Name:
Anonymous
2008-01-16 2:03
How do I differentiate:
theta = (x/25)(tan^(-1))
with respect to time??
Name:
Anonymous
2008-01-16 2:21
Chain rule. Also, you've written it incorrectly.
Name:
Anonymous
2008-01-16 17:55
>>1
Assuming x is independent of t
d(theta)/dt = 0
fix'd
Name:
Anonymous
2008-01-16 19:16
Inverse tangent of what?
If it's ß=(x/25)*arctan(t) which is what I'm guessing... (calling theta ß)
dß/dt = (x/25)*1/(1+t^2) + 0
= x/(25+25t^2)
If x is x(t) then you're gonna have a dx/dt multiplier.