Students with flu like symptoms

[caters]

Homework Statement

Students at Bacteria University begin showing flu-like symptoms. On the first day of term, 5 students have the flu. The growth of the number of infected students is modeled by

P(t) = 5e^0.3t

What is a formula for the rate of change of the number of infected students in terms of t?

To the nearest whole number, what is the rate of change in the number of infected students when t=10 days?

Homework Equations

f(x) = xⁿ
f'(x) = nx^n-1

The Attempt at a Solution

P(t) = 5e^0.3t
P'(t) = (5t*e)^-0.3t
P'(10) = 50e^-3
P'(10) = 1/50e³

Now I know that this deceleration is not right since the infection rate only starts decelerating when half of the population has been infected and it is very unlikely that half the population will be infected in 10 days.

However algebraically this makes sense since if you have a decimal exponent and you take the derivative you get a negative exponent in the derivative or in other words a fractional rate.

 

[Dr. Courtney]

Double check your equations and how you are differentiating.

 

[caters]

I am using the power rule alone to differentiate it. I am also assuming that the variable stays in the exponent when I differentiate it using the power rule.

But the 0.3t is essentially a function of its own. So really it is like P(t) = a^f(t) where a = 5e

Is that why I am getting this deceleration in the rate when the rate should be accelerating instead?

 

[SteamKing]

I am using the power rule alone to differentiate it. I am also assuming that the variable stays in the exponent when I differentiate it using the power rule.

But the 0.3t is essentially a function of its own. So really it is like P(t) = a^f(t) where a = 5e

Is that why I am getting this deceleration in the rate when the rate should be accelerating instead?

Using the power rule to differentiate an exponential function is incorrect. The power rule only applies to y = xⁿ, where x is a variable and n is a constant exponent.

The number e is a constant, not a variable.

The function y = e^x has its own derivative dy/dx. Do you know what it is?

 

[caters]

The derivative of e^x is e^x

So would the derivative be the same as the original function in this case?

 

[SteamKing]

The derivative of e^x is e^x

So would the derivative be the same as the original function in this case?

You're on the right track, but your function is not y = e^x or even y = e^t. You also must apply the chain rule.

 

[caters]

So it is like this where u=0.3t:

$5\frac{d}{dt} e^{0.3t}$
$= 5 \frac{d}{du} e^u * \frac{d}{dt} 0.3t$
$= 5e^u * 0.3$
$= \frac{3e^{0.3t}}{2}$

 

[SteamKing]

This looks better.

Why are your formulas formatted so weird?

 

[caters]

Well I used latex but didn't put the double dollar around the equal signs just the expressions. That is probably why you are seeing the equal signs in weird places.

 

[Mark44]

Well I used latex but didn't put the double dollar around the equal signs just the expressions. That is probably why you are seeing the equal signs in weird places.

Fixed. Also changed to inline Tex, which doesn't center the equations.

 

[epenguin]

Now I know that this deceleration is not right since the infection rate only starts decelerating when half of the population has been infected and it is very unlikely that half the population will be infected in 10 days.

However algebraically this makes sense since if you have a decimal exponent and you take the derivative you get a negative exponent in the derivative or in other words a fractional rate.

From your formula about 100 students would be infected after 10 days. You have given us no information to tell us this is half the students, and this number does not change anything except that's to infect 100 students there have to be at least 100 students to infect.
Do not confuse reality with models of it. There has to be a deceleration, but the model you give does not predict one!