Chapter 14: Problem 10
find \(\nabla f\) at the given point. \(\begin{array}{c}f(x, y, z)=e^{x+y} \cos z+(y+1) \sin ^{-1} x, \quad(0,0, \pi / 6)\end{array}\)
Short Answer
Expert verified
\( \nabla f(0,0, \frac{\pi}{6}) = \left( \frac{\sqrt{3}}{2} + 1, \frac{\sqrt{3}}{2}, -\frac{1}{2} \right) \).
Step by step solution
01
Understand the Gradient
The gradient of a function, denoted as \( abla f \), is a vector of partial derivatives of the function with respect to each independent variable. For a function \( f(x, y, z) \), the gradient is \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \).
02
Find the Partial Derivative with respect to x
Calculate \( \frac{\partial f}{\partial x} \) for \( f(x, y, z) = e^{x+y} \cos z + (y+1) \sin^{-1} x \). The derivative of \( e^{x+y} \cos z \) with respect to \( x \) is \( e^{x+y} \cos z \) since \( e^{u} \)'s derivative with \( u \) is \( e^u \), and \( e^{x+y} \) effectively acts as a constant multiplied by \( \cos z \). The derivative of \( (y+1) \sin^{-1} x \) with respect to \( x \) is \( (y+1) \cdot \frac{1}{\sqrt{1-x^2}} \). Thus, \( \frac{\partial f}{\partial x} = e^{x+y} \cos z + \frac{y+1}{\sqrt{1-x^2}} \).
03
Find the Partial Derivative with respect to y
Calculate \( \frac{\partial f}{\partial y} \). The derivative of \( e^{x+y} \cos z \) with respect to \( y \) is \( e^{x+y} \cos z \) (since it's similar to previously with respect to \( x \)). The derivative of \( (y+1) \sin^{-1} x \) with respect to \( y \) is \( \sin^{-1} x \) as it's linear in \( y \). Thus, \( \frac{\partial f}{\partial y} = e^{x+y} \cos z + \sin^{-1} x \).
04
Find the Partial Derivative with respect to z
Calculate \( \frac{\partial f}{\partial z} \). The derivative of \( e^{x+y} \cos z \) with respect to \( z \) is \( -e^{x+y} \sin z \) since the derivative of \( \cos z \) is \( -\sin z \). As \( (y+1) \sin^{-1} x \) is independent of \( z \), its derivative with respect to \( z \) is zero. Therefore, \( \frac{\partial f}{\partial z} = -e^{x+y} \sin z \).
05
Evaluate the Gradient at the Given Point
Substitute \( (x, y, z) = (0, 0, \frac{\pi}{6}) \) into the partial derivatives found. This gives:- \( \frac{\partial f}{\partial x} = e^{0+0} \cos \frac{\pi}{6} + \frac{0+1}{\sqrt{1-0^2}} = 1 \cdot \frac{\sqrt{3}}{2} + 1 = \frac{\sqrt{3}}{2} + 1 \).- \( \frac{\partial f}{\partial y} = e^{0+0} \cos \frac{\pi}{6} + \sin^{-1} 0 = \frac{\sqrt{3}}{2} \).- \( \frac{\partial f}{\partial z} = -e^{0+0} \sin \frac{\pi}{6} = -1 \cdot \frac{1}{2} = -\frac{1}{2} \).
06
Express the Gradient Vector
Combine the evaluated partial derivatives into the gradient vector: \( abla f(0,0, \frac{\pi}{6}) = \left( \frac{\sqrt{3}}{2} + 1, \frac{\sqrt{3}}{2}, -\frac{1}{2} \right) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
Partial derivatives are a fundamental concept in multivariable calculus. They help us understand how a function changes when we vary just one of its variables, keeping the others constant. Imagine you're climbing a hill, and you want to know how steep the path is in different directions. Partial derivatives give you the rate of change (or slope) along one direction at a time.
- With respect to \(x\): This tells us how much the function \(f(x, y, z)\) changes as \(x\) changes, while \(y\) and \(z\) are fixed.
- With respect to \(y\): Similarly, it measures the rate of change in \(f\) with \(y\) varied and \(x\), \(z\) held constant.
- With respect to \(z\): This represents the function's change when only \(z\) changes.
Multivariable Calculus
Multivariable calculus is an extension of traditional calculus to functions that have multiple variables. Picture a landscape of hills and valleys instead of a single-line curve. Instead of finding just the slope of a curve, you're deciphering the inclines and declines of a surface.
- Understanding Surfaces: Unlike single-variable calculus, where functions are curves on a plane, multivariable functions are surfaces or even higher-dimensional shapes that depend on two or more variables like \((x, y, z)\).
- Gradient Vectors: These vectors are powerful tools in multivariable calculus. They are composed of all the partial derivatives of a multivariable function. For example, if \(f(x, y, z)\) is a surface, \(abla f\) represents the steepest path 'up the hill'.
Vector Calculus
Vector calculus extends calculus to vector-valued functions, where gradient plays a central role. Picture vectors not just as arrows pointing in a direction but as dynamic quantities describing change. This field examines phenomena where direction and magnitude are involved together.
- Gradient as a Vector: The gradient \(abla f\) is a vector formed by the collection of partial derivatives \((\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z})\). It points in the direction of the greatest rate of increase of the function.
- Properties: The length of the gradient vector indicates the magnitude of the change. The direction reflects the path of maximum ascent on a surface.