The number of bass in a lake is given by
P(t) = 2800/1+6e^-0.05t
where t is the number of months that have passed since the lake was stocked with bass.
(a) How many bass were in the lake immediately after it was stocked?
? bass
(b) How many bass were in the lake 1 year after the lake was stocked? (Round your answer to the nearest whole number.)
? bass
(c) What will happen to the bass population as t increases without bound?
1.The bass population will converge to its original size.
2.The bass population will increase without bound.
3.The bass population will increase, approaching 2800.
4.The bass population will stay the same.
5.The bass population will decrease, approaching 0.
If so try:
http://www.wolframalpha.com/input/?i=plot++2800%2F%281%2B6%28e^-0.05t%29+%29
when t = 0, e^0 = 1
so 2800/7
when t-->oo
2800/( 1 +6/big) = 2800
http://www.wolframalpha.com/input/?i=plot++2800%2F%281%2B6%28e^-%280.05t%29%29+%29
note for t = 400, there are about 2800
http://www.wolframalpha.com/input/?i=plot++2800%2F%281%2B6%28e^-%280.05t%29%29+%29++for+t+%3D+400
To answer these questions, we need to substitute the given values into the formula P(t) = 2800 / (1 + 6e^(-0.05t)).
(a) To determine the number of bass in the lake immediately after stocking, we can substitute t = 0 into the formula. So, P(0) = 2800 / (1 + 6e^(-0.05*0)). Simplifying this expression, we get P(0) = 2800 / (1 + 6e^0). Since any number raised to the power of 0 is equal to 1, the expression becomes P(0) = 2800 / (1 + 6). Adding 1 and 6 gives us P(0) = 2800 / 7, which simplifies to P(0) = 400. Therefore, there were 400 bass in the lake immediately after stocking.
(b) To find the number of bass in the lake 1 year after stocking, we substitute t = 12 (since a year has 12 months) into the formula. P(12) = 2800 / (1 + 6e^(-0.05*12)). Rearranging this expression, we get P(12) = 2800 / (1 + 6e^(-0.6)). Evaluating this expression by calculating e^(-0.6) and performing the other calculations, we find that P(12) is approximately equal to 2350 (rounded to the nearest whole number). Therefore, there were 2350 bass in the lake 1 year after stocking.
(c) To determine what will happen to the bass population as t increases without bound, we analyze the formula P(t) = 2800 / (1 + 6e^(-0.05t)). As t increases without bound, the term e^(-0.05t) approaches 0 since the exponent becomes more negative. As a result, the denominator (1 + 6e^(-0.05t)) approaches 1, which means the entire fraction approaches 2800 / 1, or simply 2800. Therefore, as t increases without bound, the bass population will increase, approaching 2800. The correct answer is option 3 - "The bass population will increase, approaching 2800."
Do you mean:
P(t) = 2800/(1+6e^-0.05t)
????