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Chapter 9: Problem 18

Find \(\nabla f\), Graph some level curves \(f=\) const. Indicute \(\nabla f\) by arrows at some points of these curves. $$T=\sin x \cosh y, P:\left(\frac{1}{4} \pi, \ln 5\right)$$

Short Answer

Expert verified
\( \nabla f = 1.3 \sqrt{2} \hat{i} + 1.2 \sqrt{2} \hat{j} \), draw level curves with arrows.

Step by step solution

01

Define the Function

We have the function given as \( T = \sin x \cosh y \). To find the gradient, we must first identify the expression clearly. Here, \( T \) is a function of two variables \( x \) and \( y \).
02

Compute Partial Derivatives

To find \( abla f \), we need the partial derivatives of \( T \). The partial derivative with respect to \( x \) is \( \frac{\partial T}{\partial x} = \cos x \cosh y \), and with respect to \( y \) is \( \frac{\partial T}{\partial y} = \sin x \sinh y \).
03

Construct the Gradient Vector

The gradient \( abla f \) is a vector composed of the partial derivatives. Thus, \( abla f = \left( \cos x \cosh y, \sin x \sinh y \right) \).
04

Evaluate Gradient at Point P

Substitute \( x = \frac{1}{4} \pi \) and \( y = \ln 5 \) into the gradient. \( \cos \left(\frac{1}{4} \pi\right) = \frac{\sqrt{2}}{2} \), \( \sin \left(\frac{1}{4} \pi\right) = \frac{\sqrt{2}}{2} \), \( \cosh (\ln 5) = \frac{5 + \frac{1}{5}}{2} = \frac{26}{10} = 2.6 \), and \( \sinh (\ln 5) = \frac{5 - \frac{1}{5}}{2} = 2.4 \). The gradient vector at point \( P \) is \( \left( \frac{\sqrt{2}}{2} \times 2.6, \frac{\sqrt{2}}{2} \times 2.4 \right) = \left( 1.3 \sqrt{2}, 1.2 \sqrt{2} \right) \).
05

Graph Level Curves and Indicate Gradient

On the graph, draw several level curves of the form \( T = c \) where \( c \) is a constant value. At point \( P \), draw an arrow representing the gradient \( abla f \). This arrow should be perpendicular to the level curve passing through \( P \) and point in the direction of greatest increase of the function.

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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 allow us to understand how a function changes as we vary one of its input variables, keeping others constant. In our exercise, we examined the function \( T = \sin x \cosh y \), which depends on two variables, \( x \) and \( y \).
  • To find the partial derivative with respect to \( x \), treat \( y \) as a constant. Thus, the derivative of \( T \) with respect to \( x \) is \( \frac{\partial T}{\partial x} = \cos x \cosh y \).
  • Similarly, the partial derivative with respect to \( y \) is found by treating \( x \) as constant: \( \frac{\partial T}{\partial y} = \sin x \sinh y \).
These derivatives tell us how the function \( T \) changes as \( x \) and \( y \) are varied individually. Finding these derivatives is the first step in determining the gradient vector of a function.
Level Curves
Level curves are an essential concept when dealing with functions of two variables. They are the curves along which the function has a constant value. In essence, these curves are the set of all points \( (x, y) \) for which the function \( T \) equals a specific constant \( c \). For the function \( T = \sin x \cosh y \):
  • A level curve might be represented as \( \sin x \cosh y = c \), where \( c \) is some constant value.
  • These curves help visualize the topography of the function, similar to contour lines on a map that show areas of equal elevation.
Visualizing level curves is useful because they illustrate how a function behaves over a plane. In our exercise, plotting these curves helps us understand where and how the function remains unchanged and how its values increase or decrease.
Multivariable Calculus
Multivariable Calculus extends the concepts of single-variable calculus to functions of several variables. It introduces tools like gradients and level curves, which are fundamental in fields like physics and engineering.
  • In multivariable calculus, the gradient vector \( abla f \) is formed by the partial derivatives of a function. For \( T = \sin x \cosh y \), the gradient is \( abla T = (\cos x \cosh y, \sin x \sinh y) \).
  • This gradient points in the direction of the greatest rate of increase of the function and is perpendicular to the level curves.
Understanding multivariable calculus is crucial for optimizing functions and analyzing dynamic systems that depend on more than one variable. It provides a way to model and solve real-world problems that involve multiple changing quantities.

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