The volume of a cube is increasing at a constant rate of 824 cubic centimeters per minute. at the instant when the volume of the cube is 683683 cubic centimeters, what is the rate of change of the surface area of the cube? round your answer to three decimal places (if necessary).

DA/dt  = 75.27 cm²

Step-by-step explanation:

Cube Volume    =   V(c)   =  683  cm³

DV(c) /dt    =  824 cm³

V(c,x) = x³    

Then

DV(c,x)/ dt   =  3x² Dx/dt

( DV(c,x)/ dt )/ 3x²  =  Dx/dt       (1)

Now  as  V(c,x) = x³    when    V(c,x)  =  683 cm³     x = ∛683

x  =  8.806     ( from excel)

And by subtitution of this value in equation (1)

Dx/dt   =  ( DV(c,x)/ dt )/ 3x²   ⇒  Dx/dt  =  824 / 3*x²

  Dx/dt   = 824 /3*77.55

Dx/dt   =  824/232,64

Dx/dt   = 3,542

Then we got  Dx/dt  where x is cube edge. The area of the face is x² then

the rate of change of the suface area is  

DA/dt  = ( Dx/dt )²*6         ( 6 faces of a cube)

DA/dt  = (3.542)²*6

DA/dt  = 75.27 cm²

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this