Derivative Word Prob: rate of change in num. of mosquitoes

[xtrmk]

For 0 <= t <= 31, the rate of change of the number of mosquitoes is at time t days and is modeled by R(t) = 5 sqrt[t] cos(t/5) mosquitoes per day. There are 1000 mosquitoes at t = 0.

a) Show the number of mosquitoes is increasing at t = 6.

I took the derivative of r(t) and got -sin(t/5)(root[t)) + 5cos(t/5) / 2root(t). Then I plugged in 6 and got a negative value: -0.06 Is this right?

b) At t = 6, is the number of mosquitoes increasing at an increasing rate or increasing at a decreasing rate? Justify your answer.

Is my negative answer correct? If so, would i compare that to .. t=5 ?

Thank you!

 

[stapel]

a) Since R(t) gives the rate of change, the number would be increasing if the rate of change is positive. So plug "6" in for "t" in R(t).

b) This is asking if the rate of change, R(t), is itself increasing (so growth is accelerating) or decreasing (so growth is descelerating). This is where you'd use the derivative, R'(t).

Eliz.

 

[xtrmk]

a) Since R(t) gives the rate of change, the number would be increasing if the rate of change is positive. So plug "6" in for "t" in R(t).

b) This is asking if the rate of change, R(t), is itself increasing (so growth is accelerating) or decreasing (so growth is descelerating). This is where you'd use the derivative, R'(t).

Eliz.

Thanks!

If I have to find the maximum number of mosquitoes from 0 to 31 would I just set the derivative = to 0 ? or should I use the 2nd deriv test?

 

[stapel]

If I have to find the maximum number of mosquitoes...

Since all you have is the rate of change in the number, there is no way to find the actual number itself. In particular, while one could find the time at which the maximum occurred, one could not find the actual value of the maximum.

As mentioned previously, the derivative of the rate measures the change in the rate, not the change in the population.

Eliz.