Chapter 12: Problem 11
In Problems \(11-14\), find the gradient vector of the given function at the given point \(\mathbf{p}\). Then find the equation of the tangent plane at \(\mathbf{p}\) (see Example 1). $$ f(x, y)=x^{2} y-x y^{2}, \mathbf{p}=(-2,3) $$
Short Answer
Expert verified
The gradient vector is \((-21, 16)\); the tangent plane is \(z = -21x + 16y - 96\).
Step by step solution
01
Find the Partial Derivatives
To find the gradient vector, first calculate the partial derivatives of the function \( f(x, y) = x^2 y - x y^2 \). The partial derivative with respect to \( x \) is \( \frac{\partial f}{\partial x} = 2xy - y^2 \). The partial derivative with respect to \( y \) is \( \frac{\partial f}{\partial y} = x^2 - 2xy \).
02
Evaluate the Gradient at the Given Point
The given point is \( \mathbf{p} = (-2, 3) \). Evaluate the partial derivatives at this point: \( \frac{\partial f}{\partial x} \) at \((-2, 3)\) is \(2(-2)(3) - (3)^2 = -12 - 9 = -21\). \( \frac{\partial f}{\partial y} \) at \((-2, 3)\) is \((-2)^2 - 2(-2)(3) = 4 + 12 = 16 \). So, the gradient vector at \( \mathbf{p} \) is \( abla f(-2, 3) = (-21, 16) \).
03
Compute the Tangent Plane Equation
The equation of the tangent plane at point \( \mathbf{p} = (x_0, y_0) \) is given by \( z - f(x_0, y_0) = \frac{\partial f}{\partial x}(x - x_0) + \frac{\partial f}{\partial y}(y - y_0) \). First, find \( f(-2, 3) = (-2)^2(3) - (-2)(3)^2 = 12 - 18 = -6 \). Thus, the tangent plane equation is \( z + 6 = -21(x + 2) + 16(y - 3) \).
04
Simplify the Tangent Plane Equation
Simplify the equation: \( z + 6 = -21x - 42 + 16y - 48 \). Combine like terms to get \( z = -21x + 16y - 96 \). This is the equation of the tangent plane.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
When you encounter a multivariable function, like the one in our exercise, understanding how each variable affects the overall function requires calculating partial derivatives. Partial derivatives are like the regular derivatives from single-variable calculus, but they focus on one variable at a time, while keeping the others constant.
For the function given in the exercise, \( f(x, y) = x^2y - xy^2 \), we first need to find the partial derivatives with respect to \( x \) and \( y \).
For the function given in the exercise, \( f(x, y) = x^2y - xy^2 \), we first need to find the partial derivatives with respect to \( x \) and \( y \).
- Partial derivative with respect to \( x \): This measures how the function changes as \( x \) changes, while \( y \) remains fixed. Here, it is calculated as \( \frac{\partial f}{\partial x} = 2xy - y^2 \).
- Partial derivative with respect to \( y \): This tells us how the function changes with respect to \( y \), with \( x \) constant. For this function, it is \( \frac{\partial f}{\partial y} = x^2 - 2xy \).
Tangent Plane
The concept of a tangent plane arises in multivariable calculus when you need to approximate a surface at a certain point with a flat plane. It essentially touches the surface at that point and mirrors its slope and incline as closely as possible.
In our problem, once the gradient vector \( abla f(-2, 3) = (-21, 16) \) is evaluated, we use it to construct the equation of the tangent plane. The formula for the tangent plane to a surface \( z = f(x, y) \) at point \( (x_0, y_0) \) is:
\[ z - f(x_0, y_0) = \frac{\partial f}{\partial x}(x - x_0) + \frac{\partial f}{\partial y}(y - y_0) \]
By substituting the given values and simplifying, we derived the tangent plane's equation as \( z = -21x + 16y - 96 \). It tells us how the plane touches and approximates the surface of the function at point \((-2, 3)\). This plane can be used for linear approximations and predictions close to that point.
In our problem, once the gradient vector \( abla f(-2, 3) = (-21, 16) \) is evaluated, we use it to construct the equation of the tangent plane. The formula for the tangent plane to a surface \( z = f(x, y) \) at point \( (x_0, y_0) \) is:
\[ z - f(x_0, y_0) = \frac{\partial f}{\partial x}(x - x_0) + \frac{\partial f}{\partial y}(y - y_0) \]
By substituting the given values and simplifying, we derived the tangent plane's equation as \( z = -21x + 16y - 96 \). It tells us how the plane touches and approximates the surface of the function at point \((-2, 3)\). This plane can be used for linear approximations and predictions close to that point.
Multivariable Calculus
Multivariable calculus extends the principles of calculus to functions of several variables. This opens a path to study how different variables collectively affect a system, allowing for a richer understanding of changes and behaviors in more complex settings.
In multivariable calculus, key tools like the gradient vector and tangent planes become essential:
In multivariable calculus, key tools like the gradient vector and tangent planes become essential:
- The gradient vector combines partial derivatives, offering insights into how a function changes across multiple dimensions. It's a vector pointing towards the direction of greatest increase in function value, symbolized by \( abla f \).
- Tangent planes serve a similar role to tangent lines in single-variable calculus. They approximate surfaces at a point, providing a useful tool for predicting values and understanding local behavior.