Differential calculus question

[noahsdev]

Homework Statement

The number of termites in a colony is increasing at a rate proportional to the number present on any day. If the number of termites increases by 25% in 100 days, how much longer (to the nearest day) will it be until the population is double the initial number?

2. The attempt at a solution
$\frac{dT}{dt}$=at
T=∫at
=$\frac{at^{2}}{2}$ + c
T(x)=T
T(0)*$\frac{5}{4}$ = T(100)
=> a=$\frac{c}{20000}$

I think I am approaching this wrong. Help is appreciated,
thanks.

**UPDATE: I overlooked a something, I think I figured it out now. I got t=200, does that seem right?

 

[ManiFresh]

Your initial equation seems to be wrong. The rate of change of the population is proportional to the population (on the same day).

 

[Ray Vickson]

Homework Statement

The number of termites in a colony is increasing at a rate proportional to the number present on any day. If the number of termites increases by 25% in 100 days, how much longer (to the nearest day) will it be until the population is double the initial number?

2. The attempt at a solution
$\frac{dT}{dt}$=at
T=∫at
=$\frac{at^{2}}{2}$ + c
T(x)=T
T(0)*$\frac{5}{4}$ = T(100)
=> a=$\frac{c}{20000}$

I think I am approaching this wrong. Help is appreciated,
thanks.

**UPDATE: I overlooked a something, I think I figured it out now. I got t=200, does that seem right?

Your expression $T(t) = a t^2/2 + c$ is incorrect; it does NOT satisfy the condition that the derivative $dT/dt$ is proportional to $T$ itself.

 

[HallsofIvy]

Homework Statement

The number of termites in a colony is increasing at a rate proportional to the number present on any day. If the number of termites increases by 25% in 100 days, how much longer (to the nearest day) will it be until the population is double the initial number?

2. The attempt at a solution
$\frac{dT}{dt}$=at

ManiFresh and Ray Vickson's point is that this is not the correct equation. Saying that "the rate is proportional to the number present" is $\frac{dT}{dt}= aT$.
The "T" on the right is the number of termites, not the time in days.

T=∫at
=$\frac{at^{2}}{2}$ + c
T(x)=T
T(0)*$\frac{5}{4}$ = T(100)
=> a=$\frac{c}{20000}$

I think I am approaching this wrong. Help is appreciated,
thanks.

**UPDATE: I overlooked a something, I think I figured it out now. I got t=200, does that seem right?

 

[noahsdev]

So how do I solve it? I mean, are you saying the rate of change is dT/dt = aT?

 

[Ray Vickson]

So how do I solve it? I mean, are you saying the rate of change is dT/dt = aT?

I cannot see how to answer this question without doing your problem for you. I suggest you look through your textbook or course notes to find an answer. You could also do a Google search under "growth models", for example.

 

[Chestermiller]

What is the derivative of ln(y) with respect to t, if y is a function of t?

Chet

 

[HallsofIvy]

You need to get both "dT" and "T" on the same side of the equation, "dt" on the other side.