# Thread: Factorizing -Where to next?

1. ## 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.

Thanks
Kayne

2. 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.

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

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

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

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

I assume that:

1. You have a domain for the solutions ....
2. You don't want exact solutions.

3. 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

4. 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.

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

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.

..

6. Thanks again, for the detailed explanation. I can see where i have been going wrong now. i will work on this now and finish the problem.

kayne