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dR/dh = 0,
where R is 2 (hH-h2 )1/2
d [2 (hH-h2 )1/2 ] /dh = 0
2 x 1/2 (hH-h2)-1/2 x (H-2h) = 0
(H-2h)/rt (hH-h) = 0
H-2h=0
h = H/2
this is a part of a physics question and any help is appreciated.
I get it till the 2 x1/2 part, but how did they get the (h-2h) part?
If someone could explain this to me, I'd be super-grateful!
R=2(hH-h^2)^1/2
Let hH-h^2=t
R=2t^1/2
dR/dt=2x(1/2)xt^-(1/2)
=t^-(1/2)
t=hH-h^2
dt/dh=H-2h
dR/dh=(dR/dt)x(dt/dh)= t^-(1/2)x(H-2h)
=(H-2h)/(hH-h^2)^1/2
If we put dR/dh=0
dR/dh=(H-2h)/(hH-h^2)^1/2=0
First thing is that denominator should not be zero hence hH-h^2 is non zero
Therefore h can't be zero and can't be equal to H
numerator should be zero, thus H=2h
Hello, Booksrock!
The two h's are confusing.
I assume thatis a constant . . . Let's call it
Quote:
^{\frac{1}{2}}. \quad \text{Solve: }\,\frac{dR}{dh} \,=\,0)
Chain Rule: .