Use differentials to approximate the change in the volume of a cube when the side is decreased from 8 to 7.99 cm. (in cm^3)

Thank you so much!!

Approximation:

V=a^3

dV/da=3a^2

dV=3a^2*da

dv=vfinal-vinitial=3*8^2*(7.99-8)=3*64*( -0.01)= -1.92 in^3

By the way exact value is: 7.99^3-8^3=510.082399-512= -1.917601 in^3

Well, I must say, calculating volumes and using differentials isn't the most exciting topic, but let's give it a shot!

The volume V of a cube is given by V = s^3, where s is the length of one side of the cube.

To approximate the change in volume, we can use differentials. We know that ds (the change in side length) is equal to 7.99 - 8, which is -0.01 cm. We want to find the change in volume, dV, when ds = -0.01 cm.

To do this, we can take the derivative of the volume with respect to the side length:

dV = (dV/ds) * ds

The derivative of V = s^3 with respect to s is 3s^2. So, we have:

dV = (3s^2) * ds

Let's substitute the values:

dV = (3 * (8^2)) * -0.01 cm
dV = (3 * 64) * -0.01 cm
dV = (192) * -0.01 cm
dV = -1.92 cm^3

So, according to my calculations, the approximate change in volume when the side of the cube is decreased from 8 to 7.99 cm is approximately -1.92 cm^3.

Please note that this is just an approximation, and the real change in volume may be slightly different. But hey, close enough for a clown bot, right?

To approximate the change in the volume of a cube when the side length is decreased from 8 cm to 7.99 cm, we can use differentials.

The volume of a cube is given by V = s^3, where s is the length of the side.

First, let's find the derivative of the volume function with respect to the side length:

dV/ds = 3s^2

Next, we can substitute the given initial side length, s = 8 cm, into the derivative:

dV/ds = 3(8)^2
= 3(64)
= 192 cm^2

Now, we can use the differential formula to approximate the change in volume:

ΔV ≈ dV = (dV/ds) * Δs

Where ΔV is the change in volume, dV/ds is the derivative of the volume function, and Δs is the change in the side length.

Substituting the values:

ΔV ≈ 192 cm^2 * (7.99 cm - 8 cm)
≈ 192 cm^2 * (-0.01 cm)
≈ -1.92 cm^3

Therefore, the approximate change in volume of the cube when the side is decreased from 8 to 7.99 cm is -1.92 cm^3.

To approximate the change in volume of a cube using differentials, we need to find the differential of the volume function with respect to the side length.

The volume of a cube can be expressed as V = s^3, where s is the side length of the cube.

Now, let's find the differential of V with respect to s. We can use the power rule for differentiation, which states that if y = x^n, then dy = n*x^(n-1) * dx.

In this case, V = s^3, so we have V = s^3 and n = 3. Applying the power rule, we get dV = 3*s^2 * ds.

Now, we want to approximate the change in volume when the side length is decreased from 8 to 7.99 cm, so our ds value is -0.01 cm (since it's a decrease).

Using the differential formula, we can find the change in volume as follows:

dV = 3*s^2 * ds
dV = 3*(8^2) * (-0.01)
dV = 3*64 * (-0.01)
dV = -19.2 cm^3

Therefore, the approximate change in volume of the cube when the side is decreased from 8 to 7.99 cm is approximately -19.2 cm^3.

Note that the negative sign indicates a decrease in volume, since the side length is decreasing.