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hello, help with related rates problem. ._. (attached below)

Mathematics
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@Paynesdad It's posted ^_^

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Other answers:

am i able to use the chain rule for this problem?
Ok so take the derivative with respect of the equation using implicit differentiation. Since both x and y are variables that depend on t you will have to use the chain rule.
okay so i began doing this: DQ/DT= (2X)(5Y)
\[\frac{dQ}{dt} = -2 = 2x\frac{dx}{dt}5y^{4}\frac{dy}{dt}\]
got it.
im getting there.
no you can't do the derivative of x and y and the same time. \[\frac{ dQ }{ dt }= \frac{ d(x^2y^5) }{ dt }\]First take the derivative of the right side with respect to t of the x factor and treat y factor as a constant + the derivative of the right side with respect to t of the y factor treating the x factor as a constant.
we know that \[\frac{dx}{dt} = -2\] so when now find what \[\frac{dy}{dt} , at, x = 3, y = 1\]
So solve this \[-2 = 2(3)(-2)5(1)^{4}\frac{dy}{dt}\]
@Paynesdad I think this is implicit because they wanted it in terms of Q.
Seems like you answered it mebs...not helped with it.
@Paynesdad I think your right....my apologies I get exited when I do math.
LOL I know the feeling! I am glad I am not the only one.
Well @mathcalculus does this make sense to you.. I just subbed values that were given to me already from the question into the derivative of Q. and x and y were already defined.
sorry just doing the math
yes i see what you mean.
im not up to the answer yet but close, i just need to check
i got 48.96.... :? but it's not right..
would using the chain rule be wrong?
wait, which one's?
All of them
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You can't do what you just did in your 2nd and third step
oh no, :? why not?
Whoa whoa whoa.. Hang on here.
?
To use the product rule you do one at a time.\[\frac{dQ}{dt} = 2x\frac{dx}{dt}y^{5} + x^{2}5y^{4}\frac{dy}{dt}\]
Right ^
oh okay, i understand that.
i used it twice before.
Remember d(xy)/dt = y (dx/dt) + x (dy/dt). Also mathcalculus when you took the derivative you left off the dx/dt and dy/dt out of your equations.
You already know that \[\frac{dQ}{dt}= -2 \] You know that \[\frac{dx}{dt} = -2\] and you know that \[x = 3 , y = 1\] just substitute it into the derivative of dQ/dt and solve.
yes
So using mebs equation you should get...a fractional answer.
What did you get?
gotta start over, i'll keep you posted.
Ok... let us know.
okay, jus wondering. so am i able to substitute all of them right away after i found the derivative?
damn it. wrong again. i got -2/33 this time
Show us what you did again if you can.
ok. 1 sec
1 Attachment
Got it...Order of operations!
?
where did i miss?
Let me show you what must be done. \[-2 = 2\times 3 \times (-2) \times (1)^{5} + (3)^{2} \times 5 \times 1^{4} \frac{dy}{dx}\]
yes that i got.
I used PEMDAS
You can't add them they both don't have dy/dx dude.
you have -12 + 45*dy/dt=-2 ...You can't add -12 to 45 until you multiply 45 times dy/dt...It's like 4=-3+7x....you don't simplify that as 4 = 4x ... right
you get \[-2 = -12 + 45\frac{dy}{dx}\]
ohhhhh 45 is part of dy/dx
yea bro, step It up man dude.
oops.
lol yep... it is a coefficient...come on bro!
i wasn't looking closely.
p.s not a bro lol but thanks! ^_^
Its fine I was jking haha.hhaa
I can tell that by your handwriting haha.haha
kay, well thanks guys!!! appreciate it tons tons.
Good question @mathcalculus keep up the hard work.
@mebs you crack me up
Keep it up @mathcalculus

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