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mathcalculus
Group Title
hello, help with related rates problem.
._. (attached below)
 one year ago
 one year ago
mathcalculus Group Title
hello, help with related rates problem. ._. (attached below)
 one year ago
 one year ago

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mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
@Paynesdad It's posted ^_^
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
am i able to use the chain rule for this problem?
 one year ago

Paynesdad Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
okay so i began doing this: DQ/DT= (2X)(5Y)
 one year ago

mebs Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{dQ}{dt} = 2 = 2x\frac{dx}{dt}5y^{4}\frac{dy}{dt}\]
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
got it.
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
im getting there.
 one year ago

Paynesdad Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

mebs Group TitleBest ResponseYou've already chosen the best response.1
we know that \[\frac{dx}{dt} = 2\] so when now find what \[\frac{dy}{dt} , at, x = 3, y = 1\]
 one year ago

mebs Group TitleBest ResponseYou've already chosen the best response.1
So solve this \[2 = 2(3)(2)5(1)^{4}\frac{dy}{dt}\]
 one year ago

mebs Group TitleBest ResponseYou've already chosen the best response.1
@Paynesdad I think this is implicit because they wanted it in terms of Q.
 one year ago

Paynesdad Group TitleBest ResponseYou've already chosen the best response.1
Seems like you answered it mebs...not helped with it.
 one year ago

mebs Group TitleBest ResponseYou've already chosen the best response.1
@Paynesdad I think your right....my apologies I get exited when I do math.
 one year ago

Paynesdad Group TitleBest ResponseYou've already chosen the best response.1
LOL I know the feeling! I am glad I am not the only one.
 one year ago

mebs Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
sorry just doing the math
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
yes i see what you mean.
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
im not up to the answer yet but close, i just need to check
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i got 48.96.... :? but it's not right..
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
would using the chain rule be wrong?
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
wait, which one's?
 one year ago

mebs Group TitleBest ResponseYou've already chosen the best response.1
You can't do what you just did in your 2nd and third step
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
oh no, :? why not?
 one year ago

Paynesdad Group TitleBest ResponseYou've already chosen the best response.1
Whoa whoa whoa.. Hang on here.
 one year ago

mebs Group TitleBest ResponseYou've already chosen the best response.1
To use the product rule you do one at a time.\[\frac{dQ}{dt} = 2x\frac{dx}{dt}y^{5} + x^{2}5y^{4}\frac{dy}{dt}\]
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
oh okay, i understand that.
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i used it twice before.
 one year ago

Paynesdad Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

mebs Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

Paynesdad Group TitleBest ResponseYou've already chosen the best response.1
So using mebs equation you should get...a fractional answer.
 one year ago

Paynesdad Group TitleBest ResponseYou've already chosen the best response.1
What did you get?
 one year ago

Paynesdad Group TitleBest ResponseYou've already chosen the best response.1
@mathcalculus
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
gotta start over, i'll keep you posted.
 one year ago

Paynesdad Group TitleBest ResponseYou've already chosen the best response.1
Ok... let us know.
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
okay, jus wondering. so am i able to substitute all of them right away after i found the derivative?
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
damn it. wrong again. i got 2/33 this time
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
@mebs
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
@Paynesdad
 one year ago

Paynesdad Group TitleBest ResponseYou've already chosen the best response.1
Show us what you did again if you can.
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
ok. 1 sec
 one year ago

Paynesdad Group TitleBest ResponseYou've already chosen the best response.1
Got it...Order of operations!
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
where did i miss?
 one year ago

mebs Group TitleBest ResponseYou've already chosen the best response.1
Let me show you what must be done. \[2 = 2\times 3 \times (2) \times (1)^{5} + (3)^{2} \times 5 \times 1^{4} \frac{dy}{dx}\]
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
yes that i got.
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
I used PEMDAS
 one year ago

mebs Group TitleBest ResponseYou've already chosen the best response.1
You can't add them they both don't have dy/dx dude.
 one year ago

Paynesdad Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

mebs Group TitleBest ResponseYou've already chosen the best response.1
you get \[2 = 12 + 45\frac{dy}{dx}\]
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
ohhhhh 45 is part of dy/dx
 one year ago

mebs Group TitleBest ResponseYou've already chosen the best response.1
yea bro, step It up man dude.
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
oops.
 one year ago

Paynesdad Group TitleBest ResponseYou've already chosen the best response.1
lol yep... it is a coefficient...come on bro!
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i wasn't looking closely.
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
p.s not a bro lol but thanks! ^_^
 one year ago

mebs Group TitleBest ResponseYou've already chosen the best response.1
Its fine I was jking haha.hhaa
 one year ago

mebs Group TitleBest ResponseYou've already chosen the best response.1
I can tell that by your handwriting haha.haha
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
kay, well thanks guys!!! appreciate it tons tons.
 one year ago

mebs Group TitleBest ResponseYou've already chosen the best response.1
Good question @mathcalculus keep up the hard work.
 one year ago

Paynesdad Group TitleBest ResponseYou've already chosen the best response.1
@mebs you crack me up
 one year ago

Paynesdad Group TitleBest ResponseYou've already chosen the best response.1
Keep it up @mathcalculus
 one year ago
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