CALCULUS PROBLEM-Please help!?

Question

Hi all, I have two problems that i need help with, please explain throroughly if you can

1. find the second derivative of y=e^cos9x
2. An avian disease is decimating the bird population of a South Pacific island. The population at any time (t) is roughly approximated by the equation y=A * e^cos( π /2 * t)
where t is in years, t is between 0-2, and A is the popluation at t=0.

Find the rate at which the population is declining at t=1 (in terms of A)

Answer 1

y= e^cos(9x)
dy/dx= -9sin(9x)e^cos(9x) [ use the product rule or, d/dx e^f(x)=f'(x)e^f(x)]
second derivative= -9sin9x(e^cos9x * -9sin9x)+ e^cos9x(-81cos9x); use the product rule again.
= 81sin9x* e^cos9x - 81cos9x*e^cos9x
=e^cos9x(81sin9x- 81cos9x)


2. dy/dt= A(e^cos (π /2 * t))(-π /2*sin (π /2 * t))
= -A*π /2 *sin (π /2 * t)*e^cos( π /2 * t)
at t=1 dy/dt= -A*π /2

sin(π /2) =1 at t=1
cos (π /2)=o at t=1
e^0=1

Answer 2

for the first problem take the derivative of the function, and then take the derivative of that. as for the second one, just find the derivative of the function and then plug in 1 for t. this will give you the rate of change for population/year

Answer 3

For #1: recall that if y=e^u(x), y'=e^u(x) * u'(x)
So, y''=e^u(x) * [u'(x)]^2 + e^u(x) * u''(x)
Here u(x)=cos(9x), so u'(x)= -9sin(9x), and u''(x)= -81cos(9x)
You can substitute and simplify from here ...

For #2, y'= (A*pi/2) * (-sin(t*pi/2)) * e^cos(t*pi/2)
with t=1, y' = (A*pi/2) * (-1) * e^0 = -A*pi/2
(the "-" sign means population declining)

Answer 4

1.

y' = (cos(9x))' e^cos(9x)
= - 9sin(9x) * e^cos(9x)

y'' = [- 9sin(9x)]' * e^cos(9x) + (-9sin(9x)) * [e^cos(9x)]'
= -81cos(9x) * e^cos(9x) + (-9sin(9x)) * (-9sin(9x)) * e^cos(9x)

= 81[ -cos(9x) + sin²(9x)]* e^cos(9x)

2. there is a problem : at t = 0, y = Ae^cos0 = Ae and not A
At t = 1 y = A e^cos(pi/2) = A e^0 = A

If it was sin instead of cos, then
at t = 0, y = A
at t = 1, y = Ae

the rate is 1/e

Answer 5

y´=e^cos9x*(-sin 9x) *9
y´´= 9[e^cos9x *sin^2(9x)*9-e^cos9x*cos 9x *9]=
81e^cos9x[sin^2(9x)-cos9x ]
2) dA/dt= e^cos(pi/2*t) (-sin(pi/2 *t)*pi/2
at t = 1
dA/dt=1(-1)(pi/2 = -pi/2 (negative because declining)