WEBVTT
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Hello, everyone. This is my solution to problem
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, Tony, one of 15.8. So the problem
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ask you move A and B to find the Jacoby
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in of two different coordinate transformations. So well,
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we're going to start with the 1st 1 which will
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start right here. So we have a two variable
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coordinate transformation. When you find Jacoby in of it
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, you know where to find the Jacoby in of
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a two variable transformation, you solved the two by
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two the terminate of the two by two matrix that
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is foreign from all the partial derivatives of the coordinates
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. So dx dy you right here, Do I
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d you t x t v and do I D
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. V? So, in order, find a
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terminated two by two matrix. You take the cross
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product of the elements too far, and then that
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gives you the determinant. So the cross product of
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this matrix is dxc u times d y d v
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minus d. Y d. You times dx TV
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and I've labeled D's both one into because it will
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help us on part B. So in order to
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finish solving this, we just simply need all four
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of the partial derivatives which I've written out in this
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red box here for you. So let's go over
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real quick. So the 1st 1 dx dy you
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is just the derivative with respect to you of you
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co sign V. So when we're taking a partial
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derivative, remember, we treat all other variables except
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the one we're deriving over as a constant. And
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then we just do a normal derivative after that.
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So, since we're in a sense, retained drew
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respect to you, we are leaving this term that
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is Onley dependent on V as a constant this coastline
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v So really, we're taking Drew just you and
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derivative of you is simply one. So the whole
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partial is co signed d so similarly for do I
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d v. We have now where we're taking curative
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over v of something that something is you signed the
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So now treat the U is a constant so the
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whole thing will be multiplied by you at the end
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and then we just need the derivative of sign of
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e. So the ribbon of of scientists just co
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sign. So the whole partial, it's just you
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coast on TV and then we do a similar process
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for the next two. So d y to you
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again, just dependent on you. So it's just
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a derivative of you, which is one so that
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the whole thing is just the sign V term.
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And then for D X TV. It's the same
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thing is D Y D V. Except there's a
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slight thing. It's like thing you gotta watch out
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for. They're derogative of co sign is a negative
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sign. So the whole derivative is negative. You
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signed ah v o I'm missing a V or night
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. Let's fix that right now, everyone, There
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we go. So now that we have all four
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of these, we can just kind of plug and
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chug and then simplify our answer. So the plug
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part up here, we have co signed the times
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you co sign be so dx dy you times do
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I d v minus signed the the Times Negative you
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side VSO d y d you times dx tv So
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simplifying this gives us you co sign squared V plus
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you sine squared V and it is plus because remember
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, when we have this negative sign right here on
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the U or on the issue of the GX TV
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term that we factor out to turn the minus into
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a plus. Then we can go and factor out
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Ah, you from both terms. So we have
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you times coastlines where b plus sine squared V no
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coastlines where b plus sine squared b. As you
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can see from this trick identity I wrote over here
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this is very useful almost always in calculus, by
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the way. So please remember it is just equal
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toe one. So we can just cross that whole
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term out. And then we're left with the Jacoby
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in apart. A. That's what that little red
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subscript a is for of u V is just you
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. So now let's talk about Part B. Um
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, you could do the math again for the new
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coordinate system, But if you notice I X in
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Part B is equal to why, in part A
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. And similarly, why in part B is equal
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to X in part a. Ah, in layman's
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terms, this means the cornets have just swapped the
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court or the court axes have just swept places.
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So let's just do ourselves a favor. And instead
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of working all the math, we have all four
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of the partials we need right here again. So
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let's just rewrite this matrix from part A. So
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we're rewriting one, and I put the sub script
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. He's there. So you can remember that we're
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talking about part B here and then turned it into
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on, As you can see, I did right
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here into de y of a D u d x
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of a D u d y of a d v
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and dx of a TV. So why is this
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important? It's because when you simplify this matrix,
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right, we're sorry, not simplify. When you
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take the cross product of this, you're going to
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get question or you're gonna go get equation to again
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. But you're not gonna get equation two. You're
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gonna get negative the value of equation, too.
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So since equation to is the answer the part A
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and the answer in part B, it's just negative
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. The value of equation, too. Then the
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answer to part B is just negative. The answer
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to part A. So therefore, as I wrote
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down here there, therefore, JB of U V
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is equal to negative J of UV so JB of
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you they the is equal to just negative. New
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. Now, if you solve it, this this
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method here logically, make sure you don't just write
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down the answer. Make sure you do kind of
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either right? Some sentences to explain your logic or
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kind of do something similar to what I have done
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here by just writing out what the new matrices are
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in order to a show that you understand why the
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answer is the same. But just the inverse.
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Um, if you don't Ah, I'm I can't
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guarantee you'll gain full credit. But ah, if
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you do show your logic, you should. And
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thank you very much for watching my video. And
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I hope it is was of help to you.
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Thank you.