# calc 3

The length script l, width w, and height h of a box change with time. At a certain instant the dimensions are script l = 7 m and
w = h = 5 m,
and script l and w are increasing at a rate of 6 m/s while h is decreasing at a rate of 4 m/s. At that instant find the rates at which the following quantities are changing.
(a) The volume.
m3/s

(b) The surface area.
m2/s

(c) The length of a diagonal. (Round your answer to two decimal places.)
m/s

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1. I'll do the volume. You should be able to get the others. If not, come back and we can help you get unstuck.

v = lwh
dv/dt = wh dl/dt + lh dw/dt + lw dh/dt
plugging in the numbers,

dv/dt = 5m*5m*(6m/s) + 7m*5m*(6m/s) + 7m*5m*(-4m/s)
= 150+210-140=220 m^3/s

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2. yes i already got parts a and b i do not know how to do part c sorry i should have specified that in my post

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3. there are diagonals along each face, but thety probably want the diagonal of the brick:

d^2 = l^2 + w^2 + h^2
2d dd/dt = 2l dl/dt + 2w dw/dt + 2h hd/dt
when (l,w,h) = (7,5,5) d = sqrt(99) = 3sqrt(11)

3sqrt(11) dd/dt = 2*7*6 + 2*5*6 + 2*5*(-4)
3sqrt(11) dd/dt = 104
dd/dt = 104 / 3sqrt(11)

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