# Factorizing -Where to next?

• Oct 6th 2008, 04:26 AM
kayne
Factorizing -Where to next?
Hi everyone, having a lot of trouble with Calculus and was wondering if anyone can help.

I have d(t)= -5e^(-5t) x cos(10t)

u = -5e^(-5t)
v = cos(10t)

du/dx = 25e^(-5t)
dv/dx = -10sin10t

duv/dx = 25e^(-5t) x cos(10t) + (-10sin10t) x -5e^(-5t)

= 25e^(-5t)cos(10t) + 50e^(-5t)sin10t
= 2cos10t+sin10t

From here I am not sure where to go? Do I have the first bit correct?

0=(sin+cos=1)
0=2 x 1 x 10t
0=20t
t=1/20 ???

Or do i have to find out tan = sin(a) / cos (a)

If anyone could give me some kicks in the right direction it would be much appreciated, and i can stop wasting paper.(Speechless)

Thanks
Kayne
• Oct 6th 2008, 06:25 AM
mr fantastic
Quote:

Originally Posted by kayne
Hi everyone, having a lot of trouble with Calculus and was wondering if anyone can help.

I have d(t)= -5e^(-5t) x cos(10t)

u = -5e^(-5t)
v = cos(10t)

du/dx = 25e^(-5t)
dv/dx = -10sin10t

duv/dx = 25e^(-5t) x cos(10t) + (-10sin10t) x -5e^(-5t)

= 25e^(-5t)cos(10t) + 50e^(-5t)sin10t
= 2cos10t+sin10t

From here I am not sure where to go? Do I have the first bit correct?

0=(sin+cos=1)
0=2 x 1 x 10t
0=20t
t=1/20 ???

Or do i have to find out tan = sin(a) / cos (a)

If anyone could give me some kicks in the right direction it would be much appreciated, and i can stop wasting paper.(Speechless)

Thanks
Kayne

The derivative is $25e^{-5t} \left( \cos (10t) + 2 \sin ( {\color{red}1}0t) \right)$.

If you're trying to find when the derivative is equal to zero, you need to solve

$\cos (10t) + 2 \sin ({\color{red}1}0t) = 0 \Rightarrow \tan (10t) = - \frac{1}{2}$.

The general solution is $10 t = - \tan^{-1} \left( \frac{1}{2} \right) + n \pi \Rightarrow t = \, ....$ where n is an integer.

I assume that:

1. You have a domain for the solutions ....
2. You don't want exact solutions.
• Oct 6th 2008, 12:53 PM
kayne
Mr Fantastic,

Sorry for not being very clear, yes you are right about the equation equalling zero. The question is to find the Max and min points for a shock absorber rebound .

Once the equation = 0, i have to find the second derivative to measuse the rate of change of the curve, and subistute the x values into the original equation.

Thanks for the help.. If i get stuck again i will post a reply.

kayne
• Oct 7th 2008, 01:45 AM
kayne
Mr Fantastic,

I have been looking over your post today and i am a little lost to how you factorized the equation.

The deritive that you stated

25e^(-5t)(cos(10t) + -2sin(20t))

How did you come to this every time i do this i end up with some thing different so i must be factorizing wrong. I would like to know where the 2 and 20 come from seeing that cos(10t) = -10sin(10t)

and the next part of the equation where did the 25e^(-5t) go to as you ended up with

0 = Cos(10t) + 2sin(20t) = tan(10t) = -1/2

When i do it i end up with

0 = Cos(10t) + 2sin(10t) = 2tan(10t)

?Where have i lost my factorizing?

I have attached the graph that i am working from to find the max positive rebound and when it occurs.

• Oct 7th 2008, 02:51 AM
mr fantastic
Quote:

Originally Posted by kayne
Mr Fantastic,

I have been looking over your post today and i am a little lost to how you factorized the equation.

The deritive that you stated

25e^(-5t)(cos(10t) + -2sin(20t))

How did you come to this every time i do this i end up with some thing different so i must be factorizing wrong. Mr F says: I took out the common factor of 25 e^(-5t).

I would like to know where the 2 and 20 come from seeing that cos(10t) = -10sin(10t) Mr F says: The 20 comes from the fact that I made a typo that unfortunately got carried through when I did a copy and paste. I've edited the reply in red.

and the next part of the equation where did the 25e^(-5t) go to as you ended up with

Mr F says: ${\color{red}25e^{-5t} \left( \cos (10t) + 2 \sin ( {\color{red}1}0t) \right) = 0}$ implies that either ${\color{red}25e^{-5t} = 0}$ or ${\color{red}\cos (10t) + 2 \sin (10t) = 0}$. I thought it would be obvious that ${\color{red}25e^{-5t} = 0}$ has no real solution so I just gave ${\color{red}\cos (10t) + 2 \sin (10t) = 0}$.

0 = Cos(10t) + 2sin(20t) = tan(10t) = -1/2

When i do it i end up with

0 = Cos(10t) + 2sin(10t) = 2tan(10t) Mr F says: I'm sorry but that's totally wrong. Cos(10t) + 2sin(10t) is not the same as 2 tan(10t). Cos(10t) + 2sin(10t) = 0 => Cos(10t) = -2sin(10t) => 1 = -2 sin(10t)/cos(10t) = -2 tan(10t) => -1/2 = tan(10t).

?Where have i lost my factorizing?

I have attached the graph that i am working from to find the max positive rebound and when it occurs.