a and b are constants. additional information given was acot(theta 1) + bcot(theta 2) = 150. I know how to differentiate L, but I just can't explain why theta 1 would equal theta 2.
2007-11-14 17:43:33 · 2 answers · asked by theone 2 in Science & Mathematics ➔ Mathematics
let's use t instead of theta for simplicity.
L = acsc(t1) + bcsc(t2)
take deirvative with respect to t1,
dL/dt1 = -acot(t1)csc(t1) - bcot(t2)csc(t2) * dt2/dt1 = 0
so
acot(t1)csc(t1) + bcot(t2)csc(t2) * dt2/dt1 = 0 (1)
acot(t1) + bcot(t2) = 150
take deirvative with respect to t1,
-a(csc(t1))^2 - b(csc(t2))^2 * dt2/dt1 = 0
so dt2/dt1 = -a(csc(t1))^2/b*(csc(t2))^2
put it into (1), we get
acot(t1)csc(t1) - bcot(t2)csc(t2) * a(csc(t1))^2/b*(csc(t2))^2= 0
acsc(t1)*(cot(t1) - csc(t1)cot(t2)/csc(t2)) = 0
cot(t1) - csc(t1)cot(t2)/csc(t2) = 0
cos(t1)/sin(t1) - cos(t2)/(sin(t1) = 0
cos(t1) - cos(t2) = 0
cos(t1) = cos(t2)
so t1 = t2
2007-11-14 18:19:33 · answer #1 · answered by zsm28 5 · 0⤊ 0⤋
(Differentiating with respect to x)
The derivative of inverse cotangent is -1/ 1 + x^2
The derivative of inverse cosecant is -1/x sqrt(x^2-1)
where x^2 is "x raised to the second power) and
sqrt: is square root.
I am not sure which variable you are differentiating with respect to.
2007-11-14 17:56:06 · answer #2 · answered by ≈ nohglf 7 · 0⤊ 0⤋