Problem 1
Question
Sketch the level curve or surface passing through the indicated point. Sketch the gradient at the point. $$ f(x, y)=x-2 y ;(6,1) $$
Step-by-Step Solution
Verified Answer
The level curve is the line \( x - 2y = 4 \), and the gradient at (6,1) is the vector (1, -2).
1Step 1: Understand the Level Curve Equation
A level curve for a function of two variables, like \( f(x,y) = x - 2y \), is where \( f(x,y) = c \), with \( c \) being a constant. For the point \((6,1)\), calculate \( c = f(6,1) = 6 - 2 \times 1 = 4 \). Thus, the equation of the level curve passing through the point is \( x - 2y = 4 \).
2Step 2: Determine the Gradient Vector
The gradient vector, \( abla f(x, y) \), is composed of the partial derivatives of \( f \). Calculate them: \( \frac{\partial f}{\partial x} = 1 \) and \( \frac{\partial f}{\partial y} = -2 \). Therefore, \( abla f(x, y) = (1, -2) \). At the point \((6,1)\), the gradient vector is \( (1, -2) \).
3Step 3: Sketch the Level Curve
Since \( x - 2y = 4 \) is a linear equation, its graph is a line. For this line, find the x-intercept by setting \( y = 0 \): \( x = 4 \). Find the y-intercept by setting \( x = 0 \): \( -2y = 4 \) implies \( y = -2 \). Connect these intercepts to sketch the line.
4Step 4: Sketch the Gradient Vector
The gradient vector \( abla f(6,1) = (1, -2) \), at the point \((6,1)\), should be drawn as an arrow starting from \((6,1)\) and pointing in the direction of increasing \( x \), and decreasing \( y \). Plot the starting point at \((6,1)\), move 1 unit to the right and 2 units down to represent the gradient.
5Step 5: Interpret the Sketch
The level curve is a straight line through \((6,1)\) with equation \( x - 2y = 4 \). The gradient, represented as an arrow, indicates the direction of the steepest ascent from \((6,1)\) and points perpendicular to the level curve.
Key Concepts
Level CurvesGradient VectorPartial Derivatives
Level Curves
Level curves serve as a fascinating way to visualize functions of two variables. For a given function, a level curve is where the function maintains a constant value. When we say a level curve for the function \( f(x, y) = x - 2y \), at a point like \((6,1)\), we mean we find the value of the function at that point and look for all values \((x, y)\) that yield the same function value, which becomes a constant \(c\). In our scenario, calculating \( f(6,1) \) gets us to \( 6 - 2 \times 1 = 4 \). Hence, the equation \( x - 2y = 4 \) is what defines our level curve.
To sketch this, note it's a linear equation—a slanted line on a graph. By finding x- and y-intercepts: set \( y = 0 \), giving \( x = 4 \), and \( x = 0 \), giving \( y = -2 \); connecting these intercepts beautifully draws the line, creating a visual representation of the level curve through \((6,1)\). Level curves provide a two-dimensional picture of functions and are particularly useful to understand topography or quantities described over a plane surface.
To sketch this, note it's a linear equation—a slanted line on a graph. By finding x- and y-intercepts: set \( y = 0 \), giving \( x = 4 \), and \( x = 0 \), giving \( y = -2 \); connecting these intercepts beautifully draws the line, creating a visual representation of the level curve through \((6,1)\). Level curves provide a two-dimensional picture of functions and are particularly useful to understand topography or quantities described over a plane surface.
Gradient Vector
The gradient vector not only tells us the slope of a function at a point but also the direction where the function increases most rapidly. For the function \( f(x, y) = x - 2y \), we find the gradient vector \(abla f(x, y)\) by calculating the partial derivatives with respect to each variable.
Here, \( \frac{\partial f}{\partial x} = 1 \) and \( \frac{\partial f}{\partial y} = -2 \), which leads us to the gradient vector \( abla f(x, y) = (1, -2) \). It essentially acts as a pointer indicating how y would decrease by twice as much as x increases. At point \((6,1)\), this vector \((1, -2)\) shows the steepest ascent.
Visualize it by drawing an arrow starting at \((6,1)\)—it’ll move 1 unit to the right (as x increases) and 2 units downwards (as y declines). The gradient vector is always perpendicular to the level curve at any given point. This perpendicularity underlies why it shows the direction of maximum increase precisely.
Here, \( \frac{\partial f}{\partial x} = 1 \) and \( \frac{\partial f}{\partial y} = -2 \), which leads us to the gradient vector \( abla f(x, y) = (1, -2) \). It essentially acts as a pointer indicating how y would decrease by twice as much as x increases. At point \((6,1)\), this vector \((1, -2)\) shows the steepest ascent.
Visualize it by drawing an arrow starting at \((6,1)\)—it’ll move 1 unit to the right (as x increases) and 2 units downwards (as y declines). The gradient vector is always perpendicular to the level curve at any given point. This perpendicularity underlies why it shows the direction of maximum increase precisely.
Partial Derivatives
Partial derivatives allow us to study how a function changes as we vary one variable while keeping others constant. For our function \( f(x, y) = x - 2y \), these derivatives enable us to explore the nature of the function in the x and y directions separately.
Calculating the partial derivative with respect to x, denoted \( \frac{\partial f}{\partial x} \), means you consider y as a constant, consequently deriving 1 in this case. On the other hand, \( \frac{\partial f}{\partial y} \) involves treating x as constant, deriving \(-2\). Each partial derivative gives a piece of the gradient vector, making them essential for understanding changes in different directions.
Think of partial derivatives like finding the slope of a mountain path going either north-south or east-west—giving you an idea if you're heading up or downhill. This idea translates into real-world scenarios like optimizing conditions in economics or engineering through understanding variable dependencies uniquely.
Calculating the partial derivative with respect to x, denoted \( \frac{\partial f}{\partial x} \), means you consider y as a constant, consequently deriving 1 in this case. On the other hand, \( \frac{\partial f}{\partial y} \) involves treating x as constant, deriving \(-2\). Each partial derivative gives a piece of the gradient vector, making them essential for understanding changes in different directions.
Think of partial derivatives like finding the slope of a mountain path going either north-south or east-west—giving you an idea if you're heading up or downhill. This idea translates into real-world scenarios like optimizing conditions in economics or engineering through understanding variable dependencies uniquely.
Other exercises in this chapter
Problem 1
Evaluate \(\int_{C} G(x, y) d x, \int_{C} G(x, y) d y\), and \(\int_{C} G(x, y) d s\) on the indicated curve \(C\). $$ G(x, y)=2 x y ; x=5 \cos t, y=5 \sin t, 0
View solution Problem 1
In Problems \(1-6\), graph some representative vectors in the given vector field. $$ \mathbf{F}(x, y)=x \mathbf{i}+y \mathbf{j} $$
View solution Problem 1
Sketch some of the level curves associated with the given function. $$ f(x, y)=x+2 y $$
View solution Problem 1
In Problems, compute the gradient for the given function. $$ f(x, y)=x^{2}-x^{3} y^{2}+y^{4} $$
View solution