Question

If \(\mathbf{F}\) is defined by $$\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\mathbf{F}(r, \theta, z)=\left[\begin{array}{c} r \cos \theta \\ r \sin \theta \\ z \end{array}\right]$$ and \(\mathbf{G}\) is a branch of \(\mathbf{F}^{-1}\), find \(\mathbf{G}^{\prime}\) in terms of \(r, \theta,\) and \(z .\) HINT: See Exercise \(6.2 .14(c)\).

Short answer

Question: Find the derivative of the inverse function, \(\mathbf{G}^{\prime}\), given the function \(\mathbf{F}(r, \theta, z)\) that maps cylindrical coordinates to Cartesian coordinates. Answer: \(\mathbf{G}^{\prime} = \left[\begin{array}{ccc}\cos\theta & -\frac{\sin\theta}{r} & 0\\\sin\theta & \frac{\cos\theta}{r} & 0\\0 & 0 & 1\end{array}\right]\)

Step-by-step solution

Find the inverse function \(\mathbf{G}(x, y, z)\)

Since \(\mathbf{F}(r, \theta, z)\) maps cylindrical coordinates to Cartesian coordinates, its inverse function \(\mathbf{G}(x,y,z)\) should map Cartesian coordinates back to cylindrical coordinates. We can easily find the cylindrical coordinates \((r,\theta,z)\) from the given Cartesian coordinates \((x,y,z)\) using the following equations: 1. \(r = \sqrt{x^2 + y^2}\) 2. \(\theta = \arctan{\frac{y}{x}}\) 3. \(z = z\) Thus, the inverse function \(\mathbf{G}(x, y, z) = \left[\begin{array}{c}r\\\theta\\z\end{array}\right] = \left[\begin{array}{c}\sqrt{x^2 + y^2}\\\arctan{\frac{y}{x}}\\z\end{array}\right]\).

Compute \(\frac{\partial \mathbf{G}}{\partial x}\), \(\frac{\partial \mathbf{G}}{\partial y}\), and \(\frac{\partial \mathbf{G}}{\partial z}\)

Now we compute the partial derivatives of \(\mathbf{G}\) with respect to \(x\), \(y\), and \(z\): 1. \(\frac{\partial \mathbf{G}}{\partial x} = \left[\frac{\partial r}{\partial x}, \frac{\partial \theta}{\partial x}, \frac{\partial z}{\partial x}\right]^T = \left[\frac{x}{\sqrt{x^2+y^2}}, \frac{-y}{x^2+y^2}, 0\right]^T\) 2. \(\frac{\partial \mathbf{G}}{\partial y} = \left[\frac{\partial r}{\partial y}, \frac{\partial \theta}{\partial y}, \frac{\partial z}{\partial y}\right]^T = \left[\frac{y}{\sqrt{x^2+y^2}}, \frac{x}{x^2+y^2}, 0\right]^T\) 3. \(\frac{\partial \mathbf{G}}{\partial z} = \left[\frac{\partial r}{\partial z}, \frac{\partial \theta}{\partial z}, \frac{\partial z}{\partial z}\right]^T = \left[0, 0, 1\right]^T\)

Obtain \(\mathbf{G}^{\prime}\)

Now we can express the derivative of the inverse function \(\mathbf{G}^{\prime}\) in terms of the partial derivatives we found in Step 2: \(\mathbf{G}^{\prime} = \left[\begin{array}{ccc}\frac{\partial r}{\partial x} & \frac{\partial \theta}{\partial x} & \frac{\partial z}{\partial x}\\\frac{\partial r}{\partial y} & \frac{\partial \theta}{\partial y} & \frac{\partial z}{\partial y}\\\frac{\partial r}{\partial z} & \frac{\partial \theta}{\partial z} & \frac{\partial z}{\partial z}\end{array}\right]\) Now we plug in the partial derivatives we found in Step 2: \(\mathbf{G}^{\prime} = \left[\begin{array}{ccc}\frac{x}{\sqrt{x^2+y^2}} & \frac{-y}{x^2+y^2} & 0\\\frac{y}{\sqrt{x^2+y^2}} & \frac{x}{x^2+y^2} & 0\\0 & 0 & 1\end{array}\right]\) Now, replace \(x\) with \(r\cos\theta\) and \(y\) with \(r\sin\theta\), as given by the function \(\mathbf{F}\). \(\mathbf{G}^{\prime} = \left[\begin{array}{ccc}\frac{r\cos\theta}{r} & \frac{-r\sin\theta}{r^2} & 0\\\frac{r\sin\theta}{r} & \frac{r\cos\theta}{r^2} & 0\\0 & 0 & 1\end{array}\right]\) Finally, simplify the expressions as much as possible: \(\mathbf{G}^{\prime} = \left[\begin{array}{ccc}\cos\theta & -\frac{\sin\theta}{r} & 0\\\sin\theta & \frac{\cos\theta}{r} & 0\\0 & 0 & 1\end{array}\right]\)

Key concepts

Cylindrical to Cartesian Coordinates

When we delve into the world of multivariate calculus, the conversion between different coordinate systems becomes a crucial skill. For students studying real analysis or applied physics, understanding how to transform cylindrical coordinates to Cartesian coordinates is especially handy.

The relationship between these two coordinate systems is pretty straightforward. If you have a point specified in cylindrical coordinates \( (r, \theta, z) \), converting it to Cartesian coordinates involves a simple application of trigonometry:

• The x-coordinate is given by \( r \cos \theta \),
• The y-coordinate is \( r \sin \theta \), and
• The z-coordinate remains the same, as both systems share the same vertical axis.

These equations spring from the right-angled triangle formed by the point, the origin, and its projection onto the x-y plane in a 3D space. The radial distance \( r \) from the origin to the point in the cylindrical system becomes the hypotenuse of the triangle, while \( \theta \) is the angle formed with the positive x-axis.

It's a conceptual dance between circles and lines, angles and distances, and mastering this dance ensures a strong foundation in multivariate calculus and related fields.

Partial Derivatives

Imagine taking a hike on a mountainside; you might be interested in how the elevation changes as you walk north or maybe as you ascend directly upwards. In calculus, we have a tool for measuring these kinds of 'partial' changes in a function with multiple variables - it's called the partial derivative.

Partial derivatives show how a multivariable function changes as one variable shifts, keeping the others constant. For instance, if you have a function \( f(x, y, z) \), and you're interested in the slope of \( f \) in the x-direction at a point, you'll compute \( \frac{\partial f}{\partial x} \). Calculating the partial derivatives \( \frac{\partial \mathbf{G}}{\partial x} \), \( \frac{\partial \mathbf{G}}{\partial y} \), and \( \frac{\partial \mathbf{G}}{\partial z} \) helps in finding how each cylindrical coordinate changes with respect to each Cartesian coordinate.

This notion is as crucial in theoretical fields, like real analysis, as it is in practical applications such as engineering and physics, where it can help predict how systems behave under variable changes - just like understanding the terrain's incline guides a hiker's steps.

Jacobian Matrix

The Jacobian Matrix is like a compass for multivariable functions, guiding us through the landscape of their rates of change. In technical terms, it's a matrix consisting of all the first-order partial derivatives of a vector-valued function. For a function \( \mathbf{F} \) that maps cylindrical coordinates to Cartesian, its Jacobian matrix helps us understand how small changes in the cylindrical variables \( (r, \theta, z) \) affect the Cartesian outputs \( (x, y, z) \).

When we found the inverse of \( \mathbf{F} \) and named it \( \mathbf{G} \) (mapping Cartesian back to cylindrical), we completed the circle. Now the Jacobian of \( \mathbf{G} \) describes how small changes in the Cartesian coordinates alter the cylindrical ones. Its calculation involves taking partial derivatives of \( \mathbf{G} \) with respect to each of the Cartesian variables, x, y, and z, and then arranging these derivatives into a neat matrix.

What's captivating is that the Jacobian matrix doesn't merely tell us the rate of change; it encapsulates the entire linear approximation to the function at a point. Grasping the Jacobian is a powerful mathematics feat, akin to possessing a map that reveals the immediate terrain's steepness in any direction—a tool indispensable for both explorers and mathematicians alike.