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anonymous
 3 years ago
hello, help with related rates problem.
._. (attached below)
anonymous
 3 years ago
hello, help with related rates problem. ._. (attached below)

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Paynesdad It's posted ^_^

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0am i able to use the chain rule for this problem?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ok 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay so i began doing this: DQ/DT= (2X)(5Y)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{dQ}{dt} = 2 = 2x\frac{dx}{dt}5y^{4}\frac{dy}{dt}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0we know that \[\frac{dx}{dt} = 2\] so when now find what \[\frac{dy}{dt} , at, x = 3, y = 1\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So solve this \[2 = 2(3)(2)5(1)^{4}\frac{dy}{dt}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Paynesdad I think this is implicit because they wanted it in terms of Q.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Seems like you answered it mebs...not helped with it.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Paynesdad I think your right....my apologies I get exited when I do math.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0LOL I know the feeling! I am glad I am not the only one.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well @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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry just doing the math

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes i see what you mean.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0im not up to the answer yet but close, i just need to check

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i got 48.96.... :? but it's not right..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0would using the chain rule be wrong?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You can't do what you just did in your 2nd and third step

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Whoa whoa whoa.. Hang on here.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0To use the product rule you do one at a time.\[\frac{dQ}{dt} = 2x\frac{dx}{dt}y^{5} + x^{2}5y^{4}\frac{dy}{dt}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh okay, i understand that.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i used it twice before.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Remember 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So using mebs equation you should get...a fractional answer.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0gotta start over, i'll keep you posted.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay, jus wondering. so am i able to substitute all of them right away after i found the derivative?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0damn it. wrong again. i got 2/33 this time

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Show us what you did again if you can.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Got it...Order of operations!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Let me show you what must be done. \[2 = 2\times 3 \times (2) \times (1)^{5} + (3)^{2} \times 5 \times 1^{4} \frac{dy}{dx}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You can't add them they both don't have dy/dx dude.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you get \[2 = 12 + 45\frac{dy}{dx}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ohhhhh 45 is part of dy/dx

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yea bro, step It up man dude.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0lol yep... it is a coefficient...come on bro!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i wasn't looking closely.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0p.s not a bro lol but thanks! ^_^

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Its fine I was jking haha.hhaa

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I can tell that by your handwriting haha.haha

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0kay, well thanks guys!!! appreciate it tons tons.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Good question @mathcalculus keep up the hard work.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@mebs you crack me up

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Keep it up @mathcalculus
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