Teaching Calculus

Site owners

  • Nikolay Brodskiy
  • Alexander Brodsky

Transcript

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So what we ended up is with a vector ...

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... representing geometric property of ...

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... a point of the trajectory. And ...

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... a number positive or negative representing ...

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... the same or different -- well some property ...

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... of the same curve at the same point.

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And well the natural curiosity ...

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... suggests that we should relate those things.

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And we especially should relate those things to the ...

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... very simple case of the trajectory being a circle.

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Because what we ultimately want is the relation between ...

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... this Physics ...

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... idea of orthogonal component of acceleration and the idea of the radius of a circle.

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So to do that I will ...

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... look at a circle.

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Let's look at a circle ...

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... of radius ...

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... R.

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And er ...

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I will think about a particle moving along the circle.

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And the simplest way to let a particle move along a circle would be to ...

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... let it move [...] be parametrized using cosine and sine.

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R times cosine of t and R times sine of t ...

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... is the simplest way to parameterize a circle.

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My particle is going to be moving around a circle.

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So that at some time t its position is going to be ...

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... radius times cosine of t ...

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... being x-coordinate and radius times sine of t being the y-coordinate.

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So that's my position.

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Let's find the ...

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... velocity.

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Well now as I have a formula for the position I can simply differentiate.

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And I know how to differentiate cosine.

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So it's R times negative ...

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... sine t.

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And the derivative of sine is cosine. So it's R times cosine t.

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And I also find the second derivative.

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That is negative ...

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... R cosine t ...

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... and negative again ...

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... R sine t.

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And then I would like to evaluate those two formulas.

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So first I will evaluate the ...

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... Physics invariant ...

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... orthogonal component of the second derivative of f with respect to the first ...

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... divided by the magnitude of the first derivative squared.

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So what is that?

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Well let's find the magnitude of the first derivative.

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Magnitude of the velocity, that's the speed.

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So what's the magnitude of ...

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... f prime?

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It's a square root of sum of these squares.

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Right? Square root of R squared sine squared ...

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... plus ...

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... R squared cosine squared.

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And since R squared factors ...

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... the remaining part is sine squared plus cosine squared that gives you one.

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And square root of R squared is going to be simply R.

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So the number in denominator is going to be ...

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... radius of a circle squared.

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All right. What about the numerator?

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For the numerator ...

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... I'll have to take the second derivative ...

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... and subtract the projection of that second derivative on the first.

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The projection is computed as the second derivative dot the first derivative ...

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... divided by the ...

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... square of the magnitude of the first derivative times the first derivative.

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That's the formula for orthogonal component.

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And [...] that.

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What's the f double prime?

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It is minus R cosine t, minus R sine t.

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And then ...

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The dot product here ...

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... is going to be ...

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... minus minus gives plus ...

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... R squared ...

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... times sine t times cosine t ...

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... and then plus minus gives minus ...

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... R squared times cosine t times sine t.

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And what we notice here is that it is zero.

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The whole thing is zero.

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So no matter what we divide by ...

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... or multiply by ...

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... the whole thing is zero.

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So the whole formula is going to be equal to that over radius squared.

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And ...

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Well that's a vector.

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One radius cancels. So  have one over R ...

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... multiplied by the vector ...

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... minus cosine t ...

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... minus sine t.

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So what is the relation of this vector to the radius of the circle?

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Well if you look at the magnitude of the vector ...

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... the magnitude of that part ...

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... magnitude of this is going to be one.

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Cosine squared plus sine squared is equal to one.

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So what we see here is that the magnitude of orthogonal component ...

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... of f double prime with respect to f prime ...

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... divided by the ...

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... speed squared ...

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... magnitude of that vector ...

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... is equal to one over R.

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That's the relation ...

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... the exact relation between ...

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... the idea of normal acceleration that we experience in Physics ...

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... related to the radius of the circle that we see geometrically.

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All right. Now er ...

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Well basically what it say is that ...

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... from Physics point of view if we look at that vector ...

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... suggested as geometric invariant ...

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... the magnitude of that vector is one over the radius.

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Now let's look at that other formula.

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The algebraic formula involving area.

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How different is that invariant?

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So I need to compute the area on f prime and f double prime ...

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... divided by the magnitude of f prime cubed.

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And I have both vectors computed explicitly.

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So the area is going to be determinant of this matrix.

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So I have to multiply these two quantities.

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Minus times minus gives plus.

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R squared sine times sine.

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So R squared times sine squared t ...

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... minus that product.

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And minus minus gives plus.

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R squared cosine squared ...

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... plus R squared cosine squared.

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That's the area.

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And it should be divided by the magnitude of f prime which is R cubed.

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And of course sine squared plus cosine squared is one ...

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... after we factor R squared.

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And this is immediately one over the radius.

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So that is also related to the radius in exactly the same way.

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That is also one over radius.

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So somehow that formula coming from Physics ...

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... is [...] related to that formula coming from Algebra.

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But they look very different.

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Now let's figure out the relation geometrically.

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If I have ...

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... vector f double prime ...

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... and ... Well let's say f prime here.

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And f double prime there.

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Then what's the area on these two vectors?

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Well that is going to be the area of this parallelogram.

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That's the area we are talking about.

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So area on f prime and f double prime.

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Now I want to relate it to what?

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I want to relate it to the magnitude of orthogonal component.

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Now how can I see orthogonal component of f double prime with respect to f prime?

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It is going to be ...

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... this vector, right?

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So this is going to be ...

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... orthogonal component of f double prime with respect to f prime.

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Now the magnitude of that ...

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... of that vector ...

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... is going to be the height of the parallelogram.

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So this is the same as the height ...

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... of that parallelogram we ara looking at.

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And we know that the area is equal to the height ...

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... times the magnitude of the base.

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And now we see that there is no mystery in seeing that ...

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... the area divided by magnitude of f prime ...

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... is the same thing as ...

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... the magnitude of orthogonal component.

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So that's exactly why we see extra f prime in the denominator of the algebraic formula.

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So these two formulas are exactly about the same thing.

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And that same thing is one over radius.

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It is now obviously related to the radius of the circle.

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