M408M Learning Module Pages
Main page
Chapter 10: Parametric Equations
and Polar Coordinates
Chapter 12: Vectors and the Geometry of Space
Chapter 13: Vector Functions
Chapter 14: Partial Derivatives
Learning module LM 14.1:
Functions of 2 or 3 variables:
Learning module LM 14.3:
Partial derivatives:
Learning module LM 14.4:
Tangent planes and linear approximations:
Learning module LM 14.5:
Differentiability and the chain rule:
Learning module LM 14.6:
Gradients and directional derivatives:
Gradients
Gradients and hill climbing
Wind and weather
Directional derivatives
Worked problems
Learning module LM 14.7:
Local maxima and minima:
Learning module LM 14.8:
Absolute maxima and Lagrange multipliers:
Chapter 15: Multiple Integrals
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Gradients and hill climbing
Gradients and Hill Climbing
Properties of Gradients Suppose that we move along a path ${\bf
r}(t)$ in the plane. The chain rule for paths tells us that
$$\frac{df}{dt} \ = \ f_x \frac{dx}{dt} + f_y \frac{dy}{dt} \ = \ {\nabla}f \cdot \frac{d{\bf r}}{dt}
\ = \ \|{\nabla}f\| \|{\bf r}'(t)\| \cos(\theta),$$
where $\theta$ is the angle between our direction of motion and
$\nabla f$. If we move with unit speed in the direction of $\nabla f$,
then $f$ increases at a rate equal to $\| \nabla f\|$. If we move in
any other direction, $f$ increases at a slower rate, since
$\cos(\theta) < 1$. If we move perpendicular to $\nabla f$,
then $f$ doesn't change at all. And if we move in the direction of
$-\nabla f$, then $f$ decreases at a maximum rate. In other words,
$\nabla f$ points straight uphill, $-\nabla f$ points straight
downhill, and you have to go perpendicular to $\nabla f$ to stay
level.
Properties of the gradient $\nabla f$ of a function $f(x,y)$ include:
- $(\nabla f)(a,\, b)$ points in the direction of the maximum rate of
increase of $f(x,\,y)$ at $(a,\,b)$,
- $(\nabla f)(a,\, b)$ is perpendicular to the level curve through
$(a,\,b)$,
- The length of $(\nabla f)(a,\, b)$ is the maximum slope at of the surface $z=f(x,y)$ at $(a,\,b,\, f(a,\,b))$.
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Similar properties apply to the gradient of a function $f(x,y,z)$ of three variables:
- $(\nabla f)(a,\, b,\,c)$ points in the direction of the maximum rate of
increase of $f(x,\,y,\,z)$ at $(a,\,b,\,c)$,
- $(\nabla f)(a,\, b,\,c)$ is perpendicular to the level surface through
$(a,\,b,\,c)$,
- The length of $(\nabla f)(a,\, b)$ is the maximum rate of change of $f(x,\,y,\,z)$ at $(a,\,b,\, c)$ when moving with unit speed.
Example: Find the tangent plane to the surface $$x^2 + y^3 + z^4 \ = \ 18$$ at $(3,2,1)$.
Solution: The gradient is $$\langle 2x,\, 3y^2,\, 4z^3\rangle \ = \ \langle 6,\,12,\,4 \rangle\,,$$ |
so the tangent plane is
$$6(x-3) + 12 (y-2) + 4(z-1) \ = \ 0,$$or equivalently $$3x + 6y + 2z \ = \ 23.$$
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