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