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Zero directional derivative of \(f(x, y)=x y+y^{2}\) at \(P(3,2)\) equal to zero?

Short Answer

Expert verified
The directional vector \((-7, 2)\) makes the directional derivative zero at \(P(3, 2)\).

Step by step solution

01

Calculate the Gradient

To find the gradient of the function \( f(x, y) = xy + y^2 \), compute the partial derivatives with respect to \( x \) and \( y \). We have: \( f_x = y \) and \( f_y = x + 2y \). Therefore, the gradient is \( abla f = (y, x + 2y) \).
02

Evaluate the Gradient at P(3, 2)

Substitute \( x = 3 \) and \( y = 2 \) into the gradient: \( abla f(3, 2) = (2, 3 + 2(2)) = (2, 7) \).
03

Understand Zero Directional Derivative Condition

The directional derivative is zero when the gradient is orthogonal to the direction vector \( \mathbf{v} \). This is true when the dot product \( abla f \cdot \mathbf{v} = 0 \).
04

Determine the Direction Vector

Let the direction vector be \( \mathbf{v} = (a, b) \). The directional derivative formula gives \( abla f \cdot \mathbf{v} = 2a + 7b = 0 \).
05

Solve for the Direction Vector

From \( 2a + 7b = 0 \), solve for one variable: \( a = -\frac{7}{2}b \). This gives the family of vectors orthogonal to the gradient, such as \( (a, b) = (-7, 2) \).
06

Verify the Solution

Verify that \( (-7, 2) \) makes the directional derivative zero by checking \( abla f \cdot (-7, 2) = 2(-7) + 7(2) = 0 \). This confirms orthogonality.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gradient Vector
In calculus, the gradient vector is a crucial concept used to understand how a function changes at each point in space. For a function of two variables, such as \( f(x, y) = xy + y^2 \), the gradient vector \( abla f \) is constructed using the partial derivatives. These partial derivatives are the rates of change of the function concerning each of its variables. In our example:
  • \( f_x \), the partial derivative with respect to \( x \), is \( y \).
  • \( f_y \), the partial derivative with respect to \( y \), is \( x + 2y \).
Thus, the gradient vector in this case is given by \( abla f = (y, x + 2y) \). The gradient provides the direction of the steepest ascent of the function, and its magnitude shows how steep that ascent is. Evaluating the gradient at a specific point, like \( P(3, 2) \), helps in determining the behavior of the function at that instant.
Orthogonality
Orthogonality is a concept used to determine if two vectors are perpendicular to each other. Two vectors are orthogonal if their dot product is zero. In the context of directional derivatives, orthogonality plays a crucial role because if the gradient vector \( abla f \) is orthogonal to a direction vector \( \mathbf{v} \), it means that moving in the direction of \( \mathbf{v} \) does not increase or decrease the function value. This results in the directional derivative being zero.To determine orthogonality between the gradient vector and the direction vector \( \mathbf{v} = (a, b) \) at point \( P(3, 2) \), we compute the dot product:
  • \( abla f(3, 2) = (2, 7) \).
  • The dot product with the direction vector is \( 2a + 7b = 0 \).
Thus, any direction vector \( (a, b) \) that satisfies this equation, such as \( (-7, 2) \), is orthogonal to the gradient vector, ensuring that the directional derivative is zero.
Partial Derivatives
Partial derivatives are a fundamental part of studying multivariable functions. They capture how a function changes with respect to one variable while keeping the other variables constant. For the function \( f(x, y) = xy + y^2 \),
  • The partial derivative \( f_x = y \) implies that when you change \( x \) slightly, the function's change is directly proportional to \( y \).
  • Similarly, \( f_y = x + 2y \) indicates that the change in the function as \( y \) changes depends on both \( x \) and \( y \).
These derivatives are used to form the gradient vector, which helps understand the direction of maximal change of the function. By evaluating these derivatives at specific points, you gain insights into the local behavior of the function, essential for applications such as optimization and finding zero directional derivatives.

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Most popular questions from this chapter

You will explore functions to identify their local extrema. Use a CAS to perform the following steps: a. Plot the function over the given rectangle. b. Plot some level curves in the rectangle. c. Calculate the function's first partial derivatives and use the CAS equation solver to find the critical points. How do the critical points relate to the level curves plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for your answer. d. Calculate the function's second partial derivatives and find the discriminant \(f_{x x} f_{y y}-f_{x y}^{2}\) . e. Using the max-min tests, classify the critical points found in part (c). Are your findings consistent with your discussion in part (c)? $$\begin{array}{l}{f(x, y)=x^{4}+y^{2}-8 x^{2}-6 y+16,-3 \leq x \leq 3} \\\ {-6 \leq y \leq 6}\end{array}$$

Suppose that \begin{equation}w=x^{2}-y^{2}+4 z+t \quad \text { and } \quad x+2 z+t=25.\end{equation} Show that the equations \begin{equation}\frac{\partial w}{\partial x}=2 x-1 \quad \text { and } \quad \frac{\partial w}{\partial x}=2 x-2\end{equation} each give \(\partial w / \partial x,\) depending on which variables are chosen to be dependent and which variables are chosen to be independent. Identify the independent variables in each case.

Three variables Let \(w=f(x, y, z)\) be a function of three independent variables and write the formal definition of the partial derivative \(\partial f / \partial y\) at \(\left(x_{0}, y_{0}, z_{0}\right) .\) Use this definition to find \(\partial f / \partial y\) at \((-1,0,3)\) for \(f(x, y, z)=-2 x y^{2}+y z^{2}\)

In Exercises \(49-52\) , find an equation for and sketch the graph of the level curve of the function \(f(x, y)\) that passes through the given point. $$ f(x, y)=16-x^{2}-y^{2}, \quad(2 \sqrt{2}, \sqrt{2}) $$

You will explore functions to identify their local extrema. Use a CAS to perform the following steps: a. Plot the function over the given rectangle. b. Plot some level curves in the rectangle. c. Calculate the function's first partial derivatives and use the CAS equation solver to find the critical points. How do the critical points relate to the level curves plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for your answer. d. Calculate the function's second partial derivatives and find the discriminant \(f_{x x} f_{y y}-f_{x y}^{2}\) . e. Using the max-min tests, classify the critical points found in part (c). Are your findings consistent with your discussion in part (c)? $$f(x, y)=\left\\{\begin{array}{ll}{x^{5} \ln \left(x^{2}+y^{2}\right),} & {(x, y) \neq(0,0)} \\ {0,} & {(x, y)=(0,0)}\end{array}\right.$$ $$-2 \leq x \leq 2, \quad-2 \leq y \leq 2$$

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