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Since given increasing,so $N'(t)>0$ but what will be $N''(t)$ for the slow rate part?

asked in Calculus by Active (1k points) | 73 views

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Let the x-axis represents the time and the y-axis represents the number of people infected.

As given in the question the number of people getting infected is increasing with time so N'(t) > 0

Now, it is also given that the rate of increasing of the infected persons is becoming slower with time that means as time increases the value of N'(t) is decreasing so N''(t) < 0

Hence the answer will be N'(t) > 0 and N''(t) < 0

That is option B.
answered by Loyal (8.9k points)
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Doubt :For second case, the slope will be a bit less ,so it will be be positive. (What does the second derivative signify?)
The rate of change of first derivative

Second derivative of a function tells about the concavity of the graph of f and the rate of change of the graph of f'.

Go to the above link you will find that if f''(x) < 0 then the graph of f is concave down and if f''(x) > 0 then graph of f is concave up.

I am not getting what you want to say by the slope will be a bit less here the slope of the graph decreases as the value of x(time) increases.


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