In a certain chemical reaction, substance A combines with substance B to form substance 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<b and y≤a. At time t seconds after the start of the reaction, the quantity of Y present is y grams. For certain types of reactions, the rate of the reaction, in grams/sec, is given by
Rate=k(a−y)(b−y),
where k is a positive constant.
1. Sketch a graph of the rate against y. 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∈
2. Use your graph to find the value of y at which the rate of the reaction is fastest.
y=
Thank you so much. I am so clueless.
To sketch a graph of the rate against y, we need to consider the possible values of y. From the given information, we know that y≤a, so this sets an upper limit for y. The other limitation is that the rate must be nonnegative, so we need to find the values of y for which the rate is greater than or equal to zero.
To find the values of y for which the rate is nonnegative, we set the rate equation equal to zero:
0 = k(a−y)(b−y)
At this point, we can see that the rate will be nonnegative when either (a−y) and/or (b−y) is equal to zero. Therefore, we need to find the values of y that make one or both of these factors equal to zero.
1. For (a−y) = 0, we solve for y:
a−y = 0
y = a
So, the value of y at which (a−y) = 0 is y=a.
2. For (b−y) = 0, we solve for y:
b−y = 0
y = b
So, the value of y at which (b−y) = 0 is y=b.
Now, let's analyze the intervals between these values to determine the values of y for which the rate is nonnegative.
- If y ≤ a, then (a−y) will be positive or zero, and (b−y) will be positive, zero, or negative.
- If a ≤ y ≤ b, then (a−y) will be positive or zero, and (b−y) will be negative or zero.
- If y ≥ b, then (a−y) will be negative or zero, and (b−y) will be negative or zero.
However, since we are looking for the values of y for which the rate is nonnegative, we can disregard the intervals where (b−y) is negative. This leaves us with two intervals: (-∞, a] U (a, b].
Therefore, the answer to the first question is: y ∈ (-∞, a] U (a, b].
To find the value of y at which the rate of the reaction is fastest, we need to identify the maximum point on the graph. Since the rate is given by a quadratic equation, we can determine the maximum point by finding the vertex of the parabola.
To find the vertex, we can use the formula: y = -b/(2a), where a is the coefficient of the y^2 term and b is the coefficient of the y term.
In this case, the equation is k(a−y)(b−y). The coefficient of the y^2 term is -k and the coefficient of the y term is k(a+b). To find the value of y that maximizes the rate, we substitute these values into the vertex formula:
y = -(k(a+b))/(2(-k))
y = -(a+b)/2
So, the value of y at which the rate is fastest is y = -(a+b)/2.
Therefore, the answer to the second question is: y=-(a+b)/2.
I hope this explanation helps you understand how to sketch the graph of the rate and find the value of y at which the rate is fastest in this chemical reaction.