Problem 24 - Calculus Early Transcendentals

Question

Find the following derivatives. \(z_{s}\) and \(z_{t},\) where \(z=x y-2 x+3 y, x=\cos s,\) and \(y=\sin t\)

Step-by-Step Solution

Answer

Question: Find the partial derivatives of \(z(x,y) = xy - 2x + 3y\) with respect to \(s\) and \(t\), given that \(x = x(s) = \cos s\) and \(y = y(t) = \sin t\). Answer: The partial derivatives of \(z\) with respect to \(s\) and \(t\) are: \(z_s = -(y - 2)\sin s\) \(z_t = (x + 3)\cos t\)

Step 1: Find the partial derivatives of z with respect to x and y

First, we need to find the partial derivatives \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\). To do this, we can differentiate z with respect to x and y while treating the other variable as constant: \(\frac{\partial z}{\partial x} = \frac{\partial (xy - 2x + 3y)}{\partial x} = y - 2\) \(\frac{\partial z}{\partial y} = \frac{\partial (xy - 2x + 3y)}{\partial y} = x + 3\)

Step 2: Find the derivatives of x and y with respect to s and t

We are given that \(x(s) = \cos s\) and \(y(t) = \sin t\). Now, we can find the derivatives of x with respect to s and y with respect to t: \(\frac{\partial x}{\partial s} = \frac{\partial (\cos s)}{\partial s} = -\sin s\) \(\frac{\partial y}{\partial t} = \frac{\partial (\sin t)}{\partial t} = \cos t\)

Step 3: Apply the chain rule for \(z_s\) and \(z_t\)

Now, we can apply the chain rule for the derivatives of z with respect to s and t: \(z_s = \frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial s}\) \(z_s = (y - 2)(-\sin s) + (x + 3)(0) = -(y - 2)\sin s\) For \(z_t\), note that \(\frac{\partial y}{\partial s} = \frac{\partial (\sin t)}{\partial s} = 0\) and \(\frac{\partial x}{\partial t} = \frac{\partial (\cos s)}{\partial t} = 0\). Now, apply the chain rule: \(z_t = \frac{\partial z}{\partial t} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial t}\) \(z_t = (y - 2)(0) + (x + 3)(\cos t) = (x + 3)\cos t\)

Step 4: Final Solution

So the derivatives of z with respect to s and t are: \(z_s = -(y - 2)\sin s\) \(z_t = (x + 3)\cos t\)

Key Concepts

Chain Rule in Calculus

The chain rule is a fundamental technique used in calculus, particularly when working with composite functions. It helps us differentiate functions that are composed of other functions. Imagine two functions combined, like layers of an onion. The chain rule helps peel back these layers intricately. This is particularly useful in multivariable calculus.
For functions of several variables, when a variable depends indirectly on other variables via a chain of dependencies, the chain rule provides a method to compute the overall derivative efficiently.
In our problem, the function \(z\) depends on \(x\) and \(y\), which in turn depend on \(s\) and \(t\) respectively. The chain rule allows us to compute \(\frac{\partial z}{\partial s}\) and \(\frac{\partial z}{\partial t}\) by combining the individual derivatives:

• \( z_s = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial s} \)
• \( z_t = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial t} \)

By following these steps, we can untangle the dependencies and find the derivative of \(z\) with respect to each parameter.

Understanding Derivatives

Derivatives measure how a function changes as its input changes. Think of a derivative as the slope of a curve, showing how steep a graph is at any point. In multivariable functions, partial derivatives examine how a function changes with respect to one variable while keeping others fixed.
Calculating partial derivatives involves differentiating a function concerning each variable separately. It helps determine how a single input variable influences the whole function. This becomes crucial when dealing with functions of more than one variable, as in our problem, where we have \(z = xy - 2x + 3y\).
In this context:

• \(\frac{\partial z}{\partial x} = y - 2\)
• \(\frac{\partial z}{\partial y} = x + 3\)

These equations tell us how \(z\) changes when \(x\) or \(y\) varies independently. Understanding these basics lays the groundwork for applying the chain rule.

Trigonometric Functions in Calculus

Trigonometric functions like \(\cos\) and \(\sin\) frequently appear in calculus and modeling situations due to their periodic nature and properties. These functions describe repeating patterns and rotations, making them indispensable in various scientific fields.
When differentiating trigonometric functions, we rely on specific derivative rules. For instance, the derivative of \(\cos(s)\) with respect to \(s\) is \(-\sin(s)\). Similarly, the derivative of \(\sin(t)\) with respect to \(t\) is \(\cos(t)\).
These derivatives are pivotal in our problem:

• \(\frac{\partial x}{\partial s} = -\sin s\)
• \(\frac{\partial y}{\partial t} = \cos t\)

With these rules, we can effectively apply the chain rule in multivariable calculus scenarios.