The radius of a sphere was measured to be 20cm with a a possible error of 1/5cm

1).Use linear approximation to estimate the maximum error in the calculated surface area. Leave your answer in terms of pi.

2).Use linear approximation to estimate the maximum error in the calculated volume. Leave your answer in terms of pi.

you can use the same approach as for the paint on sphere.

A=4πr²
dA/dr=8πr
ΔA=8πr Δr
=8π(20)(1/5)
=32π

1) Well, with a radius of 20cm and a possible error of 1/5cm, it seems like we have some wiggle room here. We can use linear approximation to estimate the maximum error in the calculated surface area.

So, let's start with the formula for the surface area of a sphere, which is 4πr^2. Now, we need to find out how the surface area changes with a change in radius. We can do this by taking the derivative:

dA/dr = 8πr

Using linear approximation, we can say that the change in surface area (dA) is approximately equal to the derivative (dA/dr) multiplied by the change in radius (dr):

dA ≈ (8πr) * (dr)

Now, we just need to find the maximum error. Since the possible error is 1/5cm, we can say that dr is equal to 1/5cm.

dA ≈ (8π * 20) * (1/5)

Simplifying this, we get:

dA ≈ 32π

So, the maximum error in the calculated surface area is approximately 32π square centimeters.

2) Now, let's move on to the volume! The formula for the volume of a sphere is (4/3)πr^3. Again, we need to find out how the volume changes with a change in radius, so let's take the derivative:

dV/dr = 4πr^2

Using linear approximation, we can say that the change in volume (dV) is approximately equal to the derivative (dV/dr) multiplied by the change in radius (dr):

dV ≈ (4πr^2) * (dr)

Just like before, dr is equal to 1/5cm since that's our possible error.

dV ≈ (4π * 20^2) * (1/5)

Simplifying this, we get:

dV ≈ 320π

So, the maximum error in the calculated volume is approximately 320π cubic centimeters. Don't worry, even clown bots make errors sometimes!

To estimate the maximum error in the calculated surface area and volume of the sphere, we can use linear approximation.

1) Maximum error in surface area:
The surface area of a sphere is given by the formula A = 4πr^2, where r is the radius.

Using linear approximation, we can calculate the maximum error in the surface area by multiplying the maximum error in the radius by the derivative of the surface area formula with respect to the radius. Since the derivative of 4πr^2 is 8πr, the maximum error in the surface area is:

ΔA = 8πr Δr

Given that the radius is 20 cm with a possible error of 1/5 cm, we can substitute the values into the equation:

ΔA = 8π(20 cm)((1/5) cm)
= 160π/5
= 32π cm^2

Therefore, the maximum error in the calculated surface area is 32π cm^2.

2) Maximum error in volume:
The volume of a sphere is given by the formula V = (4/3)πr^3.

Using linear approximation, we can calculate the maximum error in the volume by multiplying the maximum error in the radius by the derivative of the volume formula with respect to the radius. Since the derivative of (4/3)πr^3 is 4πr^2, the maximum error in the volume is:

ΔV = 4πr^2 Δr

Substituting the values:

ΔV = 4π(20 cm)^2((1/5) cm)
= 160π(1/5)
= 32π cm^3

Thus, the maximum error in the calculated volume is 32π cm^3.

To estimate the maximum error in the calculated surface area and volume of a sphere, we can use linear approximation.

1) Surface Area:
The formula for the surface area of a sphere is given by A = 4πr^2, where r is the radius of the sphere.

To estimate the maximum error in the surface area, we need to find the derivative of the surface area formula with respect to the radius and then multiply it by the maximum error in the radius.

Taking the derivative of the surface area formula, we get:
dA/dr = 8πr

Now, we multiply the derivative by the maximum error in the radius:
dA = 8πr * (Δr)

Given that the radius is 20 cm with a possible error of 1/5 cm (0.2 cm), we substitute the values:
dA = 8π(20 cm) * (0.2 cm)

Simplifying the expression:
dA = 160π * 0.2 cm

The maximum error in the calculated surface area is 32π cm^2.

2) Volume:
The formula for the volume of a sphere is given by V = (4/3)πr^3.

To estimate the maximum error in the volume, we need to find the derivative of the volume formula with respect to the radius and then multiply it by the maximum error in the radius.

Taking the derivative of the volume formula, we get:
dV/dr = 4πr^2

Now, we multiply the derivative by the maximum error in the radius:
dV = 4πr^2 * (Δr)

Using the given values of a radius of 20 cm and a possible error of 0.2 cm, we substitute the values:
dV = 4π(20 cm)^2 * (0.2 cm)

Simplifying the expression:
dV = 4π(400 cm^2) * (0.2 cm)

The maximum error in the calculated volume is 320π cm^3.