Do ONE of the following:(a) Bacterial population A is growing exponentially while B is shrinking exponentially.At time t-0, population A has 3000 cells and population B has 9000 cells. If the twopopulations have the same number of cells at t=5 what is the ratio of population A topopulation B at t=15?(b) Find the volume bound by the curvesx = 2y? – 3y – 3 and a == -y? +3if they are rotated around the liney = -2.(c) Evaluate:/sec" (a)dz . Do ONE of the following:(a) Bacterial population A is growing exponentially while B is shrinking exponentially.At time t-0, population A has 3000 cells and population B has 9000 cells. If the twopopulations have the same number of cells at t=5 what is the ratio of population A topopulation B at t=15?(b) Find the volume bound by the curvesx = 2y? – 3y – 3 and a == -y? +3if they are rotated around the liney = -2.(c) Evaluate:/sec" (a)dz .

Math

Advanced Math

(c) Evaluate: sec" (x)dz.

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter5: Inverse, Exponential, And Logarithmic Functions

Section: Chapter Questions

Problem 18T

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Question

Do ONE of the following:
(a) Bacterial population A is growing exponentially while B is shrinking exponentially.
At time t-0, population A has 3000 cells and population B has 9000 cells. If the two
populations have the same number of cells at t=5 what is the ratio of population A to
population B at t=15?
(b) Find the volume bound by the curves
x = 2y? – 3y – 3 and a =
= -y? +3
if they are rotated around the line
y = -2.
(c) Evaluate:
/sec" (a)dz .

Expert Solution

Step 1

(c)

reduction formula for sec x is given as,

$I_{n} (n - 1) = s e c^{n - 2} x \tan x + (n - 2) I_{n - 2} I_{n} = \int s e c^{n} x d x$ where

taking $n = 5$

$I_{5} \times 4 = s e c^{3} x \tan x + 3 I_{3}$

Also $I_{3}$ can be resolved in the same way as

$I_{3} \times 2 = s e c x \tan x + I_{1}$

Plugging this value above

$4 I_{5} = s e c^{3} x \tan x + 3 (\frac{1}{2} s e c x \tan x + \frac{1}{2} I_{1})$

Further,

$I_{5} = \frac{1}{4} s e c^{3} x \tan x + \frac{3}{8} (s e c x \tan x + \int s e c x d x)$

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