Problem 1
Question
Find the linearization \(L(x, y)\) of the function \(f(x, y)=\sqrt{161-9 x^{2}-4 y^{2}}\) at (-4,-2) \(L(x, y)=\) ____________. Note: Your answer should be an expression in \(x\) and \(y ;\) e.g. " \(3 x-5 y\) \(+9 "\)
Step-by-Step Solution
Verified Answer
\(L(x, y) =5-4(x+4)-2(y+2)\)
1Step 1: Calculate Partial Derivatives
We need to find the partial derivatives of the function with respect to both \(x\) and \(y\).
The partial derivative with respect to \(x\) is:
\[
\frac{\partial f(x, y)}{\partial x}=\frac{-18x}{2\sqrt{161-9x^{2}-4y^{2}}}
\]
And the partial derivative with respect to \(y\) is:
\[
\frac{\partial f(x, y)}{\partial y}=\frac{-8y}{2\sqrt{161-9x^{2}-4y^{2}}}
\]
2Step 2: Evaluate Partial Derivatives at Given Point
Next, plug in the point (-4, -2) into the partial derivatives to find the slope of the tangent line at this point.
For \(\frac{\partial f}{\partial x}\), we get
\[
\frac{\partial f(-4, -2)}{\partial x}=\frac{-18*(-4)}{2\sqrt{161-9*(-4)^{2}-4*(-2)^{2}}} = -4
\]
For \(\frac{\partial f}{\partial y}\), we get
\[
\frac{\partial f(-4, -2)}{\partial y}=\frac{-8*(-2)}{2\sqrt{161-9*(-4)^{2}-4*(-2)^{2}}} = -2
\]
3Step 3: Find the Value of the Function at the Given Point
Before we can form the linear approximation, we also need to find the value of the function at the point (-4, -2). Plugging in the values, we get
\[
f(-4, -2) = \sqrt{161-9*(-4)^2-4*(-2)^2} = 5
\]
4Step 4: Formulate The Linear Approximation
Now that we have the values of the partial derivatives and the function at the given point, we can formulate the linear approximation, \(L(x, y)\).
The formula for the linear approximation in two variables is:
\[
L(x, y)=f(a, b)+\frac{\partial f(a, b)}{\partial x}(x-a)+\frac{\partial f(a, b)}{\partial y}(y-b)
\]
Substituting the values we have \(a = -4\), \(b = -2\), \(\frac{\partial f(-4, -2)}{\partial x} = -4\), \(\frac{\partial f(-4, -2)}{\partial y} = -2\), and \(f(-4,-2) = 5\), we get:
\[
L(x,y)=5-4(x+4)-2(y+2)
\]
Thus, the linearization of the function \(f(x, y)=\sqrt{161-9 x^{2}-4 y^{2}}\) at (-4, -2) is \(L(x, y) =5-4(x+4)-2(y+2)\).
Key Concepts
Partial DerivativesLinearizationTangent Plane ApproximationGradient Vector
Partial Derivatives
In multivariable calculus, understanding partial derivatives is crucial as they extend the concept of derivatives to functions of several variables. A partial derivative represents how a function changes as one particular variable changes while keeping other variables constant.
For a function like \( f(x, y) \), the partial derivative with respect to \( x \) is denoted as \( \frac{\partial f}{\partial x} \), while the partial derivative with respect to \( y \) is \( \frac{\partial f}{\partial y} \). This means:
In the given problem, we calculate these derivatives for the function \( f(x, y)=\sqrt{161-9x^{2}-4y^{2}} \), giving insight into the rate of change along the \( x \)-axis and the \( y \)-axis.
For a function like \( f(x, y) \), the partial derivative with respect to \( x \) is denoted as \( \frac{\partial f}{\partial x} \), while the partial derivative with respect to \( y \) is \( \frac{\partial f}{\partial y} \). This means:
- \( \frac{\partial f}{\partial x} \): Change in \( f \) as \( x \) changes, with \( y \) constant.
- \( \frac{\partial f}{\partial y} \): Change in \( f \) as \( y \) changes, with \( x \) constant.
In the given problem, we calculate these derivatives for the function \( f(x, y)=\sqrt{161-9x^{2}-4y^{2}} \), giving insight into the rate of change along the \( x \)-axis and the \( y \)-axis.
Linearization
Linearization is a powerful technique to approximate complex functions using simpler linear functions.
It provides a way to estimate the behavior of a function around a given point by fitting a tangent plane. At a particular point \((a, b)\), the linearization of a function \( f(x, y) \) is given by the formula:\[l(x, y) = f(a, b) + \frac{\partial f}{\partial x}(x - a) + \frac{\partial f}{\partial y}(y - b)\]
This formula allows us to take advantage of both the partial derivatives and the value of function at the specific point.
Linearization is particularly useful when you need an easy computation method for evaluating complex functions at points near \((a, b)\). In this context, the calculated linearization \( L(x, y) = 5 - 4(x + 4) - 2(y + 2) \) shows how the function \( f(x, y) \) behaves closely around the point (-4, -2). The terms involving \( x \) and \( y \) describe the respective slope or change for the \( x \) and \( y \)-directions.
It provides a way to estimate the behavior of a function around a given point by fitting a tangent plane. At a particular point \((a, b)\), the linearization of a function \( f(x, y) \) is given by the formula:\[l(x, y) = f(a, b) + \frac{\partial f}{\partial x}(x - a) + \frac{\partial f}{\partial y}(y - b)\]
This formula allows us to take advantage of both the partial derivatives and the value of function at the specific point.
Linearization is particularly useful when you need an easy computation method for evaluating complex functions at points near \((a, b)\). In this context, the calculated linearization \( L(x, y) = 5 - 4(x + 4) - 2(y + 2) \) shows how the function \( f(x, y) \) behaves closely around the point (-4, -2). The terms involving \( x \) and \( y \) describe the respective slope or change for the \( x \) and \( y \)-directions.
Tangent Plane Approximation
The tangent plane approximation in multivariable calculus is an extension of the tangent line concept from single-variable calculus. It provides the best linear approximator of a surface at a point. When a surface described by a function \( f(x, y) \) is interested at a point \((a, b)\), the tangent plane makes calculations easier by representing the surface as a plane in the neighborhood of that point.
The function's graph is approximated using the tangent plane, whose equation is based on the value of the function and its partial derivatives at \((a, b)\):\[L(x, y) = f(a, b) + \frac{\partial f}{\partial x}(x - a) + \frac{\partial f}{\partial y}(y - b)\]
The function's graph is approximated using the tangent plane, whose equation is based on the value of the function and its partial derivatives at \((a, b)\):\[L(x, y) = f(a, b) + \frac{\partial f}{\partial x}(x - a) + \frac{\partial f}{\partial y}(y - b)\]
- The term \( f(a, b) \) provides the constant surface height at the point.
- \( \frac{\partial f}{\partial x}(x - a) \): Represents how steep the slope is in the x-direction.
- \( \frac{\partial f}{\partial y}(y - b) \): Represents how steep the slope is in the y-direction.
Gradient Vector
The gradient vector is a vector that encapsulates the partial derivatives of a multivariable function and points in the direction of the greatest rate of increase of the function. In two dimensions, for a function \( f(x, y) \), the gradient vector is represented as:\[abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)\]
The magnitude of the gradient vector represents the rate of the steepest ascent. In the linearization process, understanding the gradient allows to align the approximation with the actual function's behavior.
In the problem, values \(-4\) and \(-2\) represent the slopes computed by the gradient at the point \((-4,-2)\), forming the core of the linearized function. This helps to highlight how the original function aligns with its approximation at the given point.
- The \( \frac{\partial f}{\partial x} \) component indicates the rate of change along the \( x \)-axis.
- The \( \frac{\partial f}{\partial y} \) component indicates the rate of change along the \( y \)-axis.
The magnitude of the gradient vector represents the rate of the steepest ascent. In the linearization process, understanding the gradient allows to align the approximation with the actual function's behavior.
In the problem, values \(-4\) and \(-2\) represent the slopes computed by the gradient at the point \((-4,-2)\), forming the core of the linearized function. This helps to highlight how the original function aligns with its approximation at the given point.
Other exercises in this chapter
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