side of the cube increasing at rate of 6 cm/s. find the rate of increase of the volume when the length of a side is 9 cm.[ans 1458cm^3/s]
my answer:
~729x6=4374cm^3/s
i need some1 show me the calculation if im wrong.
V = x^3
dV/dx = 3 x^2
so
dV/dt = 3 x^2 dx/dt
dV/dt = 3 * 81 * 6 = 1458 cm^3/s
A side of a cube is constantly increasing at a rate of 6 cm/sec. Find the rate of increase in volume and surface area when the one side of the cube is 12 cm.
To find the rate of increase of the volume when the length of a side is 9 cm, we can use the formula for the volume of a cube.
The volume of a cube is given by V = s^3, where s is the length of a side.
Given that the side of the cube is increasing at a rate of 6 cm/s, we can also express this as ds/dt = 6 cm/s, where ds/dt is the rate of change of the side length.
We want to find dV/dt, the rate of change of the volume, when s = 9 cm.
To do this, we differentiate V = s^3 with respect to t:
dV/dt = d(s^3)/dt
Using the power rule for differentiation, we get:
dV/dt = 3s^2 * ds/dt
Substituting the values we know:
dV/dt = 3(9^2) * 6 cm^2/s
dV/dt = 3(81) * 6 cm^2/s
dV/dt = 243 * 6 cm^3/s
dV/dt = 1458 cm^3/s
Therefore, the rate of increase of the volume when the length of a side is 9 cm is 1458 cm^3/s.
To find the rate of increase of the volume, we can use the formula for the volume of a cube: V = s³, where V is the volume and s is the length of a side.
We are given that the side of the cube is increasing at a rate of 6 cm/s. This means that ds/dt = 6 cm/s. We want to find the rate of increase of the volume, which is dV/dt.
To find this, we can differentiate both sides of the volume formula with respect to time:
dV/dt = d/dt (s³)
Using the chain rule, we can rewrite this as:
dV/dt = 3s² * ds/dt
Now we can substitute the given values: s = 9 cm and ds/dt = 6 cm/s.
dV/dt = 3(9²) * 6
= 3(81) * 6
= 243 * 6
= 1458 cm³/s
Therefore, the rate of increase of the volume when the length of a side is 9 cm is 1458 cm³/s.