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[music] Here, I've got an orange. I'd like to measure say, the radius of the

orange, thinking of this orange as a perfect sphere.

One way to do this would be to measure, say, the circumference of the orange

around its equator. So here if I measure that it looks like

it's about 24 centimeters. So the circumference of the orange is 24

centimeters. What does that tell me about, say the

radius of the orange? Well that circumference Will be 2 pi times

the radius of the orange. So if I divide both sides by 2 pi.

I get that the radius of the orange is about 12 over pi centimeters.

How much rind is there on this orange? That could be formulated as a linear

approximation problem. Well let's fit this up./g So, I just

calculated the orange radius to be about 12 over pi centimeters.

I want to figure out how thick the rind is.

Let me. Let me peel off a bit of the orange here.

Just want to measure how thick this is with, say my little ruler here and yeah,

maybe three millimeters. So, point three centimeters will be the

rind thickness. Now how does knowing the rind thickness

going to help me to calculate, at least approximate, the volume of the rind in

this whole orange. Well I know the volume of a sphere, which

by modeling the orange to be, is 4 3rds pi r q.

And the rind volume, is at least approximately how much the volume changes

when I take this formula, and replace r with r minus h.

Right? When I take that sphere and I make it just

a little bit smaller. That change in volume is the volume of the

rind. And to understand how wiggling the input

effects the output, I'm going to use the derivative.

The formula that I want to be differentiating here is this volume of a

sphere of radius r. Which is 4 thirds pi r cubed [NOISE]And if

I differentiate this with respect to r. The r cubed differentiates to 3r squared.

And the 3 in the denominator here will cancel this 3 by the power rule.

So the derivative will be 4 pi r squared. Now what I'm trying to approximate is the

rind volume which is really the difference between a sphere of radius r and a

slightly smaller sphere where I've shrunk down just past the rind, right.

That difference will be the volume of the rind.

And approximately this h times the derivitive of v at r.

Right, because this is asking how does the output change when the input goes from R

minus H to R. And it's approximately the input change

times the derivative. Now in this case what do I know?

The orange's radius is 12 over Pi, so I'll use that for R.

And the rind's thickness is point three centimeters so I'm going to use that for

H. So this is point three centimeters Times 4

pi, times this radius, which is 12 over pi centimeters squared.

So this is cubic centimeters. Well that's good because I'm trying to

calculate a volume. And I'll just keep calculating here.

So it's point 3 times 4 pi times 144 over pi squared.

Cubic centimeters. And 4 times 144 is 576 times .3, the pi

divided the pi squared is going to be a pi in the denominator.

So I've got .3 times 576 over pi Cubic centimeters.

That's about 55 cubic centimeters. So the volume of the rind is approximately

55 cubic centimeters. Now admittedly I could have calculated the

volume of the rind just by calculating the volume of the whole thing.

Calculating the volume of the whole thing with the radius reduced by h, and taking a

difference. But the neat thing here is that you can

approximate that quantity by using calculus, right?

By thinking about this linear approximation business.

I also want to see now how close the linear approximation was.

Alright, so I've taken my orange and peeled off all the rind, and I've taken

the rind and smooched it down into this ball, and I'll try to calculate the volume

now of this rind ball. I want to compute the volume of a rind,

and I made that rind ball here, let me Let me compute the circumference of this rind

ball. That's about 15 centimeters.

So the circumference of my ball of rind is about 15 centimeters, that means my rind

ball has a radius. 15 over 2 pi centimeters.

And here, assuming my Rind ball is a perfect sphere.

This would be the volume of the Rind ball. But in here I gotta put the radius, which

is 15 over. 2 pie.

So, 4 3rds pie the radius cubed, and that's about 57 cubic centimeters.

So, yeah, I mean, I actually took all the rind off the orange and the value that we

got is pretty close to what the linear approximation told us the volume of rind

should be. This perspective sheds some light on what

might otherwise be seen as just some random coincidence.

Now what you notice is that the volume of a sphere which is 4 3rds pi r cubed and

the surface area of a sphere which is 4 pi r squared.

These two formulas are related right. If I differentiate this volume formula

with respect to r, I get the surface area. Formula.

Again we're seeing that calculus isn't really about solving specific problems.

I don't care about the volume of the rind of this orange.

Right? If that was what calculus was good for,

who would care? Right.

What calculus is really good at Is providing a general framework and with

that general framework we can begin to see the patterns between different concepts.

Like this fact, the surface area of a sphere is the derivative of the volume of

a sphere with respect to the radius. That's not really a coincidence, right?

Calculus is providing an explanation for that supposed coincidence.