The size of an undisturbed fish population has been modeled by the formula p n + 1 = b p n a + p n where p n is the fish population after n years and a and b are positive constants that depend on the species and its environment. Suppose that the population in year 0 is P 0 > 0. (a) Show that if { p n } is convergent, then the only possible values for its Limit are 0 and b − a. (b) Show that P n +1 < ( b/a ) p n. (c) Use part (b) to show that if a > b, then lim n →∞ p n = 0; in other words, the population dies out. (d) Now assume that a < b. Show that if P 0 < b − a, then { p n } is increasing and 0 < p n < b − a. Show also that if P 0 > b − a, then { p n } is decreasing and p n > b − a. Deduce that if a < b, then lim n →∞ p n = b − a.
The size of an undisturbed fish population has been modeled by the formula p n + 1 = b p n a + p n where p n is the fish population after n years and a and b are positive constants that depend on the species and its environment. Suppose that the population in year 0 is P 0 > 0. (a) Show that if { p n } is convergent, then the only possible values for its Limit are 0 and b − a. (b) Show that P n +1 < ( b/a ) p n. (c) Use part (b) to show that if a > b, then lim n →∞ p n = 0; in other words, the population dies out. (d) Now assume that a < b. Show that if P 0 < b − a, then { p n } is increasing and 0 < p n < b − a. Show also that if P 0 > b − a, then { p n } is decreasing and p n > b − a. Deduce that if a < b, then lim n →∞ p n = b − a.
Solution Summary: The author explains that if leftp_nright is convergent, the only possible values for the limit are 0 and b-a.
The size of an undisturbed fish population has been modeled by the formula
p
n
+
1
=
b
p
n
a
+
p
n
where pn is the fish population after n years and a and b are positive constants that depend on the species and its environment. Suppose that the population in year 0 is P0 > 0.
(a) Show that if {pn} is convergent, then the only possible values for its Limit are 0 and b − a.
(b) Show that Pn+1 < (b/a)pn.
(c) Use part (b) to show that if a > b, then limn→∞pn = 0; in other words, the population dies out.
(d) Now assume that a < b. Show that if P0 < b − a, then {pn} is increasing and 0 < pn < b − a. Show also that if P0> b − a, then {pn} is decreasing and pn > b − a. Deduce that if a < b, then limn→∞pn = b − a.
This question builds on an earlier problem. The randomized numbers may have changed, but have your work for the previous problem available to help with this one.
A 4-centimeter rod is attached at one end to a point A rotating counterclockwise on a wheel of radius 2 cm. The other end B is free to move back and forth along a horizontal bar that goes through the center of the wheel. At time t=0 the rod is situated as in the diagram at the left below. The
wheel rotates counterclockwise at 1.5 rev/sec. At some point, the rod will be tangent to the circle as shown in the third picture.
A
B
A
B
at some instant, the piston will be tangent to the circle
(a) Express the x and y coordinates of point A as functions of t:
x= 2 cos(3πt)
and y= 2 sin(3t)
(b) Write a formula for the slope of the tangent line to the circle at the point A at time t seconds:
-cot(3πt)
sin(3лt)
(c) Express the x-coordinate of the right end of the rod at point B as a function of t: 2 cos(3πt) +411-
4
-2 sin (3лt)
(d)…
5. [-/1 Points]
DETAILS
MY NOTES
SESSCALCET2 6.5.AE.003.
y
y= ex²
0
Video Example
x
EXAMPLE 3
(a) Use the Midpoint Rule with n = 10 to approximate the integral
कर
L'ex²
dx.
(b) Give an upper bound for the error involved in this approximation.
SOLUTION
8+2
1
L'ex² d
(a) Since a = 0, b = 1, and n = 10, the Midpoint Rule gives the following. (Round your answer to six decimal places.)
dx Ax[f(0.05) + f(0.15) + ... + f(0.85) + f(0.95)]
0.1 [0.0025 +0.0225
+
+ e0.0625 + 0.1225
e0.3025 + e0.4225
+ e0.2025
+
+ e0.5625 €0.7225 +0.9025]
The figure illustrates this approximation.
(b) Since f(x) = ex², we have f'(x)
=
0 ≤ f'(x) =
< 6e.
ASK YOUR TEACHER
and f'(x) =
Also, since 0 ≤ x ≤ 1 we have x² ≤
and so
Taking K = 6e, a = 0, b = 1, and n = 10 in the error estimate, we see that an upper bound for the error is as follows. (Round your final
answer to five decimal places.)
6e(1)3
e
24(
=
≈
2. [-/1 Points]
DETAILS
MY NOTES
SESSCALCET2 6.5.015.
Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n. (Round your answers to six decimal places.)
ASK YOUR TEACHER
3
1
3 +
dy, n = 6
(a) the Trapezoidal Rule
(b) the Midpoint Rule
(c) Simpson's Rule
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