5c-14.5-3.math

14.5 39

The length ℓ, width w, and height h of a box change with time. At a certain instant the dimensions are ℓ = 8 m and 
w = h = 1 m, and ℓ and w are increasing at a rate of 2 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.
l = 8 w = 1 h =1
dl = 2 dw = 2 dh = -4

v=lwh
dv = wh dl + lh dw + lw dh
dv = 1(1)(2) + 8(1)(2) + 8(1)(-4)
dv = 2+16-32= -16+2= -14

s=2(lw+lh+wh)
ds = 2(w+h)dl+2(l+h)dw+2(l+w)dh
ds = 2(1+1)(2)+2(8+1)(2)+2(8+1)(-4)
ds = -28

a=sqrt(l^2+w^2+h^2)
da = 2ldl/2sqrt(l^2+w^2+h^2)+ 2wdw/2sqrt(l^2+w^2+h^2)+ 2hdh/2sqrt(l^2+w^2+h^2)
da = 8*2/sqrt(64+1+1) + 2/sqrt(66)-4/sqrt(66)
da = 14/sqrt(66)
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14.5 43

A= xysin(theta)/2
dA = ysin(theta)dx/2 + xsin(theta)dy/2 + xycos(theta)dtheta/2
A is constant => dA = 0
0 = ysin(theta)dx/2 + xsin(theta)dy/2 +xycos(theta)dtheta/2
0 = 34sin(pi/6)9/2 + 20sin(pi/6)*-5/2 + 20*34cos(pi/6)dtheta/2
=>
dtheta = -103/(340sqrt(3))