The Central Idea
For a differentiable scalar field
\[ f(x,y), \]the gradient is the vector field
\[ \boxed{ \nabla f = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right\rangle }. \]At each point, \(\nabla f\) points in the direction of greatest local increase of \(f\).
Its magnitude tells us the greatest local rate of increase per unit distance.
From Scalar Field to Vector Field
The original function
\[ f(x,y) \]assigns one number to every point.
The gradient
\[ \nabla f(x,y) \]assigns one vector to every point where the required partial derivatives exist.
Thus the gradient converts a scalar field into a vector field describing local change.
The Gradient Symbol
The symbol
\[ \nabla \]is called nabla or del.
When applied to a scalar field,
\[ \nabla f, \]it means to collect the appropriate partial derivatives into a vector.
Partial Derivatives Measure Coordinate Change
For a scalar field
\[ f(x,y), \]the partial derivative
\[ f_x=\frac{\partial f}{\partial x} \]measures the local rate of change as \(x\) varies while \(y\) is held fixed.
Similarly,
\[ f_y=\frac{\partial f}{\partial y} \]measures the local rate of change as \(y\) varies while \(x\) is held fixed.
The Gradient Combines Coordinate Rates
The gradient packages the coordinate-direction rates into one vector:
\[ \nabla f = \langle f_x,f_y\rangle. \]The first component describes local change in the \(x\)-direction.
The second component describes local change in the \(y\)-direction.
Together, they determine how the field changes in every planar direction.
A First Gradient Example
Let
\[ f(x,y)=x^2+y^2. \]Differentiate with respect to \(x\):
\[ f_x=2x. \]Differentiate with respect to \(y\):
\[ f_y=2y. \]Therefore,
\[ \boxed{ \nabla f(x,y)=\langle 2x,2y\rangle }. \]Evaluating the Gradient at a Point
For
\[ \nabla f(x,y)=\langle 2x,2y\rangle, \]evaluate at \((3,4)\):
\[ \nabla f(3,4) = \langle 6,8\rangle. \]The gradient at that point has magnitude
\[ \sqrt{6^2+8^2}=10. \]What the Gradient Direction Means
At \((3,4)\), the gradient
\[ \langle 6,8\rangle \]points in the direction in which
\[ f(x,y)=x^2+y^2 \]increases most rapidly.
Since the vector points outward from the origin, the field increases most rapidly by moving directly away from the origin.
The Gradient Magnitude
The magnitude of the gradient is
\[ \boxed{ \|\nabla f\| = \sqrt{f_x^2+f_y^2} }. \]This magnitude represents the maximum local rate of increase of the scalar field per unit distance.
Greatest Increase and Greatest Decrease
The direction of greatest local increase is
\[ \nabla f. \]The direction of greatest local decrease is the opposite direction:
\[ \boxed{-\nabla f}. \]Reversing the gradient reverses the direction while preserving the same magnitude.
Unit Direction of Greatest Increase
When
\[ \nabla f\neq \mathbf{0}, \]the unit vector pointing in the direction of greatest increase is
\[ \boxed{ \frac{\nabla f}{\|\nabla f\|} }. \]This separates direction from the size of the rate of change.
Example: Unit Gradient Direction
Suppose
\[ \nabla f(3,4)=\langle 6,8\rangle. \]Its magnitude is \(10\), so the unit direction is
\[ \frac{\langle 6,8\rangle}{10} = \boxed{ \left\langle \frac35,\frac45\right\rangle }. \]Gradients in Three Dimensions
For a scalar field
\[ f(x,y,z), \]the gradient is
\[ \boxed{ \nabla f = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right\rangle }. \]It points in the direction of greatest local increase through three-dimensional space.
Example: Three-Dimensional Gradient
Let
\[ f(x,y,z)=x^2+2y^2+3z^2. \]Then
\[ f_x=2x, \qquad f_y=4y, \qquad f_z=6z. \]Therefore,
\[ \boxed{ \nabla f = \langle 2x,4y,6z\rangle }. \]Why Directional Change Needs a Vector
A scalar field can change differently depending on which way we move.
For example,
\[ f(x,y)=x+3y \]increases by \(1\) per unit movement in the positive \(x\)-direction, but by \(3\) per unit movement in the positive \(y\)-direction.
The gradient
\[ \nabla f=\langle 1,3\rangle \]combines those directional tendencies.
Directional Derivatives
Suppose \(\mathbf{u}\) is a unit vector.
The rate of change of \(f\) in the direction \(\mathbf{u}\) is the directional derivative:
\[ \boxed{ D_{\mathbf{u}}f = \nabla f\cdot\mathbf{u} }. \]The dot product compares the chosen direction with the gradient direction.
Why the Dot Product Appears
For vectors \(\nabla f\) and \(\mathbf{u}\),
\[ \nabla f\cdot\mathbf{u} = \|\nabla f\| \|\mathbf{u}\| \cos\theta. \]If \(\mathbf{u}\) is a unit vector, then
\[ D_{\mathbf{u}}f = \|\nabla f\|\cos\theta, \]where \(\theta\) is the angle between the chosen direction and the gradient.
Maximum Directional Derivative
Because
\[ -1\leq\cos\theta\leq1, \]the directional derivative is largest when
\[ \cos\theta=1. \]This occurs when \(\mathbf{u}\) points in the same direction as \(\nabla f\).
Therefore, the maximum directional derivative is
\[ \boxed{\|\nabla f\|}. \]Minimum Directional Derivative
The directional derivative is smallest when
\[ \cos\theta=-1. \]This occurs in the direction opposite the gradient.
The minimum directional derivative is therefore
\[ \boxed{-\|\nabla f\|}. \]No First-Order Change Perpendicular to the Gradient
If a unit direction \(\mathbf{u}\) is perpendicular to the gradient, then
\[ \theta=90^\circ \]and
\[ \cos\theta=0. \]Therefore,
\[ D_{\mathbf{u}}f=0. \]To first order, the field does not change when moving perpendicular to the gradient.
Connection to Level Curves
A level curve satisfies
\[ f(x,y)=c. \]Moving along that curve keeps the field value constant.
Therefore, the direction tangent to the level curve has zero directional derivative.
The gradient must therefore be perpendicular to the level curve whenever the gradient is nonzero.
The Fundamental Geometric Relationship
At a regular point of a level curve,
\[ \boxed{ \nabla f \perp \text{level curve} }. \]In three dimensions,
\[ \boxed{ \nabla f \perp \text{level surface} }. \]The gradient is therefore a normal vector to the equal-value geometry.
Example: Circular Level Curves
Let
\[ f(x,y)=x^2+y^2. \]Its level curves are circles:
\[ x^2+y^2=c. \]The gradient is
\[ \nabla f=\langle 2x,2y\rangle. \]This is parallel to the position vector \(\langle x,y\rangle\), so it points radially outward, perpendicular to the circles.
Example: Linear Level Curves
Let
\[ f(x,y)=2x+3y. \]The level curves are
\[ 2x+3y=c. \]The gradient is
\[ \nabla f=\langle 2,3\rangle. \]Every level line has the same normal vector because all of the level lines are parallel.
Contour Spacing and Gradient Magnitude
When contour intervals are equal, close contours indicate rapid spatial change.
This corresponds to a larger gradient magnitude.
Widely spaced contours indicate slower spatial change and therefore a smaller gradient magnitude.
The gradient gives a quantitative version of what contour spacing shows visually.
The Gradient Points across Contours
Because the gradient is perpendicular to level curves, it points across contours rather than along them.
It points from lower field values toward higher field values.
The negative gradient points from higher values toward lower values.
Reading Gradient Direction from a Contour Map
To estimate gradient direction from a contour map:
- locate the point;
- identify the local contour direction;
- choose a perpendicular direction;
- use the contour labels to determine which perpendicular direction points toward larger values.
That direction approximates the direction of the gradient.
Reading Gradient Magnitude Qualitatively
If the contour interval is fixed, the gradient magnitude is larger where contours are more tightly packed.
It is smaller where contours are farther apart.
A contour map therefore gives qualitative information about both gradient direction and gradient strength.
Gradient of a Constant Field
If
\[ f(x,y)=7, \]then
\[ f_x=0, \qquad f_y=0. \]Therefore,
\[ \boxed{ \nabla f=\langle 0,0\rangle }. \]A constant field has no local increase or decrease.
Critical Points
A point where
\[ \nabla f=\mathbf{0} \]is often called a critical point, provided the required derivatives exist.
At such a point, the gradient does not identify a unique direction of greatest first-order increase.
A Critical Point Need Not Be a Maximum or Minimum
Consider
\[ f(x,y)=x^2-y^2. \]The gradient is
\[ \nabla f=\langle 2x,-2y\rangle. \]At the origin,
\[ \nabla f(0,0)=\langle 0,0\rangle. \]Yet the origin is neither a local maximum nor a local minimum. It is a saddle-type point.
Gradient at a Local Peak
Consider
\[ f(x,y)=10-x^2-y^2. \]The gradient is
\[ \nabla f=\langle -2x,-2y\rangle. \]At the peak \((0,0)\),
\[ \nabla f(0,0)=\mathbf{0}. \]Every immediate direction leads downward to first nonzero order beyond the stationary point.
Gradient at a Local Basin
Consider
\[ f(x,y)=x^2+y^2. \]Again,
\[ \nabla f(0,0)=\mathbf{0}. \]Here the origin is a local and global minimum.
The zero gradient alone does not tell us whether a point is a peak, basin, or saddle.
Gradient and Tangent Directions
If a level curve passes through a point and has tangent vector
\[ \mathbf{t}, \]then
\[ \boxed{ \nabla f\cdot\mathbf{t}=0 }. \]This gives an algebraic way to find tangent directions from the gradient.
Finding a Tangent Line from the Gradient
Suppose
\[ f(x,y)=x^2+y^2 \]and consider the level curve through \((3,4)\).
The gradient there is
\[ \nabla f(3,4)=\langle 6,8\rangle. \]This is a normal vector to the tangent line.
Therefore, a tangent-line equation is
\[ 6(x-3)+8(y-4)=0. \]Simplifying the Tangent-Line Equation
Starting from
\[ 6(x-3)+8(y-4)=0, \]divide by \(2\):
\[ 3(x-3)+4(y-4)=0. \]Expand:
\[ 3x+4y=25. \]This is tangent to the circle
\[ x^2+y^2=25 \]at \((3,4)\).
Normal Lines to Level Curves
Because the gradient is perpendicular to the level curve, it points along the normal line.
At a point \((x_0,y_0)\), a normal line may be parameterized as
\[ \boxed{ \mathbf{r}(t) = \langle x_0,y_0\rangle + t\nabla f(x_0,y_0) }. \]Normal Vectors to Level Surfaces
In three dimensions, if
\[ f(x,y,z)=c, \]then
\[ \nabla f(x_0,y_0,z_0) \]is normal to the level surface at a regular point \((x_0,y_0,z_0)\).
This provides a direct bridge from multivariable calculus to equations of tangent planes.
Tangent Plane from the Gradient
If
\[ \nabla f(x_0,y_0,z_0) = \langle A,B,C\rangle, \]then a tangent plane to the level surface \(f=c\) at that point is
\[ \boxed{ A(x-x_0)+B(y-y_0)+C(z-z_0)=0 }. \]Example: Tangent Plane
Let
\[ f(x,y,z)=x^2+y^2+z^2. \]At \((1,2,2)\),
\[ \nabla f=\langle 2x,2y,2z\rangle \]gives
\[ \nabla f(1,2,2)=\langle 2,4,4\rangle. \]A tangent plane to the sphere through that point is
\[ 2(x-1)+4(y-2)+4(z-2)=0. \]Gradient of a Linear Field
For
\[ f(x,y)=ax+by+c, \]the gradient is constant:
\[ \boxed{ \nabla f=\langle a,b\rangle }. \]Therefore, the direction of greatest increase is the same everywhere.
Gradient of a Radial Field
Let
\[ f(x,y)=g(x^2+y^2), \]where \(g\) is differentiable.
By the chain rule,
\[ \nabla f = 2g'(x^2+y^2)\langle x,y\rangle. \]The gradient is therefore radial wherever the scalar factor is nonzero.
Gradient of Distance
Consider distance from the origin:
\[ d(x,y)=\sqrt{x^2+y^2}. \]Away from the origin,
\[ \nabla d = \left\langle \frac{x}{\sqrt{x^2+y^2}}, \frac{y}{\sqrt{x^2+y^2}} \right\rangle. \]This is the outward radial unit vector.
The Gradient of Distance Has Magnitude One
For
\[ d(x,y)=\sqrt{x^2+y^2}, \]we have
\[ \|\nabla d\| = \sqrt{ \frac{x^2}{x^2+y^2} + \frac{y^2}{x^2+y^2} } = 1 \]wherever \(d\) is differentiable.
Distance increases at one unit of distance per unit movement directly outward.
Gradient of Temperature
If
\[ T(x,y,z) \]describes temperature, then
\[ \nabla T \]points toward the direction of greatest local temperature increase.
The vector
\[ -\nabla T \]points toward greatest local temperature decrease.
Gradient of Elevation
If
\[ h(x,y) \]is terrain elevation, then
\[ \nabla h \]points in the direction of steepest uphill ascent.
Its negative points in the steepest downhill direction.
Moving along an ideal elevation contour produces no first-order change in elevation.
Gradient and Steepest Ascent
The phrase steepest ascent refers to the direction of greatest local increase of the scalar field.
Mathematically, that direction is
\[ \boxed{\nabla f}. \]The steepness itself is measured by
\[ \boxed{\|\nabla f\|}. \]Gradient and Steepest Descent
The direction of steepest local descent is
\[ \boxed{-\nabla f}. \]This idea is central in optimization, where one often seeks to reduce a function by moving opposite its gradient.
Detailed optimization methods appear elsewhere, but the geometric principle begins here.
Directional Derivative Example
Let
\[ f(x,y)=x^2+y^2. \]At \((1,2)\),
\[ \nabla f(1,2)=\langle 2,4\rangle. \]Find the directional derivative in the unit direction
\[ \mathbf{u} = \left\langle \frac35,\frac45\right\rangle. \]Then
\[ D_{\mathbf{u}}f = \langle 2,4\rangle \cdot \left\langle \frac35,\frac45\right\rangle = \frac65+\frac{16}{5} = \boxed{\frac{22}{5}}. \]The Direction Vector Must Be a Unit Vector
The standard directional derivative formula
\[ D_{\mathbf{u}}f = \nabla f\cdot\mathbf{u} \]assumes
\[ \|\mathbf{u}\|=1. \]If a direction is given by a nonunit vector \(\mathbf{v}\), normalize it first:
\[ \mathbf{u} = \frac{\mathbf{v}}{\|\mathbf{v}\|}. \]Example: Normalize Before Differentiating
Suppose the desired direction is
\[ \mathbf{v}=\langle 3,4\rangle. \]Its magnitude is \(5\), so the corresponding unit vector is
\[ \mathbf{u} = \left\langle \frac35,\frac45\right\rangle. \]That unit vector, not \(\langle 3,4\rangle\) itself, is used in the standard directional derivative formula.
Moving along a Parametric Path
Suppose a path is
\[ \mathbf{r}(t) = \langle x(t),y(t)\rangle. \]The scalar field along the path becomes
\[ f(\mathbf{r}(t)) = f(x(t),y(t)). \]The multivariable chain rule gives
\[ \boxed{ \frac{d}{dt}f(\mathbf{r}(t)) = \nabla f(\mathbf{r}(t)) \cdot \mathbf{r}'(t) }. \]Gradient and Motion through a Field
If an object moves with velocity
\[ \mathbf{v}(t)=\mathbf{r}'(t), \]then the rate at which it experiences change in a scalar field is
\[ \frac{df}{dt} = \nabla f\cdot\mathbf{v}. \]This formula connects spatial change with motion through space.
Moving along a Level Curve
If a path stays entirely on the level curve
\[ f(x,y)=c, \]then
\[ f(\mathbf{r}(t))=c. \]Differentiating gives
\[ 0 = \nabla f\cdot\mathbf{r}'(t). \]Thus the path tangent is perpendicular to the gradient.
Why the Gradient Is Normal to Level Sets
The perpendicular relationship is not merely a visual rule.
It follows from the fact that the scalar field does not change while moving along a level set.
Zero change along every tangent direction means that the gradient must be orthogonal to those tangent directions.
Gradient Fields
A vector field that can be written as
\[ \nabla f \]for some scalar field \(f\) is often called a gradient field.
Gradient fields are important because they connect vector behavior to an underlying scalar potential.
The later lesson on potential and conservative fields develops this idea in greater depth.
Not Every Vector Field Is Introduced as a Gradient
A general vector field may be given directly:
\[ \mathbf{F}(x,y) = \langle P(x,y),Q(x,y)\rangle. \]Whether such a field comes from some scalar potential is a separate question.
The gradient operation begins with a scalar field and constructs a vector field from it.
Gradient and Units
Gradient components have units of field value divided by spatial distance.
If temperature is measured in degrees Celsius and position in meters, then
\[ \nabla T \]has units such as
\[ ^\circ\text{C}/\text{m}. \]The gradient magnitude has the same units.
Example: Units of an Elevation Gradient
Suppose \(h(x,y)\) is measured in meters while \(x\) and \(y\) are measured in kilometers.
Then
\[ h_x \quad\text{and}\quad h_y \]have units of meters of elevation change per kilometer of horizontal displacement.
Unit consistency matters when interpreting gradient magnitude physically.
Common Error: Treating the Gradient as a Scalar
If
\[ f(x,y)=x^2+y^2, \]then
\[ \nabla f=\langle 2x,2y\rangle, \]not
\[ 2x+2y. \]The gradient is a vector whose components are partial derivatives.
Common Error: Forgetting a Partial Derivative
For
\[ f(x,y,z), \]a three-dimensional gradient needs three components:
\[ \nabla f = \langle f_x,f_y,f_z\rangle. \]Omitting one component loses directional information.
Common Error: Using a Nonunit Direction
If a problem asks for the directional derivative in the direction of
\[ \mathbf{v}=\langle 6,8\rangle, \]first normalize:
\[ \mathbf{u} = \left\langle \frac35,\frac45\right\rangle. \]Otherwise the dot product includes unwanted scaling from the length of \(\mathbf{v}\).
Common Error: Saying the Gradient Points Along a Contour
The gradient does not point along a regular level curve.
It points perpendicular to it.
Moving along the contour keeps the scalar value constant, while moving in the gradient direction increases the field most rapidly.
Common Error: Assuming a Zero Gradient Proves an Extremum
A zero gradient identifies a critical point, but it does not by itself classify the point.
The point may be:
- a local maximum;
- a local minimum;
- a saddle point;
- or another degenerate critical point.
Additional analysis is required.
Common Error: Confusing Gradient Magnitude with Field Value
A scalar field value
\[ f(x,y) \]and the gradient magnitude
\[ \|\nabla f(x,y)\| \]measure different things.
The first gives the amount of the field at the point. The second describes how rapidly the field can change nearby.
A Reliable Procedure for Finding a Gradient
- Identify the scalar field.
- Differentiate with respect to each spatial coordinate.
- Place the partial derivatives in coordinate order.
- Simplify the resulting vector field.
- Evaluate at the requested point if needed.
- Compute the magnitude if the maximum rate of increase is requested.
- Normalize if a unit direction is requested.
A Reliable Procedure for a Directional Derivative
- Find \(\nabla f\).
- Evaluate the gradient at the point.
- Identify the requested direction vector.
- Normalize that direction if necessary.
- Compute the dot product \(\nabla f\cdot\mathbf{u}\).
- Interpret the sign and magnitude of the result.
Interpreting the Sign of a Directional Derivative
If
\[ D_{\mathbf{u}}f>0, \]the field increases in direction \(\mathbf{u}\).
If
\[ D_{\mathbf{u}}f<0, \]the field decreases.
If
\[ D_{\mathbf{u}}f=0, \]there is no first-order change in that direction.
Example: Full Gradient Workflow
Let
\[ f(x,y)=3x^2+xy-2y^2. \]Compute the partial derivatives:
\[ f_x=6x+y, \] \[ f_y=x-4y. \]Therefore,
\[ \boxed{ \nabla f = \langle 6x+y,x-4y\rangle }. \]At \((1,2)\),
\[ \nabla f(1,2) = \langle 8,-7\rangle. \]Example: Maximum Rate of Increase
For
\[ \nabla f(1,2)=\langle 8,-7\rangle, \]the maximum local rate of increase is
\[ \|\nabla f(1,2)\| = \sqrt{8^2+(-7)^2} = \boxed{\sqrt{113}}. \]The direction of greatest increase is the direction of \(\langle 8,-7\rangle\).
Example: Unit Direction of Greatest Increase
The unit direction corresponding to
\[ \langle 8,-7\rangle \]is
\[ \boxed{ \left\langle \frac{8}{\sqrt{113}}, \frac{-7}{\sqrt{113}} \right\rangle }. \]Example: Directional Derivative from a Given Vector
Let
\[ \nabla f(1,2)=\langle 8,-7\rangle \]and suppose the requested direction is
\[ \mathbf{v}=\langle 3,4\rangle. \]Normalize:
\[ \mathbf{u} = \left\langle \frac35,\frac45\right\rangle. \]Then
\[ D_{\mathbf{u}}f = \langle 8,-7\rangle \cdot \left\langle \frac35,\frac45\right\rangle = \frac{24}{5}-\frac{28}{5} = \boxed{-\frac45}. \]The negative sign means the field decreases in that direction.
Practice 1: Find a Gradient
Let
\[ f(x,y)=x^2+3y^2. \]Find \(\nabla f\).
Practice 2: Evaluate a Gradient
For
\[ f(x,y)=x^2+3y^2, \]find \(\nabla f(2,-1)\).
Practice 3: Maximum Rate of Increase
If
\[ \nabla f(1,2)=\langle 5,12\rangle, \]find the maximum local rate of increase.
Practice 4: Direction of Greatest Increase
If
\[ \nabla f(1,2)=\langle 5,12\rangle, \]find a unit vector in the direction of greatest increase.
Practice 5: Direction of Greatest Decrease
If
\[ \nabla f(3,4)=\langle 6,8\rangle, \]give a vector pointing in the direction of greatest decrease.
Practice 6: Directional Derivative
Let
\[ \nabla f(2,1)=\langle 4,3\rangle \]and
\[ \mathbf{u}=\left\langle \frac35,\frac45\right\rangle. \]Find \(D_{\mathbf{u}}f\).
Practice 7: Perpendicular Direction
If
\[ \nabla f=\langle 2,5\rangle, \]give one nonzero vector tangent to the corresponding level curve.
Practice 8: Find a Critical Point
Let
\[ f(x,y)=x^2+y^2-4x+6y. \]Find where
\[ \nabla f=\mathbf{0}. \]Practice 9: Three-Dimensional Gradient
Let
\[ f(x,y,z)=x^2+yz+z^2. \]Find \(\nabla f\).
Practice 10: Explain the Contour Relationship
Explain why the gradient at a regular point is perpendicular to the level curve through that point.
Practice Answers
Practice 1
\[ f_x=2x, \qquad f_y=6y. \]Therefore,
\[ \boxed{ \nabla f=\langle 2x,6y\rangle }. \]Practice 2
\[ \nabla f(2,-1) = \langle 4,-6\rangle. \]Therefore,
\[ \boxed{\nabla f(2,-1)=\langle 4,-6\rangle}. \]Practice 3
\[ \|\nabla f\| = \sqrt{5^2+12^2} = 13. \]Therefore, the maximum local rate of increase is
\[ \boxed{13}. \]Practice 4
\[ \boxed{ \frac{\nabla f}{\|\nabla f\|} = \left\langle \frac5{13}, \frac{12}{13} \right\rangle }. \]Practice 5
The direction of greatest decrease is opposite the gradient:
\[ \boxed{\langle -6,-8\rangle}. \]Any positive scalar multiple of this vector points in the same direction.
Practice 6
\[ D_{\mathbf{u}}f = \langle 4,3\rangle \cdot \left\langle \frac35,\frac45\right\rangle = \frac{12}{5}+\frac{12}{5} = \boxed{\frac{24}{5}}. \]Practice 7
A vector perpendicular to
\[ \langle 2,5\rangle \]is
\[ \boxed{\langle -5,2\rangle}, \]since
\[ 2(-5)+5(2)=0. \]Practice 8
\[ \nabla f = \langle 2x-4,2y+6\rangle. \]Set both components equal to zero:
\[ 2x-4=0, \qquad 2y+6=0. \]Therefore,
\[ \boxed{(x,y)=(2,-3)}. \]Practice 9
\[ f_x=2x, \qquad f_y=z, \qquad f_z=y+2z. \]Therefore,
\[ \boxed{ \nabla f = \langle 2x,z,y+2z\rangle }. \]Practice 10
Moving tangent to a level curve keeps \(f\) constant, so the directional derivative in every tangent direction is zero. Since
\[ D_{\mathbf{u}}f=\nabla f\cdot\mathbf{u}, \]the gradient has zero dot product with the tangent direction. Therefore, the gradient is perpendicular to the level curve.
What to Remember
The gradient converts local scalar-field change into a vector:
\[ \boxed{ \nabla f = \langle f_x,f_y\rangle } \]in two dimensions, and
\[ \boxed{ \nabla f = \langle f_x,f_y,f_z\rangle } \]in three dimensions.
- the gradient points toward greatest local increase;
- the negative gradient points toward greatest local decrease;
- the gradient magnitude is the maximum local rate of increase;
- directional derivatives are computed with \(\nabla f\cdot\mathbf{u}\);
- the gradient is perpendicular to regular level curves and level surfaces;
- close contours usually correspond to larger gradient magnitude when contour intervals are fixed;
- a zero gradient identifies a critical point but does not classify it by itself.
The central geometric idea is simple: level sets run along equal values, while gradients point across them toward greater values.
Where We Go Next
Gradients describe local directions of change in scalar fields. The next step is to study how vector fields organize possible motion through space.
The next lesson, Flow Lines and Trajectories, develops curves that follow the directions prescribed by a vector field and connects field geometry to motion.