a force in Newtons but really Newtons are units of kgm/s^2

Answer 13

Yes, that's true, but that's because your denominator was y to begin with.
When you have
\[\Large \frac{a}b=\frac{c}d\] You can only conclude that a = c if b=d to begin with.

Answer 14

which I dont think we can do so I dont think we can say dx/dt=dy/dt/dy/dx

Answer 15

Have it your way, but it really follows quite logically -_-
\[\Large \frac{dx}{dt}= \frac{\frac1{dt}}{\frac1{dx}}\]

Answer 16

\[\Large \frac{dx}{dt}= \frac{\frac1{dt}}{\frac1{dx}}=\frac{\frac1{dt}}{\frac1{dx}}\cdot \frac{dy}{dy}\]

Answer 17

\[\Large \frac{dx}{dt}= \frac{\frac1{dt}}{\frac1{dx}}=\frac{\frac1{dt}}{\frac1{dx}}\cdot \frac{dy}{dy}=\frac{\frac{dy}{dt}}{\frac{dy}{dx}}\]

Answer 18

how did you change them to one over the original doesnt that require another change?

Answer 19

@agent0smith

Answer 20

Is it or is it not true that
\[\Large \frac{a}b = \frac{^1/_b}{^1/_a}\]

Answer 21

ok yes sorry

Answer 22

Then... everything follows.
\[\Large \frac{dx}{dt}= \frac{\frac1{dt}}{\frac1{dx}}=\frac{\frac1{dt}}{\frac1{dx}}\cdot \frac{dy}{dy}=\frac{\frac{dy}{dt}}{\frac{dy}{dx}}\]

Answer 23

alright Im seeing it now thank you

Answer 24

so I have three on top and just take the derivative of the original equation and place it on bottom

Answer 25

and evaluate that derivative when x = -7 and y = 5, of course.

Answer 26

alright thank you sir I apologize for being difficult at first

Answer 27

That's okay. As long as we learn ^.^