Question

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

Answer 1

??

Answer 2

@Paynesdad It's posted ^_^

Answer 3

am i able to use the chain rule for this problem?

Answer 4

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.

Answer 5

okay so i began doing this: DQ/DT= (2X)(5Y)

Answer 6

\[\frac{dQ}{dt} = -2 = 2x\frac{dx}{dt}5y^{4}\frac{dy}{dt}\]

Answer 7

got it.

Answer 8

im getting there.

Answer 9

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.

Answer 10

we know that \[\frac{dx}{dt} = -2\]

so when now find what \[\frac{dy}{dt} , at, x = 3, y = 1\]

Answer 11

So solve this \[-2 = 2(3)(-2)5(1)^{4}\frac{dy}{dt}\]

Answer 12

@Paynesdad I think this is implicit because they wanted it in terms of Q.

Answer 13

Seems like you answered it mebs...not helped with it.

Answer 14

@Paynesdad I think your right....my apologies I get exited when I do math.

Answer 15

LOL I know the feeling! I am glad I am not the only one.

Answer 16

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.

Answer 17

sorry just doing the math

Answer 18

yes i see what you mean.

Answer 19

im not up to the answer yet but close, i just need to check

Answer 20

i got 48.96.... :? but it's not right..

Answer 21

would using the chain rule be wrong?

Answer 22

wait, which one's?

Answer 23

All of them

Answer 24

You can't do what you just did in your 2nd and third step

Answer 25

oh no, :? why not?

Answer 26

Whoa whoa whoa.. Hang on here.

Answer 27

?

Answer 28

To use the product rule you do one at a time.\[\frac{dQ}{dt} = 2x\frac{dx}{dt}y^{5} + x^{2}5y^{4}\frac{dy}{dt}\]

Answer 29

Right ^

Answer 30

oh okay, i understand that.

Answer 31

i used it twice before.

Answer 32

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.

Answer 33

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.

Answer 34

yes

Answer 35

So using mebs equation you should get...a fractional answer.

Answer 36

What did you get?

Answer 37

@mathcalculus

Answer 38

gotta start over, i'll keep you posted.

Answer 39

Ok... let us know.

Answer 40

okay, jus wondering. so am i able to substitute all of them right away after i found the derivative?

Answer 41

damn it. wrong again. i got -2/33 this time

Answer 42

@mebs

Answer 43

@Paynesdad

Answer 44

Show us what you did again if you can.

Answer 45

ok. 1 sec

Answer 46

Got it...Order of operations!

Answer 47

?

Answer 48

where did i miss?

Answer 49

Let me show you what must be done.

\[-2 = 2\times 3 \times (-2) \times (1)^{5} + (3)^{2} \times 5 \times 1^{4} \frac{dy}{dx}\]

Answer 50

yes that i got.

Answer 51

I used PEMDAS

Answer 52

You can't add them they both don't have dy/dx dude.

Answer 53

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

Answer 54

you get \[-2 = -12 + 45\frac{dy}{dx}\]

Answer 55

ohhhhh 45 is part of dy/dx

Answer 56

yea bro, step It up man dude.

Answer 57

oops.

Answer 58

lol yep... it is a coefficient...come on bro!

Answer 59

i wasn't looking closely.

Answer 60

p.s not a bro lol but thanks! ^_^

Answer 61

Its fine I was jking haha.hhaa

Answer 62

I can tell that by your handwriting haha.haha

Answer 63

kay, well thanks guys!!! appreciate it tons tons.

Answer 64

Good question @mathcalculus keep up the hard work.

Answer 65

@mebs you crack me up

Answer 66

Keep it up @mathcalculus