What percentage is likely to be infected in week 4 if N(3) = 8 and N′(3) = 1.2?

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Homework Statement


Let N(t) be the percentage of a state population infected with a flu virus on week t of an epidemic. What percentage is likely to be infected in week 4 if N(3) = 8 and N′(3) = 1.2?

Homework Equations


Derivative EQs?


The Attempt at a Solution


My first(and failed) idea was t^3, since 2^3=8. but then this derivative didn't equal 1.2. I couldn't get the answer with my guess and check method...So is there another way I could try it?
 
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You do understand that the derivative is a "rate of change" don't you? If the number of infected in week three was 8 and was increasing by 1.2 per week, how many do you think will be infected in one more week?

That is exactly the same as using the "tangent line" approximatin: Y(t)= N'(3)(t- 3)+ N(3).
 
thank you so much, I don't know why I didnt think of that!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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