M408M Learning Module Pages

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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



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 $(a,\,b,\, f(a,\,b))$.

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 $$3x + 6y + 2z \ = \ 23.$$