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Problem 21 Medium Difficulty

Find the Jacobian $\partial(x, y) / \partial(u, v)$ of the transformation
\begin{equation}
\begin{array}{l}{\text { a. } x=u \cos v, \quad y=u \sin v} \\ {\text { b. } x=u \sin v, \quad y=u \cos v.}\end{array}
\end{equation}

Answer

The Jacobian of a two variable transformation is found from the determinant of the following matrix.
\begin{vmatrix}

\frac{\delta x}{\delta \mathcal{u}} & \frac{\delta x}{\delta \mathcal{v}}\\
\frac{\delta y}{\delta \mathcal{u}} & \frac{\delta y}{\delta \mathcal{v}}
\end{vmatrix}

Using the definition of the determinant we find

\begin{equation}

\begin{vmatrix}
\frac{\delta x}{\delta \mathcal{u}} & \frac{\delta x}{\delta \mathcal{v}}\\
\frac{\delta y}{\delta \mathcal{u}} & \frac{\delta y}{\delta \mathcal{v}}
\end{vmatrix} = \frac{\delta x}{\delta u}\frac{\delta y}{\delta v} - \frac{\delta y}{\delta u}\frac{\delta x}{\delta v}

\end{equation}

Next Calculate all the partial derivatives.

\begin{align}

\frac{\delta x}{\delta u} = \frac{\delta}{\delta u}u\cos(v) = \cos(v) \\
\frac{\delta y}{\delta v} = \frac{\delta}{\delta v}u\sin(v) = u\cos(v) \\
\frac{\delta y}{\delta u} = \frac{\delta}{\delta u}u\sin(v) = \sin(v)\\
\frac{\delta x}{\delta v} = \frac{\delta}{\delta v}u\cos(v) = -u\cos(v)\\

\end{align}

Now substitute and calculate the answer.

\begin{equation*}

\frac{\delta x}{\delta u}\frac{\delta y}{\delta v} - \frac{\delta y}{\delta v}\frac{\delta x}{\delta v} = u\cos(v)\cos(v) -[-u\sin(v)\sin(v)]

\end{equation*}

Simplify using the trig identity
\begin{equation}
\cos^2(x) + \sin^2(x) = 1
\end{equation}

\begin{equation}
u[\cos^2(v) + \sin^2(v)] = u
\end{equation}

\[
\boxed{\mathcal{J}(u,v) = u}
\]

Following this same process for part b) gives us the following

\begin{align}

\frac{\delta x}{\delta u} = \frac{\delta}{\delta u}u\sin(v) = \sin(v) \\
\frac{\delta y}{\delta v} = \frac{\delta}{\delta v}u\cos(v) = -u\sin(v) \\
\frac{\delta y}{\delta u} = \frac{\delta}{\delta u}u\cos(v) = \cos(v)\\
\frac{\delta x}{\delta v} = \frac{\delta}{\delta v}u\sin(v) = u\cos(v)\\

\end{align}

\begin{equation*}

\frac{\delta x}{\delta u}\frac{\delta y}{\delta v} - \frac{\delta y}{\delta v}\frac{\delta x}{\delta v} = -[u\sin(v)\sin(v)] -[u\cos(v)\cos(v)]

\end{equation*}

\begin{equation}
-u[\sin^2(v)+\cos^2(v)] = u
\end{equation}

\[
\boxed{\mathcal{J}(u,v) = -u}
\]

Discussion

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Video Transcript

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

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