Wondering if someone could quickly confirm the answer to this

You gotta compute the derivatives and then manipulate them to get the expression of [math] da/dt [/math].

>>8521167
Ok so da and dt here are just da/dN and dt/dN
So just compute derivatives of a and t with respect to N ? Is it normal to write da and dt and it meanda/dN, dt/dN? I mean it should be obvious I'm differentiating with respect to N since its the only variable but it triggers my autism a little

>>8521158
Well you're given a in terms of n, so just take the derivative with respect to time (I'm assuming A is constant):
da/dt=(A/2)*(sin(n))*(dn/dt)
Now, to solve for dn/dt you need to take derivative of the time equation with respect to time:
dt/dt=(A/2)*(dn/dt-cos(n)dn/dt)
dt/dt is equal to 1, so solving for dn/dt we get (2/A)/(1-cos(n)). Plugging this back into da/dt we have
da/dt=sin(n)/(1-cos(n))

>>8521181
So calculating da/dN and dt/dN then dividing da/dN by dt/dN would be wrong?

>>8521181
I got this answer as well

>>8521236
>>8521181
Okay then thanks a lot for showing me how to get the answer

>>8521252
da/dt = (da/dn)(dn/dt)
use the equation for a and take the derivative with respect to n.
da/dn=(A/2)sin n
Then use the equation for t and calculate dt/dn.
dt/dn = a
dn/dt = 1/a = (2/A)/(1-cos n)
da/dt = [(A/2)sin n][ (2/A) / (1-cos n) ]
da/dt = sin n / (1- cos n)

>>8521262
I wasnt being sarcasting btw i was thanking the other guy for showing it and you for confirming it, but thanks

>>8521275
Sorry, couldn't tell the difference