Problem 243 - Calculus 3

Question

Let \(z=\frac{x}{y}, x=2 \cos u\) and \(y=3 \sin v .\) Find \(\frac{\partial z}{\partial u}\) and \(\frac{\partial z}{\partial v}\)

Step-by-Step Solution

Answer

\(\frac{\partial z}{\partial u} = \frac{-2 \sin u}{3 \sin v}\) and \(\frac{\partial z}{\partial v} = \frac{-2 \cos u \cos v}{3 \sin^2 v}\).

Step 1: Compute the Expression for z

First, substitute the given expressions for \(x\) and \(y\) into \(z = \frac{x}{y}\). Substitute \(x = 2 \cos u\) and \(y = 3 \sin v\) to get \(z = \frac{2 \cos u}{3 \sin v}\).

Step 2: Compute Partial Derivative of z with Respect to u

Using the quotient rule for differentiation, \(\frac{\partial}{\partial u} \left( \frac{2 \cos u}{3 \sin v} \right)\): \[\frac{\partial z}{\partial u} = \frac{(3 \sin v)(-2 \sin u) - (2 \cos u)(0)}{(3 \sin v)^2} = \frac{-6 \sin u \sin v}{9 \sin^2 v}\]Simplify to get \(\frac{\partial z}{\partial u} = \frac{-2 \sin u}{3 \sin v}\).

Step 3: Compute Partial Derivative of z with Respect to v

Again, using the quotient rule for \(\frac{\partial}{\partial v}\):\[\frac{\partial z}{\partial v} = \frac{(3 \sin v) (0) - (2 \cos u)(3 \cos v)}{(3 \sin v)^2} = \frac{-6 \cos u \cos v}{9 \sin^2 v}\]Simplify to get \(\frac{\partial z}{\partial v} = \frac{-2 \cos u \cos v}{3 \sin^2 v}\).

Key Concepts

Quotient Rule

The quotient rule is a crucial technique in calculus, especially helpful when working with functions expressed in the form of a ratio. When dealing with partial derivatives of a quotient where both numerator and denominator are functions, the quotient rule provides a systematic way to find the derivative.

• The general formula for the derivative of a quotient \( \frac{f(u)}{g(v)} \) is: \[ \frac{d}{dx} \left( \frac{f(u)}{g(v)} \right) = \frac{g(v)f'(u) - f(u)g'(v)}{(g(v))^2} \]
• It's vital to remember that the differentiation requires keeping the denominator constant initially and differentiating the numerator, then vice versa.
• In the given exercise, the function \( z = \frac{2 \cos u}{3 \sin v} \) was differentiated by applying the quotient rule to find \( \frac{\partial z}{\partial u} \) and \( \frac{\partial z}{\partial v} \).
• Notice how the product of derivatives and original functions were carefully placed and simplified.

Breaking down the operations step-by-step according to the rule ensures that all components are accounted for and greatly minimizes errors.

Trigonometric Functions

Trigonometric functions such as sine and cosine are fundamental in understanding relationships involving angles and distances. In calculus, they frequently appear due to their cyclical nature and their derivatives, which are themselves trigonometric functions.

• For cosine functions, the derivative is: \[ \frac{d}{du}(\cos u) = -\sin u \]
• For sine functions, the derivative is: \[ \frac{d}{dv}(\sin v) = \cos v \]
• In our given problem, \(x=2\cos u\) and \(y=3\sin v\) use these trigonometric functions to express values for \(x\) and \(y\).
• Knowing these derivatives is key when applying the quotient rule to find partial derivatives, as it directly affects how the function is simplified.

The interplay between trigonometric functions and their derivatives is what makes them versatile and indispensable tools in calculus.

Chain Rule

The chain rule is another powerful tool in calculus, especially when dealing with composite functions. It allows us to differentiate a function concerning an intermediate variable through another function, which is particularly useful when multiple variables are involved.

• The chain rule states that if a variable \( z \) depends on \( y \), which in turn depends on \( x \), then the derivative of \( z \) with respect to \( x \) is: \[ \frac{dz}{dx} = \frac{dz}{dy} \cdot \frac{dy}{dx} \]
• This rule was applied implicitly in the process of solving the original exercise, although not explicitly mentioned. The differentiation of cyclic x and y functions like \(x=2\cos u\) and \(y=3\sin v\) partially attribute to chain reactions between trigonometric derivatives and the function of interest.
• In multi-variable calculus contexts, like partial derivatives, observing how changes in one variable impact another through their compounded effect is essential.
• A proper understanding and application of the chain rule enable us to correctly manipulate and simplify functions to extract derivative expressions.

Remember, practice in using the chain rule helps ensure smooth handling of complex equations involving multiple layers of functions.