Asolution initially contains 200 bacteria. 1. assuming the number y increases at a rate proportional to the number present, write down a differential equation connecting y and the time t. 2. if the rate of increase of the number is initially 100 per hour, how many bacteria are there after 2 hours? solution:

1.

2.543.6

Step-by-step explanation:

We are given that

y(0)=200

Let y be the number of bacteria at any time

=Number of bacteria per unit time

Where k=Proportionality constant

2.,y'(0)=100

Integrating on both sides then, we get

We have y(0)=200

Substitute the values then , we get

Substitute the value of C then we get

Differentiate w.r.t

Substitute the given condition then, we get

Substitute t=2

Then, we get

e=2.718

Hence, the number of bacteria after 2 hours=543.6

1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.

9987

step-by-step explanation: