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mathcalculus
 2 years ago
hello, help with related rates problem.
._. (attached below)
mathcalculus
 2 years ago
hello, help with related rates problem. ._. (attached below)

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

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

Paynesdad
 2 years ago
Best ResponseYou've already chosen the best response.1Ok 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
 2 years ago
Best ResponseYou've already chosen the best response.0okay so i began doing this: DQ/DT= (2X)(5Y)

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

Paynesdad
 2 years ago
Best ResponseYou've already chosen the best response.1no 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
 2 years ago
Best ResponseYou've already chosen the best response.1we know that \[\frac{dx}{dt} = 2\] so when now find what \[\frac{dy}{dt} , at, x = 3, y = 1\]

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

mebs
 2 years ago
Best ResponseYou've already chosen the best response.1@Paynesdad I think this is implicit because they wanted it in terms of Q.

Paynesdad
 2 years ago
Best ResponseYou've already chosen the best response.1Seems like you answered it mebs...not helped with it.

mebs
 2 years ago
Best ResponseYou've already chosen the best response.1@Paynesdad I think your right....my apologies I get exited when I do math.

Paynesdad
 2 years ago
Best ResponseYou've already chosen the best response.1LOL I know the feeling! I am glad I am not the only one.

mebs
 2 years ago
Best ResponseYou've already chosen the best response.1Well @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
 2 years ago
Best ResponseYou've already chosen the best response.0sorry just doing the math

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

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

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

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

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0wait, which one's?

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

mathcalculus
 2 years ago
Best ResponseYou've already chosen the best response.0oh no, :? why not?

Paynesdad
 2 years ago
Best ResponseYou've already chosen the best response.1Whoa whoa whoa.. Hang on here.

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

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

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

Paynesdad
 2 years ago
Best ResponseYou've already chosen the best response.1Remember 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
 2 years ago
Best ResponseYou've already chosen the best response.1You 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.

Paynesdad
 2 years ago
Best ResponseYou've already chosen the best response.1So using mebs equation you should get...a fractional answer.

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

mathcalculus
 2 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?

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

Paynesdad
 2 years ago
Best ResponseYou've already chosen the best response.1Show us what you did again if you can.

Paynesdad
 2 years ago
Best ResponseYou've already chosen the best response.1Got it...Order of operations!

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

mebs
 2 years ago
Best ResponseYou've already chosen the best response.1You can't add them they both don't have dy/dx dude.

Paynesdad
 2 years ago
Best ResponseYou've already chosen the best response.1you 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
 2 years ago
Best ResponseYou've already chosen the best response.1you get \[2 = 12 + 45\frac{dy}{dx}\]

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

mebs
 2 years ago
Best ResponseYou've already chosen the best response.1yea bro, step It up man dude.

Paynesdad
 2 years ago
Best ResponseYou've already chosen the best response.1lol yep... it is a coefficient...come on bro!

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

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

mebs
 2 years ago
Best ResponseYou've already chosen the best response.1Its fine I was jking haha.hhaa

mebs
 2 years ago
Best ResponseYou've already chosen the best response.1I can tell that by your handwriting haha.haha

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

mebs
 2 years ago
Best ResponseYou've already chosen the best response.1Good question @mathcalculus keep up the hard work.

Paynesdad
 2 years ago
Best ResponseYou've already chosen the best response.1@mebs you crack me up

Paynesdad
 2 years ago
Best ResponseYou've already chosen the best response.1Keep it up @mathcalculus
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