Problem 22 - Calculus with Concepts in Calculus

Question

Find the first partial derivatives of \(f\) at the given point. $$ f(x, y, z)=x y^{2} \sin z ;(-1,2,0) $$

Step-by-Step Solution

Answer

\(\frac{\partial f}{\partial x} = 0, \frac{\partial f}{\partial y} = 0, \frac{\partial f}{\partial z} = -4\) at \((-1, 2, 0)\).

Step 1: Understanding the Function

We are given the function \(f(x, y, z) = x y^2 \sin z\). We need to find the first partial derivatives of this function with respect to \(x\), \(y\), and \(z\) at the point \((-1, 2, 0)\).

Step 2: Partial Derivative with respect to x

To find the partial derivative of \(f\) with respect to \(x\), we treat \(y\) and \(z\) as constants. The derivative of \(xf\) becomes \(y^2 \sin z\). Thus, \(\frac{\partial f}{\partial x} = y^2 \sin z\).

Step 3: Calculate \(\frac{\partial f}{\partial x}\) at \((-1, 2, 0)\)

Substitute \(y = 2\) and \(z = 0\) into the derivative \(\frac{\partial f}{\partial x} = y^2 \sin z\). We get \(2^2 \sin(0) = 4\cdot0 = 0\).

Step 4: Partial Derivative with respect to y

For the partial derivative with respect to \(y\), treat \(x\) and \(z\) as constants. Differentiate \(x y^2 \sin z\) with respect to \(y\) to get \(2xy \sin z\). So, \(\frac{\partial f}{\partial y} = 2xy \sin z\).

Step 5: Calculate \(\frac{\partial f}{\partial y}\) at \((-1, 2, 0)\)

Substitute \(x = -1\), \(y = 2\), and \(z = 0\) into the derivative \(\frac{\partial f}{\partial y} = 2xy \sin z\). We get \(2(-1)(2)\sin(0) = -4 \cdot 0 = 0\).

Step 6: Partial Derivative with respect to z

For the partial derivative with respect to \(z\), treat \(x\) and \(y\) as constants. The derivative of \(x y^2 \sin z\) with respect to \(z\) is \(x y^2 \cos z\). So, \(\frac{\partial f}{\partial z} = xy^2 \cos z\).

Step 7: Calculate \(\frac{\partial f}{\partial z}\) at \((-1, 2, 0)\)

Substitute \(x = -1\), \(y = 2\), and \(z = 0\) into the derivative \(\frac{\partial f}{\partial z} = xy^2 \cos z\). We find \((-1)\cdot (2)^2 \cdot \cos(0) = -4\cdot 1 = -4\).

Key Concepts

Multivariable Calculus

Multivariable calculus is a branch of calculus that deals with functions of multiple variables. Unlike single-variable calculus, where we deal with functions of one variable, multivariable calculus considers functions that have two or more variables like the one in our exercise: \(f(x, y, z) = x y^2 \sin z\). In this context, we explore changes in functions with respect to changes in each of their variables.

Key concepts in multivariable calculus include:

• **Partial Derivatives**: These represent the rate of change of the function with respect to one variable, while keeping the other variables constant.
• **Gradient**: This is a vector that points in the direction of the greatest rate of increase of the function, consisting of all its partial derivatives.

Understanding how to compute and interpret partial derivatives is crucial in fields ranging from physics to economics, where we often need to understand how changes in multiple factors affect a system.

Chain Rule

The chain rule is a fundamental concept in calculus used to differentiate compositions of functions. When dealing with multivariable functions, the chain rule becomes essential because variables often depend on each other, and a change in one can affect the others.

For instance, if we had a scenario where \(z\) depended on \(t\), and we needed to find how \(f\) changes with a change in \(t\), we would use the chain rule. It allows us to break down complex relationships and find derivatives in steps.

The chain rule in multivariable calculus looks like this: if \(z = g(x, y)\) and \(y = h(t)\), then the derivative of \(z\) with respect to \(t\) is:

\[ \frac{dz}{dt} = \frac{\partial z}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial z}{\partial y} \cdot \frac{dy}{dt} \]

This rule helps simplify the process of dealing with derivative calculations involving multiple related variables.

Sin and Cos Functions

The sine (\(\sin\)) and cosine (\(\cos\)) functions are fundamental elements of trigonometry, and they appear frequently in calculus. In the exercise, we specifically saw how these functions interact with derivatives.

The derivatives of sine and cosine functions are particularly important:

• The derivative of \(\sin z\) with respect to \(z\) is \(\cos z\).
• The derivative of \(\cos z\) with respect to \(z\) is \(-\sin z\).

When working with partial derivatives in multivariable calculus, recognizing these derivatives is essential.

In our problem, this knowledge helped us find the partial derivative of \(f = x y^2 \sin z\) with respect to \(z\), which resulted in \(xy^2 \cos z\). Remembering these simple trigonometric derivatives can simplify many calculation processes, particularly in physics and engineering contexts, where periodic functions are common.