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stuck with some derivative..

1. The problem statement, all variables and given/known data

i need to find the first and second of derivative of this
#952;(t) = A exp#8722;#955;t cos(#969;t) with (-lambda*t) being the power of exp

in order to substitute into here
I(theta) = #8722;#956;#952;(1st deriv of theta) + #947;#952;(2nd deriv of theta)

and then prove that
#955; = #8722;#947;/2I

2. Relevant equations
#952;(t) = A exp#8722;#955;t cos(#969;t)3. The attempt at a solution

have tried deriving with wolfram alpha myself and have gotten very stuck:

1st deriv : Calcula...=42amp;w=342amp;h=36

2nd deriv: Calcul...=11amp;w=270amp;h=61

with l representing lambda.

any help would be much appreciated!

Regards,
Mitch
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution  

    This looks like a book keeping nightmare. The first derivative is pretty straight forward. I would start by letting U=-lambda*t*Cos(w*t) determine dU/dt. d(theta)/dt=d(theta)/dU*dU/dt.  Make the same U substitution for the second derivative. It looks like it will get messy.  

                                   Originally Posted by RTW69                   This looks like a book keeping nightmare. The first derivative is pretty straight forward. I would start by letting U=-lambda*t*Cos(w*t) determine dU/dt. d(theta)/dt=d(theta)/dU*dU/dt.  Make the same U substitution for the second derivative. It looks like it will get messy.                  
very very messy...
this is my first lab for 2nd year uni physics. oh what joy lays ahead no it should be fine, this is one of the harder labs.

thanks for you advice!

mitch  

    What do #956;#952; and #947;#952; represent ?  

                                   Originally Posted by SammyS                   What do #956;#952; and #947;#952; represent ?                  
thanks for trying to help Sammy, I have since solved the problem. By letting each sin and cos equal zero, they can be eliminated and the equation will still hold. It's hard to explain here, but it worked out well.
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