1. ## Differentiation Help?

A globe with a diameter of 30 cm is to be painted with a uniform layer of paint which is 4 mm thick. Use the method of differentiation to find the approximate volume of paint needed.

Help anyone?

2. Hello, xje4bv!

If these "differential approximations" are confusing to you, don't give up.
It took me the looongest time to understand them.

A globe with a diameter of 30 cm is to be painted
with a uniform layer of paint which is 4 mm thick.
Use the method of differentiation to find the approximate volume of paint needed.

We have a sphere with radius 15 cm.
It has a volume according to the formula: . $V \:=\:\frac{4}{3}\pi r^3$

By adding a layer of paint, the radius is increased by $4\text{ mm } = 0.4\text{ cm}.$

The question is: by how much is the volume increased?

We could do the actual arithmetic,
. . but we can approximate the answer with differentials.

We have: . $V \:=\:\frac{4}{3}\pi r^3$

Take differentials: . $dV \:=\:4\pi r^2\,dr$

If we understand what each part means, we're all set!

. . $\underbrace{dV}_{\text{change in volume}} \;=\;4\pi\cdot\underbrace{r^2}_{\text{radius}^2}\c dot\underbrace{dr}_{\text{change in radius}}$

We know: . $r = 15,\;dr = 0.4$

Therefore: . $dV \;=\;4\pi(15^2)(0.4) \;=\;360\pi\text{ cm}^3$

The volume of the paint needed is approximately 1131 cm³

3. Originally Posted by Soroban
Hello, xje4bv!

If these "differential approximations" are confusing to you, don't give up.
It took me the looongest time to understand them.

We have a sphere with radius 15 cm.
It has a volume according to the formula: . $V \:=\:\frac{4}{3}\pi r^3$

By adding a layer of paint, the radius is increased by $4\text{ mm } = 0.4\text{ cm}.$

The question is: by how much is the volume increased?

We could do the actual arithmetic,
. . but we can approximate the answer with differentials.

We have: . $V \:=\:\frac{4}{3}\pi r^3$

Take differentials: . $dV \:=\:4\pi r^2\,dr$

If we understand what each part means, we're all set!

. . $\underbrace{dV}_{\text{change in volume}} \;=\;4\pi\cdot\underbrace{r^2}_{\text{radius}^2}\c dot\underbrace{dr}_{\text{change in radius}}$

We know: . $r = 15,\;dr = 0.4$

Therefore: . $dV \;=\;4\pi(15^2)(0.4) \;=\;360\pi\text{ cm}^3$

The volume of the paint needed is approximately 1131 cm³

Thanks Soroban, your answer only reaffirms my answer which was also 360 pi cm cube. However, the answer I got from the textbook was 1440 pi cm cube. Well, I guess the textbook's wrong then.