In a chemical reaction, substance A combines with substance B to form subtance Y. At the start of the reaction, the quantity of A present is a grams, and the quantity of B present is b grams. Assume a is less than b. At time t seconds after the start of the reaction, the quantity of Y present is y grams. For certain types of reaction, the rate of the reaction, in grams/sec, is given by
Rate = k(a-y)(b-y), k is a positive constant.
a. For what values of y is the rate nonnegative?
Give your answer as a union of intervals, e.g., (-infinity,-a] U (a, 2b)
y E_______________________
b. Find the value of y at which the rate of the reaction is fastest.
y= _________________________
I thought that in part A all nonnegative values were going to be anything less than a and everything larger than b so I typed (my homework is online) (-INF, a] U [b, INF) and in b the answer I got was (1/2)(a+b) but they are both wrong... please help....
Calculus - Jai, Sunday, April 10, 2011 at 12:48am
for (a) since a < b , for a certain value of y , the value of (a - y) will become negative first, and thus the Rate becomes negative,, then after some time (b - y) will become negative too, and the Rate becomes positive. let's look at some points:
at 0 <= y < a , Rate > 0
at y = a , Rate = 0
at a < y < b , Rate < 0
at y = b , Rate = 0
at y > b , Rate > 0
thus, rate is non-negative (but may be equal to zero) at values of y which is
[0 , a] U [b , +infinity)
*note that we start at 0 since quantity/mass can never be negative. another, the +infinity will only possible if a and be is continuously supplied or fed to the reactor. otherwise, at a finite value of a and b, y will only reach a certain maximum value (y,max) when the reaction is complete (or at infinite time)
for (b), we take the derivative of Rate = k(a-y)(b-y) with respect to y, and equate Rate to zero since maximum rate (slope is zero):
R = k(a-y)(b-y)
R = k(ab - by - ay + y^2)
0 = k[-b - a + 2y]
0 = -b - a + 2y
y = (a+b)/2
*we got the same answer. are you sure it's wrong?
hope this helps~
Calculus - Jai, Sunday, April 10, 2011 at 1:24am
ahh i think i know why y = (a+b)/2 is wrong,, it's actually the MINIMUM, not the maximum~ ^^;
i tried assigning some values to the variables,, and from the graph, rate -> infinity at y -> inifinity , or at a finite value of a and b, when the reaction is at completion (time at infinity), the rate is max at y = y,max , provided that this y,max is greater than b.
Sorry to post this one again but there are things that I still don't understand... Thanks for the explanation btw, I don't know why it didn't occur to me that mass can't be negative =P... I now get why the answer for B can't be (a+b)/2 but still don't get what the max is, sorry.... and I typed the new answer for part A and for some reason it still says it's wrong.
a) y E (0,a)
b) y = 0
Why are you posting under so many different names -- and answering your own question??
??
If we expect the rate to be nonnegative, we must have 0 ≤ y ≤ a and 0 ≤ y ≤ b.
Since we assume a < b, we restrict y to 0 ≤ y ≤ a.
In fact, the expression for the rate is nonnegative for y greater than b, but these
values of y are not meaningful for the reaction.
Rate
0
(b) From the graph, we see that the maximum rate occurs when y = 0; that is, at the
start of the reaction.
Hi, I wasn't posting under different names, I have just used my name and I was just posting the explanation I received from other tutor, I didn't understand some of it so I posted it again because right now it must be on page 5 or something like that... I'm not answering my own questions... I was truly confused...
Thank you helplessHelper...
Well, it seems like you're having some trouble with this problem. Let me sprinkle some humor into the mix and see if that helps clarify things for you.
a. For what values of y is the rate nonnegative?
Well, since mass can't be negative (I mean, who would want negative mass?), we need to find the values of y where the rate is greater than or equal to zero. According to my calculations, the rate is nonnegative when y is in the interval [0 , a] U [b , +infinity). So, if y is anywhere between zero and a, or anywhere between b and positive infinity, you're good to go!
b. Find the value of y at which the rate of the reaction is fastest.
Ah, the need for speed! To find the value of y where the rate is at its maximum, we need to take the derivative of the rate expression and set it equal to zero. After some mathematical acrobatics, we find that y = (a+b)/2. But hold your horses! This is actually the minimum, not the maximum. The rate keeps on increasing as y approaches infinity or, if we have finite values of a and b, when the reaction is complete.
I hope this helps you understand the problem a bit better. And remember, even if you're feeling confused, just keep clowning around and eventually it'll all make sense!
Let's break down the problem step by step:
a. To find the values of y for which the rate is nonnegative, we need to find the values of y that make the expression (a - y)(b - y) nonnegative.
We know that (a - y) and (b - y) are nonnegative when y is less than both a and b, as well as when y is greater than both a and b.
So the values of y that make the expression (a - y)(b - y) nonnegative are in the intervals [0, a] and [b, +infinity).
Therefore, the answer for part a is: y E [0, a] U [b, +infinity).
b. To find the value of y at which the rate is fastest, we need to find the y-value that maximizes the function Rate = k(a - y)(b - y).
To maximize this function, we can take its derivative with respect to y and set it equal to zero:
Rate' = k(-b - a + 2y) = 0
Solving for y, we get:
2y = b + a
y = (b + a)/2
Therefore, the value of y at which the rate is fastest is y = (b + a)/2.
I hope this clarifies the solution for you.