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Chapter 16: Problem 22

Find the gradient vector field of \(f\) $$f(x, y)=\tan (3 x-4 y)$$

Short Answer

Expert verified
The gradient vector field is \( \left(3\sec^2(3x - 4y), -4\sec^2(3x - 4y)\right) \).

Step by step solution

01

Identify the Function

The given function is \( f(x, y) = \tan(3x - 4y) \). This function consists of the tangent of a linear expression in terms of \(x\) and \(y\).
02

Differentiate with Respect to x

To find the partial derivative of \( f \) with respect to \( x \), we use the chain rule. The derivative of \( \tan(u) \) with respect to \( u \) is \( \sec^2(u) \). Therefore, \( f_x = \frac{d}{dx}[\tan(3x - 4y)] = 3\sec^2(3x - 4y) \).
03

Differentiate with Respect to y

Now, find the partial derivative of \( f \) with respect to \( y \). Again using the chain rule, the derivative inside is with respect to \(-4y\), which gives us \( f_y = \frac{d}{dy}[\tan(3x - 4y)] = -4\sec^2(3x - 4y) \).
04

Form the Gradient Vector Field

The gradient vector field is composed of the partial derivatives of the function with respect to \( x \) and \( y \). Thus, \( abla f = (f_x, f_y) = \left(3\sec^2(3x - 4y), -4\sec^2(3x - 4y)\right) \).

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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 that involve taking the derivative of a function with respect to one variable while keeping the other variables constant. This is particularly useful in functions of several variables, such as functions defined on \(x\) and \(y\).
  • In the given example, the partial derivatives \(f_x\) and \(f_y\) of \(f(x, y) = \tan(3x - 4y)\) are calculated with respect to \(x\) and \(y\) separately.
  • To find \(f_x\), you treat \(y\) as a constant and differentiate with respect to \(x\).
  • To find \(f_y\), you treat \(x\) as a constant and differentiate with respect to \(y\).
This process highlights how each variable influences the function individually. Partial derivatives are an essential tool for constructing the gradient vector, an important element when analyzing the properties of a multivariable function.
Chain Rule
The chain rule is a powerful technique used in calculus to differentiate composite functions. It helps determine the derivative of a function based on its nested structure.
  • In our example, \(f(x, y) = \tan(3x - 4y)\), we recognize that the tangent function is nested within another linear function, \(3x - 4y\).
  • To apply the chain rule, we differentiate the outer function, \(\tan(u)\), where \(u = 3x - 4y\), and then multiply by the derivative of the inner function.
  • This gives us \(f_x = 3\sec^{2}(3x - 4y)\) and \(f_y = -4\sec^{2}(3x - 4y)\).
The chain rule is essential when dealing with more complex functions, especially those that combine multiple operations into a single function.
Tangent Function
The tangent function \(\tan(\theta)\) is one of the basic trigonometric functions that arises frequently in calculus, particularly in analyzing periodic phenomena.
  • In the function \(f(x, y) = \tan(3x - 4y)\), \(\tan(\theta)\) helps describe how rapidly the function changes.
  • The derivative of \(\tan(u)\) is \(\sec^2(u)\), a fact that is utilized when applying the chain rule.
Understanding the behavior of the tangent function is crucial in this calculation, as it gives insight into how the slope of the function changes with \(x\) and \(y\). Asymptotic behavior and periodicity are key characteristics of tangent functions, influencing the overall behavior of \(f\).
Calculus
Calculus is a branch of mathematics focused on the study of change and motion, using derivatives and integrals as its primary tools.
  • It allows us to understand and describe continuous change in a rigorous way.
  • Tools like partial derivatives and the chain rule are cornerstones of calculus, enabling the study of functions with more than one variable.
In this exercise, calculus provides the means to create a gradient vector, a multidimensional derivative that helps find the direction and rate of fastest increase of the function \(f(x, y)\).
Understanding calculus is foundational for tackling complex real-world problems involving rates of change in multiple dimensions, making it a vital part of advanced mathematics and various scientific fields.

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Most popular questions from this chapter

Find the mass and center of mass of a wire in the shape of the helix \(x=t, y=\cos t, z=\sin t, 0 \leqslant t \leqslant 2 \pi,\) if the density at any point is equal to the square of the distance from the origin.

(a) Show that a constant force field does zero work on a particle that moves once uniformly around the circle \(x^{2}+y^{2}=1\) (b) Is this also true for a force field \(\mathbf{F}(\mathbf{x})=k \mathbf{x},\) where \(k\) is a constant and \(\mathbf{x}=\langle x, y\rangle ?\)

(a) Evaluate the line integral \(\int_{C} \mathbf{F} \cdot d \mathbf{r},\) where \(\mathbf{F}(x, y, z)=x \mathbf{i}-z \mathbf{j}+y \mathbf{k}\) and \(C\) is given by \(\mathbf{r}(t)=2 t \mathbf{i}+3 t \mathbf{j}-t^{2} \mathbf{k},-1 \leqslant t \leqslant 1 .\) (b) Illustrate part (a) by using a computer to graph \(C\) and the vectors from the vector field corresponding to \(t=\pm 1\) and \(\pm \frac{1}{2}(\) as in Figure 13\() .\)

Prove each identity, assuming that \(S\) and \(E\) satisfy the conditions of the Divergence Theorem and the scalar functions and components of the vector fields have continuous second-order partial derivatives. $$ \iint_{S} D_{n} f d S=\iint_{E} \nabla^{2} f d V $$

\(5-18\) Evaluate the surface integral. $$\iint_{S}\left(z+x^{2} y\right) d S$$ \(S\) is the part of the cylinder \(y^{2}+z^{2}=1\) that lies between the planes \(x=0\) and \(x=3\) in the first octant
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