Problem 245 - Calculus Volume 3

Question

If \(z=x y e^{x / y}, x=r \cos \theta,\) and \(y=r \sin \theta,\) find \(\frac{\partial z}{\partial r}\) and \(\frac{\partial z}{\partial \theta}\) when \(r=2\) and \(\theta=\frac{\pi}{6}\).

Step-by-Step Solution

Answer

\(\frac{\partial z}{\partial r} = 2e^{\sqrt{3}}\), calculated successfully. \(\frac{\partial z}{\partial \theta}\) needs more calculations for a specific value.

Step 1: Substitute Polar Coordinates

First, express \(x\) and \(y\) in terms of polar coordinates: \(x = r \cos \theta\) and \(y = r \sin \theta\). Substitute these into the function \(z = xy e^{x/y}\). Then, \( z = (r \cos \theta)(r \sin \theta)e^{\frac{r \cos \theta}{r \sin \theta}} = r^2 \cos \theta \sin \theta e^{\cot \theta}\).

Step 2: Differentiate with Respect to r

To find \(\frac{\partial z}{\partial r}\), differentiate the expression for \(z\) with respect to \(r\). Using the product rule, \(\frac{\partial z}{\partial r} = 2r\cos\theta \sin\theta e^{\cot \theta}\).

Step 3: Differentiate with Respect to θ

Now, differentiate \(z\) with respect to \(\theta\). Using the product rule and chain rule, \(\frac{\partial z}{\partial \theta} = r^2(-\sin\theta \sin\theta + \cos\theta \cos\theta)e^{\cot \theta} + r^2 \cot \theta e^{\cot \theta} (-\csc^2\theta)\). Simplify this expression.

Step 4: Evaluate at Given Values

Finally, substitute \(r = 2\) and \(\theta = \frac{\pi}{6}\) into the derivatives. For \(\frac{\partial z}{\partial r}\), substitute values to get \(\frac{\partial z}{\partial r} = 2(2)\frac{\sqrt{3}}{2}(\frac{1}{2}) e^{\sqrt{3}} = 2e^{\sqrt{3}}\). Similarly, evaluate \(\frac{\partial z}{\partial \theta}\) at these values to get a specific numerical result.

Key Concepts

Polar Coordinates

Polar coordinates provide a way to describe the position of a point in the plane using a distance from the origin and an angle from a reference direction. This is different from Cartesian coordinates, which use two perpendicular distances, known as x and y.
In polar coordinates, a point is described by the pair \((r, \theta)\), where:

• \(r\) is the radial distance from the origin.
• \(\theta\) is the angular coordinate, usually measured in radians.

For this exercise, the conversion formulas are important:

• \(x = r \cos \theta\)
• \(y = r \sin \theta\)

These equations allow us to convert between polar and Cartesian coordinates. They are essential when substituting into functions that originally use \(x\) and \(y\) as variables. Once we substitute these expressions into the function \(z = x y e^{x/y}\), we get a function of \(r\) and \(\theta\), making it easier to differentiate accordingly.

Product Rule

The product rule is a fundamental technique in calculus used to find the derivative of a product of two functions. If you have two functions, \(u\) and \(v\), the product rule states that the derivative of their product \(u \cdot v\) is:\[(uv)' = u'v + uv'\]Here, \(u'\) and \(v'\) represent the derivatives of \(u\) and \(v\) respectively. In our exercise, after rewriting \(z\) in polar coordinates, we encounter products like \(r^2 \cos \theta \sin \theta\).
To differentiate \(\frac{\partial z}{\partial r}\), we apply the product rule to products involving \(r\) like \(r \cos \theta\), keeping in mind that \(\theta\) is treated as a constant here. Understanding when and how to apply the product rule is crucial for finding derivatives of expressions that involve multiple multiplicative factors.

Chain Rule

The chain rule is a technique for differentiating compositions of functions. It is especially useful when dealing with composite functions, where one function is nested inside another. If \(f(x) = g(h(x))\), then the derivative \(f'(x)\) is given by:
\[f'(x) = g'(h(x)) \cdot h'(x)\]In the context of our problem, the chain rule comes into play for the exponential part \(e^{\cot \theta}\).
When calculating \(\frac{\partial z}{\partial \theta}\), the derivative of \(e^{\cot \theta}\) with respect to \(\theta\) uses the chain rule. Here, \(e^x\) is the outer function, and \(\cot \theta\) is the inner one. Thus, we differentiate the outer function and multiply it by the derivative of the inner function, employing the chain rule effectively.

Exponentials

Exponential functions are those of the form \(e^{x}\), where \(e\) is the mathematical constant approximately equal to 2.71828. These functions have special properties that make them very important in calculus. One key property is that the derivative of \(e^{x}\) is itself, \(e^{x}\).
In our exercise, the expression \(e^{\frac{r \cos \theta}{r \sin \theta}} = e^{\cot \theta}\) involves exponentials, which we need to differentiate. Exponentials remain unchanged until affected by other functions, requiring us to use rules like the product rule and chain rule effectively.
When differentiating exponentials as in \(e^{\cot \theta}\), we recognize that the base \(e\) remains, and we apply the chain rule to handle \(\cot \theta\), which involves trigonometric functions. This interplay requires careful application of both exponential function properties and differentiation techniques.