cssftw
Member
- Joined
- Jun 19, 2009
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- HSC
- 2011
Q. A block of ice in the form of a CUBE has one edge 10cm long. It is melting so that its dimensions decrease at the rate of 1mm/s (the block always remains a cube).
At what rate is the DIAGONAL decreasing:
(a) initially? (when edgelength = 10cm)?
(b) when the edge is 5cm long??
My solution:
Let edge length = x
Let diagonal = h
we know dx/dt = -1mm/s
dh/dt = dx/dt * dh/dx
to find dh/dx -- we need to find an expression of h in terms of x.
Using pythagoras on the square (side face of cube):
x^2 + x^2 = h^2
2x^2 = h^2
therefore h = sqrt(2)*x (h>0, x>0)
therefore dh/dx = sqrt(2)
therefore dh/dt = -1(sqrt(2))
dh/dt = -sqrt(2) mm/s --> which would imply that the rate is constant.
So is the answer -sqrt(2) mm/s i.e -0.141... cm/s??
Could someone please check the answer for me? thanks
At what rate is the DIAGONAL decreasing:
(a) initially? (when edgelength = 10cm)?
(b) when the edge is 5cm long??
My solution:
Let edge length = x
Let diagonal = h
we know dx/dt = -1mm/s
dh/dt = dx/dt * dh/dx
to find dh/dx -- we need to find an expression of h in terms of x.
Using pythagoras on the square (side face of cube):
x^2 + x^2 = h^2
2x^2 = h^2
therefore h = sqrt(2)*x (h>0, x>0)
therefore dh/dx = sqrt(2)
therefore dh/dt = -1(sqrt(2))
dh/dt = -sqrt(2) mm/s --> which would imply that the rate is constant.
So is the answer -sqrt(2) mm/s i.e -0.141... cm/s??
Could someone please check the answer for me? thanks