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.
In part a, to find the values of y for which the rate is nonnegative, we need to consider the different intervals where the rate could potentially be negative or zero.
We are given that a < b, which means that the initial quantity of substance A is less than the initial quantity of substance B.
Let's analyze the different intervals:
1. For 0 <= y < a, the value of (a - y) is positive, and (b - y) is positive as well. Therefore, the rate equation is positive in this interval: Rate > 0.
2. At y = a, we have (a - y) = 0, so the rate equation becomes Rate = k * 0 * (b - y) = 0. The rate is zero at this point.
3. For a < y < b, the value of (a - y) becomes negative while (b - y) is still positive. This makes the rate equation negative: Rate < 0.
4. At y = b, (b - y) = 0, so the rate equation becomes Rate = k * (a - y) * 0 = 0. The rate is zero at this point.
5. For y > b, both (a - y) and (b - y) become negative, resulting in a positive rate again: Rate > 0.
Putting all these intervals together, the values of y for which the rate is nonnegative can be expressed as [0, a] U [b, +infinity).
Now in part b, we want to find the value of y at which the rate is fastest. The highest rate of the reaction corresponds to the point of maximum slope on the rate graph, which occurs when the derivative of the rate with respect to y is zero.
Taking the derivative of the rate equation with respect to y:
d(Rate)/dy = k * (-1) * (b - y) + k * (a - y) * (-1)
d(Rate)/dy = -k * (b - y) - k * (a - y)
d(Rate)/dy = -k * (b - y - a + y)
d(Rate)/dy = -k * (-b - a + 2y)
d(Rate)/dy = k * (b + a - 2y)
Setting d(Rate)/dy equal to zero:
0 = k * (b + a - 2y)
Solving for y:
2y = b + a
y = (b + a)/2
The value of y that corresponds to the maximum rate of the reaction is (b + a)/2.
I apologize for the misunderstanding in my previous response. The correct answer for part b is y = (b + a)/2.