/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 46 Find a unit vector \(\mathbf{u}\... [FREE SOLUTION] | 91Ó°ÊÓ

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Find a unit vector \(\mathbf{u}\) that is normal at \(P(2,-3)\) to the level curve of \(f(x, y)=3 x^{2} y-x y\) through \(P\).

Short Answer

Expert verified
The unit normal vector is \( \left(\frac{-33}{\sqrt{1189}}, \frac{10}{\sqrt{1189}}\right) \).

Step by step solution

01

Understand the Problem

We need to find a unit vector that is normal to the level curve of the function \( f(x, y) = 3x^2y - xy \) at the point \( P(2, -3) \). A normal vector to the curve at a given point can be obtained from the gradient of the function evaluated at that point.
02

Compute the Gradient

The gradient vector \( abla f(x, y) \) is obtained by computing the partial derivatives of \( f \) with respect to \( x \) and \( y \).\[ f_x = \frac{\partial}{\partial x}(3x^2y - xy) = 6xy - y \]\[ f_y = \frac{\partial}{\partial y}(3x^2y - xy) = 3x^2 - x \]So, \( abla f(x, y) = (6xy - y, 3x^2 - x) \).
03

Evaluate the Gradient at Point P

Substitute \( x = 2 \) and \( y = -3 \) into the gradient:\[ abla f(2, -3) = (6(2)(-3) - (-3), 3(2)^2 - 2) \] Calculate:\[ = (-36 + 3, 12 - 2) \]\[ = (-33, 10) \] Thus, the gradient at \( P \) is \( (-33, 10) \). This is normal to the level curve at \( P \).
04

Calculate the Magnitude of the Gradient

Find the magnitude of the gradient vector \( (-33, 10) \):\[ \text{Magnitude} = \sqrt{(-33)^2 + 10^2} = \sqrt{1089 + 100} \]\[ = \sqrt{1189} \]
05

Normalize the Gradient Vector

To find the unit vector, divide each component of the gradient by its magnitude:\[ \mathbf{u} = \left(\frac{-33}{\sqrt{1189}}, \frac{10}{\sqrt{1189}}\right) \] This is the unit vector normal to the level curve at the point \( P \).

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Key Concepts

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

Gradient Vector
The gradient vector is a critical concept in calculus and vector analysis. It helps in understanding the behavior of functions. Think of the gradient vector as a direction indicator. It points toward the direction of steepest ascent of a function at a given point. For a function \( f(x, y) \), the gradient, denoted by \( abla f \), is a vector composed of its partial derivatives. It is given by:
  • \( abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \)
The gradient provides valuable information about the rate of change of the function. When applied to a surface, it becomes perpendicular to the level curves. Hence, any change is steepest in the direction of the gradient.
When solving problems involving gradients, evaluating the gradient at a specific point helps to understand how the surface behaves precisely at that point. The vector you get helps in deriving normal vectors to surfaces, which we used in our solution to find a normal unit vector.
Partial Derivatives
Partial derivatives are foundational for multivariable calculus, allowing examination of how a function changes as one variable varies, keeping other variables constant. For a function \( f(x, y) \), the partial derivative with respect to \( x \) means we're considering how \( f \) changes as \( x \) changes, with \( y \) fixed.
  • \( f_x = \frac{\partial}{\partial x} f(x, y) \)
  • \( f_y = \frac{\partial}{\partial y} f(x, y) \)
These derivatives are like slices through the function, showing changes along each axis. In the context of the exercise, partial derivatives helped form the gradient vector, \( abla f(x, y) = (6xy - y, 3x^2 - x) \).
By performing these calculations, we identified how much and in what manner the function's value shifts around the chosen point \( P(2,-3) \). Mastering partial derivatives is vital because they are a stepping-stone to more complex concepts like the gradient, divergence, and curvature.
Level Curve
Level curves are an exciting concept in multivariable calculus. They are like contour lines on a map that help visualize 3D surfaces in 2D. For a function \( f(x, y) \), a level curve is defined by setting \( f(x, y) = c \), where \( c \) is a constant. This means a level curve comprises all points \( (x, y) \) where the function takes the same value \( c \).
  • Level curves provide a slice of a 3D surface at given heights and are useful for understanding the function's topography.
  • In problems involving constraints or optimizing functions, level curves are an essential aspect of visualization.
In our exercise, the level curve at \( P(2, -3) \) is where we sought a normal vector. The gradient at \( P \), \( (-33, 10) \), acts perpendicular to this level curve, indicating how the function's value is changing most rapidly there. Understanding level curves helps visualize where the function stays constant, aiding in solving problems involving optimization and constraints.

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Most popular questions from this chapter

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