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Ice Cube Melting
- Thread starter luis1123
- Start date
An Ice Cube stays in shape of a cube as it melts. If the sides of the cube are decreasing at the rate of one inch per hour, find the rate of change of the volume of the cube when the volume is eight cubic inches .
V = x^3
dx/dt = -1
Your turn.
wjm11 said:An Ice Cube stays in shape of a cube as it melts. If the sides of the cube are decreasing at the rate of one inch per hour, find the rate of change of the volume of the cube when the volume is eight cubic inches .
V = x^3
dx/dt = -1
Your turn.
X=8
dv/dt =3x^2 I think
dv/dt =3x^2 I think
Not quite. We are not differentiating with respect to x, but rather to t.
V = x^3
dV/dt = (3x^2)(dx/dt)
wjm11 said:dv/dt =3x^2 I think
Not quite. We are not differentiating with respect to x, but rather to t.
V = x^3
dV/dt = (3x^2)(dx/dt)
So do I plug the variables into dv/dt= (3x^2)(dx/dt) or is their something I'm missing.
D
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luis1123 said:wjm11 said:dv/dt =3x^2 I think
Not quite. We are not differentiating with respect to x, but rather to t.
V = x^3
dV/dt = (3x^2)(dx/dt)
So do I plug the variables into dv/dt= (3x^2)(dx/dt) or is their something I'm missing.
You also need to find the value of 'x' given V = 8
Then plug everything in.