Application of the first derivative biology?

[cmnalo]

The number of people p(t) in hundreds infected t days after a epidemic begins is approximated by p(t) 2+50t-5t^2 / 2

P'(x)= (2d)/dx(2+50t-5t^2) -(2+50t-5t^2) d/dx (2) / 2^2

P'(x)= (2)(50-10t) - (2+50t-5t^2)(0) / 2^2

P'(x)= 100-20t / 4

P'(x)= 25-5t

25-5t=0

t=4

The answer is: After 10 days.

Where am I going wrong? Isn't 4 the critical point?

 

[stapel]

What is the actual question? What are you needing to find? (You've posted the explanation of what the function stands for, but not what "after ten days" might mean, is why I ask.)

Thank you.

Eliz.

 

[cmnalo]

Sorry about that. I'm trying to find when the number of people infected will start to decline. When P(t) is decreasing?

 

[skeeter]

uhhh ... P'(t) = 50 - 5t

 

[cmnalo]

How did you get 50. Can you see where I might have gone wrong?

 

[skeeter]

P(t) = 2 + 50t - (5t^2)/2
P'(t) = 0 + 50 - 5t

the derivative of 2 is 0 ... the derivative of 50t is 50 ... the derivative of (5t^2)/2 is 5t

 

[cmnalo]

So your appling the Quotient rule to -5t^2/2?

(2)d/dx(-5t^2) -(-5t^2)d/dx(2) / 2^2

(2)(-10t) -(-5t^2)(0) / 2^2

-20t / 4 = -5t

Correct?

 

[skeeter]

the quotient rule is not necessary ...

-5t²/2 = (-5/2)t²

derivative is 2(-5/2)t = -5t

 

[stapel]

The number...is approximated by p(t) 2+50t-5t^2 / 2

I would expect that what you posted means one of the following:

. . . . .\(\displaystyle \L p(t)\, =\, 2\, +\, 50t\, -\, \frac{5t^2}{2}\, =\, 2\, +\, 50t\, -\, \frac{5}{2}\, t^2\)

. . . . .\(\displaystyle \L p(t)\, =\, \frac{2\, +\, 50t\, -\, 5t^2}{2}\, =\, 1\, +\, 25t\, -\, \frac{5}{2}\, t^2\)

But, in either case, the Quotient Rule is unnecessary.

Eliz.