A particle moves on the x-axis in such a way that its position at time t, t>0, is given by x(t) = (ln x)^2. At what value of t does the velocity of the particle attain maximum?
Suppose a population of bears grows according to the logistic differential equation
dP/dt = 2P - 0.01P^2
where P is the number of bears at time t in years. Which of the following statements are true?
I. The growth rate of the bear population is greatest at 100
II. If P > 200, the population of bears is decreasing.
III. lim t -> inf P(t) = 200
A conical tank is being filled with water at the rate of 16 ft^3/min. The rate of change of the depth of the water is 4 times the rate of change of the radius of the water surface. At the moment when the depth is 8ft and the radius of the surface is 2ft, the area of the surface is changing at what rate?
Out of 28 different questions, I couldn't figure out how to do these.
Thanks for your help
For the conical tank:
A handy thing to know is that the rate of change of the volume equals the cross-sectional area at that instant times the rate of change of the height.
In other words, ...
The area of the surface when r=2 would be
But remember, .
Now, can you finish?.
I. The growth rate is at its greatest when
I get values of P = 0, 100, 200. The question is, which gives the greatest growth rate? Plugging the numbers in I get P = 100. So this one is true.
II. When P > 200
so this one is true.
III. If there is a limit for P at very large times then we know that
This corresponds to a P equal to P = 0 or P = 200. If P = 0 at t = 0 then P = 0 always. That is the only way I can get the P = 0 solution to work at all. Thus the answer has to be P = 200. So III is correct also.
top, thanks so much for your help, I understand it now. But I. is true as well since the d2p/dt2 = 2 - 0.02p, 0 = 2 - 0.02p making p = 100 and it increases pre 100 and decrease after 100, so it is greatest at P = 100. I don't understand why you said I. isnt true, mistake maybe?