Problem 18
Question
\(f(x, y)=e^{x} \cos y+e^{y} \sin x ; P(1,0), Q(-3,3)\)
Step-by-Step Solution
Verified Answer
For P(1,0): \( e + \sin(1) \); for Q(-3,3): \( e^{-3} \cos(3) - e^{3} \sin(3) \)
1Step 1 - Evaluate the function at point P(1,0)
Substitute the coordinates of point P into the function: \[ f(1, 0) = e^{1} \cos(0) + e^{0} \sin(1) \] We know that \( e^1 = e \), \( \cos(0) = 1 \), and \( e^0 = 1 \). Also, \( \sin(1) \) remains as it is. Therefore, \[ f(1, 0) = e \cdot 1 + 1 \cdot \sin(1) = e + \sin(1) \]
2Step 2 - Evaluate the function at point Q(-3,3)
Substitute the coordinates of point Q into the function: \[ f(-3, 3) = e^{-3} \cos(3) + e^{3} \sin(-3) \] We can simplify the terms: \( \sin(-3) = -\sin(3) \). Therefore, \[ f(-3, 3) = e^{-3} \cos(3) + e^{3} ( -\sin(3) ) \] \[ f(-3, 3) = e^{-3} \cos(3) - e^{3} \sin(3) \]
3Step 3 - Simplify the expressions
Combine the results from Steps 1 and 2 for better visibility: For point P(1,0): \[ f(1,0) = e + \sin(1) \] For point Q(-3,3): \[ f(-3,3) = e^{-3} \cos(3) - e^{3} \sin(3) \]
Key Concepts
Function EvaluationExponential FunctionsTrigonometric Functions
Function Evaluation
Function evaluation is the process of finding the value of a function at specific points. In the given exercise, the function is evaluated at two points: P(1,0) and Q(-3,3). To evaluate a function, we simply substitute the given coordinates into the function's formula.
For example, with the function \( f(x, y) = e^{x} \cos(y) + e^{y} \sin(x) \) and point P(1,0), we substitute 1 for x and 0 for y. This gives us:
\[ f(1, 0) = e^{1} \cos(0) + e^{0} \sin(1) \]
Each term is then simplified based on known values of exponential and trigonometric functions, such as \( e^1 = e \), \( \cos(0) = 1 \), and \( e^0 = 1 \).
Finally, we sum up the simplified terms to get the function's value at that point. Making the substitution process clearer and ensuring the correct application of function properties is crucial for accurate function evaluation.
For example, with the function \( f(x, y) = e^{x} \cos(y) + e^{y} \sin(x) \) and point P(1,0), we substitute 1 for x and 0 for y. This gives us:
\[ f(1, 0) = e^{1} \cos(0) + e^{0} \sin(1) \]
Each term is then simplified based on known values of exponential and trigonometric functions, such as \( e^1 = e \), \( \cos(0) = 1 \), and \( e^0 = 1 \).
Finally, we sum up the simplified terms to get the function's value at that point. Making the substitution process clearer and ensuring the correct application of function properties is crucial for accurate function evaluation.
Exponential Functions
Exponential functions are mathematical functions of the form \( e^{x} \), where \( e \) is the base of natural logarithms, approximately equal to 2.71828. These functions grow rapidly as the exponent increases and are crucial in many fields such as calculus, physics, and finance.
In the given function \( f(x, y) = e^{x} \cos(y) + e^{y} \sin(x) \), the terms \( e^{x} \) and \( e^{y} \) are exponential functions. At point P(1,0), \( e^1 \) simplifies to \( e \), and \( e^0 \) simplifies to 1.
At point Q(-3,3), the terms \( e^{-3} \) and \( e^3 \) are used. An important property of exponential functions is that \( e^{-a} \) is the reciprocal of \( e^{a} \), meaning that \( e^{-3} = \frac{1}{e^3} \). Understanding these properties helps in simplifying and evaluating the function correctly.
In the given function \( f(x, y) = e^{x} \cos(y) + e^{y} \sin(x) \), the terms \( e^{x} \) and \( e^{y} \) are exponential functions. At point P(1,0), \( e^1 \) simplifies to \( e \), and \( e^0 \) simplifies to 1.
At point Q(-3,3), the terms \( e^{-3} \) and \( e^3 \) are used. An important property of exponential functions is that \( e^{-a} \) is the reciprocal of \( e^{a} \), meaning that \( e^{-3} = \frac{1}{e^3} \). Understanding these properties helps in simplifying and evaluating the function correctly.
Trigonometric Functions
Trigonometric functions like \( \sin \) and \( \cos \) are vital in various fields including mathematics, physics, and engineering due to their periodic nature. In this exercise, we encounter \( \cos(0) \), \( \cos(3) \), \( \sin(1) \), and \( \sin(-3) \). Knowing the fundamental values of these functions can help simplify the evaluation process.
For instance, \( \cos(0) = 1 \), and the sine and cosine of negative angles relate to those of positive angles: \( \sin(-x) = -\sin(x) \). Therefore, \( \sin(-3) = -\sin(3) \).
When evaluating at point Q(-3,3), substituting these values into the function gives us:
\[ f(-3, 3) = e^{-3} \cos(3) + e^{3} \sin(-3) \]
This simplifies to:
\[ f(-3, 3) = e^{-3} \cos(3) - e^{3} \sin(3) \]
Mastery of these trigonometric properties and their values greatly aids in working with complex functions involving multiple variables.
For instance, \( \cos(0) = 1 \), and the sine and cosine of negative angles relate to those of positive angles: \( \sin(-x) = -\sin(x) \). Therefore, \( \sin(-3) = -\sin(3) \).
When evaluating at point Q(-3,3), substituting these values into the function gives us:
\[ f(-3, 3) = e^{-3} \cos(3) + e^{3} \sin(-3) \]
This simplifies to:
\[ f(-3, 3) = e^{-3} \cos(3) - e^{3} \sin(3) \]
Mastery of these trigonometric properties and their values greatly aids in working with complex functions involving multiple variables.
Other exercises in this chapter
Problem 18
Prove that the box having the largest volume that can be placed inside a sphere is in the shape of a cube.
View solution Problem 18
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\(\int_{C}(2 x \cos y-3) d x-\left(x^{2} \sin y+z^{2}\right) d y-(2 y z-2) d z ; A\) is \((-1,0,3)\) and \(B\) is \((1, \pi, 0) ;\) Exercise 9
View solution Problem 19
In the following exercises, determine if the vector is a gradient. If it is, find a function having the given gradient \(z \tan y \mathbf{i}+x z \sec ^{2} y \ma
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