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.
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.
4 answers
a) y E (0,a)
b) y = 0
b) y = 0
Why are you posting under so many different names -- and answering your own question??
??
??
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...
Thank you helplessHelper...
3 answers