anonymous
  • anonymous
In an autocatalytic chemical reaction, the product formed is a catalyst for the reaction. If Qo is the amount of the original substance and x is the amount of catalyst formed, the rate of chemical reaction is Q'(x)= kx(Q0 − x) For what value of x will the rate of chemical reaction be greatest?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
ok, so I have no idea how to deal with this. i should find where Q''(x)=0, right? but how do you take the derivative with all those variables (still on my 1st semester).
inkyvoyd
  • inkyvoyd
Those variables that are constant are treated as constants. Those variables that are variables are treated as functions of whatever you are taking the derivative to the respect to. Use the chain rule. I'm pretty sure only the first sentence applies to you, however.
anonymous
  • anonymous
so is the derivative -2kx+Qo

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More answers

inkyvoyd
  • inkyvoyd
I think so.
inkyvoyd
  • inkyvoyd
I've never taken a formal calculus course, so you probably should wait for another to answer, like @satellite73
anonymous
  • anonymous
or @UnkleRhaukus or @Zarkon ? any of you out there?
anonymous
  • anonymous
i know nothing about this, but you are given the derivative, it is \[Q'(x)= kx(Q_0 − x)\]
anonymous
  • anonymous
you want the max, set it equal to zero and solve
anonymous
  • anonymous
ok that is wrong, you want the max of the rate take the derivative of this mess, set it equal to zero and solve
anonymous
  • anonymous
you have \[Q'(x)=Q_0kx-kx^2\] max is at the vertex, namely \[-\frac{b}{2a}=-\frac{Q_0k}{2\times -k}=\frac{A_0}{2}\]
anonymous
  • anonymous
\[-\frac{b}{2a}=-\frac{Q_0k}{2\times -k}=\frac{Q_0}{2}\]
anonymous
  • anonymous
my typing is not so good
anonymous
  • anonymous
ok, that makes more sense. I'm with ya there
inkyvoyd
  • inkyvoyd
satellite, they are asking for the rate of the rate, which is a double derivative
anonymous
  • anonymous
so would I solve for Qo in 0=kx-kx^2 and then put it into Qo/2 to get the max?
anonymous
  • anonymous
Sorry, 0=Qo(kx)-kx^2
inkyvoyd
  • inkyvoyd
hick, they want when the rate is the greatest.
inkyvoyd
  • inkyvoyd
That means that the rate of the rate must be 0
inkyvoyd
  • inkyvoyd
0=-2kx+Qo
anonymous
  • anonymous
ok, so x=Qo/2k
inkyvoyd
  • inkyvoyd
yup.
anonymous
  • anonymous
but they're asking me for an x value
anonymous
  • anonymous
and they don't tell you Qo or k
anonymous
  • anonymous
that is the x value
anonymous
  • anonymous
and that is what you are asked for, so you are done
anonymous
  • anonymous
my homework won't accept that answer
anonymous
  • anonymous
am I missing something important here? I gave you everything they gave me for the question. I wonder what else they could be looking for?
UnkleRhaukus
  • UnkleRhaukus
The rate of chemical reaction will be greatest when \[x=\frac {Q_0}{2}\]
anonymous
  • anonymous
yay! it took that! i wonder where the k went?
anonymous
  • anonymous
Thanks a bunch!
UnkleRhaukus
  • UnkleRhaukus
I just read and re-read the post above made by satellite73, i dont think they are asking for the rate of the rate

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