Problem 4
Question
Let \(g(x, y, z)=\sqrt{x \cos y}+z^{2}\). Find each value. (a) \(g(4,0,2)\) (b) \(g(-9, \pi, 3)\) (c) \(g(2, \pi / 3,-1)\) (d) \(g(3,6,1.2)\)
Step-by-Step Solution
Verified Answer
(a) 6, (b) 12, (c) 2, (d) approximately 3.137
1Step 1: Substitute and Solve for g(4,0,2)
Substitute the values into the function: \(g(4,0,2) = \sqrt{4 \cos 0} + 2^2\).Since \(\cos 0 = 1\), the equation becomes \(\sqrt{4 \times 1} + 4\).Thus, we have \(\sqrt{4} + 4 = 2 + 4 = 6\).
2Step 2: Substitute and Solve for g(-9,π,3)
Substitute the values into the function: \(g(-9, \pi, 3) = \sqrt{-9 \cos \pi} + 3^2\).Since \(\cos \pi = -1\), the equation becomes \(\sqrt{-9 \times -1} + 9\).Thus, we have \(\sqrt{9} + 9 = 3 + 9 = 12\).
3Step 3: Substitute and Solve for g(2,π/3,-1)
Substitute the values into the function: \(g(2, \pi/3, -1) = \sqrt{2 \cos(\pi/3)} + (-1)^2\).Since \(\cos(\pi/3) = 1/2\), the equation becomes \(\sqrt{2 \times 1/2} + 1\).Thus, we have \(\sqrt{1} + 1 = 1 + 1 = 2\).
4Step 4: Substitute and Solve for g(3,6,1.2)
Substitute the values into the function: \(g(3,6,1.2) = \sqrt{3 \cos 6} + (1.2)^2\).Using the value of \(\cos 6\) (which is approximately 0.960), compute: \(\sqrt{3 \times 0.960} + 1.44\).Thus, we have \(\sqrt{2.88} + 1.44\), and finding the square root gives approximately \(1.697\).Add \(1.44\) to get approximately \(3.137\).
Key Concepts
Function EvaluationTrigonometric FunctionsPartial Derivatives
Function Evaluation
Function evaluation involves finding the value of a function when particular numbers are substituted for its variables.
In multivariable calculus, functions can depend on more than one variable, like the function given: \[ g(x, y, z) = \sqrt{x \cos y} + z^2 \]This function depends on three variables: \(x\), \(y\), and \(z\). To evaluate this function at a specific point, like \((4, 0, 2)\), substitute these numbers into the function:
Function evaluation is an important skill in calculus, allowing one to interpret functions graphically and analytically by determining how changes in variables affect the function's value.
In multivariable calculus, functions can depend on more than one variable, like the function given: \[ g(x, y, z) = \sqrt{x \cos y} + z^2 \]This function depends on three variables: \(x\), \(y\), and \(z\). To evaluate this function at a specific point, like \((4, 0, 2)\), substitute these numbers into the function:
- \(x = 4\)
- \(y = 0\)
- \(z = 2\)
Function evaluation is an important skill in calculus, allowing one to interpret functions graphically and analytically by determining how changes in variables affect the function's value.
Trigonometric Functions
Trigonometric functions are fundamental in calculus and help in evaluating expressions involving angles and periodic phenomena.
In our function \(g(x, y, z)\), the \(\cos\) function is used, which measures the cosine of an angle \(y\):\[ \sqrt{x \cos y} \]The cosine function has important properties:
This understanding helps in substituting and solving trigonometric expressions accurately, crucial for precise calculations in calculus.
In our function \(g(x, y, z)\), the \(\cos\) function is used, which measures the cosine of an angle \(y\):\[ \sqrt{x \cos y} \]The cosine function has important properties:
- It oscillates between -1 and 1.
- Common angles such as \(0\), \(\pi\), \(\pi/3\) have known values. For instance, \(\cos 0 = 1\), \(\cos \pi = -1\), and \(\cos \pi/3 = 0.5\).
This understanding helps in substituting and solving trigonometric expressions accurately, crucial for precise calculations in calculus.
Partial Derivatives
Though the exercise involves evaluating a function, understanding partial derivatives is pivotal.
Partial derivatives are used when differentiating functions of multiple variables and signify the rate of change of the function concerning one variable, while keeping others constant.
For a function \( g(x, y, z) \), the partial derivative with respect to \(x\), denoted \( \frac{\partial g}{\partial x} \), involves differentiating while treating \(y\) and \(z\) as constants. For example:\[ \frac{\partial}{\partial x} ( \sqrt{x \cos y} + z^2 ) \]
Would focus on the first part:
Partial derivatives are used when differentiating functions of multiple variables and signify the rate of change of the function concerning one variable, while keeping others constant.
For a function \( g(x, y, z) \), the partial derivative with respect to \(x\), denoted \( \frac{\partial g}{\partial x} \), involves differentiating while treating \(y\) and \(z\) as constants. For example:\[ \frac{\partial}{\partial x} ( \sqrt{x \cos y} + z^2 ) \]
Would focus on the first part:
- The derivative of \(\sqrt{x\cos y}\) with respect to \(x\), applying the chain rule, considering \(y\) constant.
Other exercises in this chapter
Problem 4
In Problems 1-8, find the directional derivative of \(f\) at the point \(\mathbf{p}\) in the direction of \(\mathbf{a}\). \(f(x, y)=x^{2}-3 x y+2 y^{2} ; \mathb
View solution Problem 4
In Problems 1-6, find dw/dt by using the Chain Rule. Express your final answer in terms of \(t\). $$ w=\ln (x / y) ; x=\tan t, y=\sec ^{2} t $$
View solution Problem 4
$$ f(x, y)=x y^{2}-6 x^{2}-3 y^{2} $$
View solution Problem 4
\(\lim _{(x, y) \rightarrow(1,2)} \frac{x^{3}-3 x^{2} y+3 x y^{2}-y^{3}}{y-2 x^{2}}\)
View solution