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Topic 11

Directional Derivatives & the Gradient

The ∇f=⟨fx,fy,fz⟩\nabla f = \langle f_x, f_y, f_z\rangle holds every rate of change at a point. In the direction of a unit vector u\mathbf u, the rate is the Duf=∇f⋅uD_{\mathbf u}f = \nabla f \cdot \mathbf u. So ff increases fastest along ∇f\nabla f, at rate ∣∇f∣|\nabla f|. In directions perpendicular to ∇f\nabla f, the rate is 00.

5 kinds of question. Open the one you are stuck on.

1

Find the directional derivative of a function in a given direction

e.g. “Find the directional derivative of ff at … in the direction of …”
  1. 1
    Compute the gradient ∇f\nabla f
  2. 2
    Evaluate it at the point
  3. 3
    Make the direction a unit vector u\mathbf u

    Divide by its length. For "toward QQ" from PP, use Q−PQ - P first.

  4. 4
    Dot them: Duf=∇f⋅uD_{\mathbf u}f = \nabla f \cdot \mathbf u

Remember: Gradient, point, unit vector, dot.

See it in a problem3 worked examples

Find the directional derivative of f(x,y)=x2y−y3f(x,y) = x^2y - y^3 at (2,1)(2, 1) in the direction of ⟨3,−4⟩\langle 3, -4\rangle.

  1. 1
    Compute the gradient ∇f\nabla f

    ∇f=⟨2xy, x2−3y2⟩\nabla f = \langle 2xy,\ x^2 - 3y^2\rangle.

  2. 2
    Evaluate it at the point

    ∇f(2,1)=⟨4,1⟩\nabla f(2, 1) = \langle 4, 1\rangle.

  3. 3
    Make the direction a unit vector u\mathbf u

    ∣⟨3,−4⟩∣=5|\langle 3, -4\rangle| = 5, so u=15⟨3,−4⟩\mathbf u = \tfrac{1}{5}\langle 3, -4\rangle.

  4. 4
    Dot them: Duf=∇f⋅uD_{\mathbf u}f = \nabla f \cdot \mathbf u

    Duf=125−45=85D_{\mathbf u}f = \tfrac{12}{5} - \tfrac{4}{5} = \tfrac{8}{5}.

Answer:

85\dfrac{8}{5}

Common mistakes3
  • Skipping the unit vector. Using ⟨3,−4⟩\langle 3, -4\rangle instead of 15⟨3,−4⟩\tfrac{1}{5}\langle 3, -4\rangle makes the answer 55 times too big.
  • Using the target point QQ as the direction. The direction is Q−PQ - P.
  • Forgetting to plug in the point. The answer is a number, not a function.
2

Find the direction of steepest ascent and the maximum rate of change

e.g. “In what direction should the hiker proceed to ascend along the steepest path? What is the slope?”
  1. 1
    Compute ∇f\nabla f at the point
  2. 2
    Pick the direction: ∇f\nabla f uphill, −∇f-\nabla f downhill

    Unit vector asked? Divide by ∣∇f∣|\nabla f|. With xx east and yy north, ⟨−1,1⟩\langle -1, 1\rangle points north-west.

  3. 3
    The rate is ∣∇f∣|\nabla f| uphill, −∣∇f∣-|\nabla f| downhill

Remember: The gradient points uphill; its length is the steepest slope.

See it in a problem3 worked examples

Find the maximum rate of change of f(x,y)=xy2−x2f(x,y) = xy^2 - x^2 at (2,1)(2, 1), and the direction in which it occurs.

  1. 1
    Compute ∇f\nabla f at the point

    ∇f=⟨y2−2x, 2xy⟩\nabla f = \langle y^2 - 2x,\ 2xy\rangle, so ∇f(2,1)=⟨−3,4⟩\nabla f(2, 1) = \langle -3, 4\rangle.

  2. 2
    Pick the direction: ∇f\nabla f uphill, −∇f-\nabla f downhill

    Direction: ∇f\nabla f, as a unit vector 15⟨−3,4⟩\tfrac{1}{5}\langle -3, 4\rangle.

  3. 3
    The rate is ∣∇f∣|\nabla f| uphill, −∣∇f∣-|\nabla f| downhill

    Rate: ∣⟨−3,4⟩∣=5|\langle -3, 4\rangle| = 5.

Answer:

Maximum rate 55, in the direction 15⟨−3,4⟩\tfrac{1}{5}\langle -3, 4\rangle

Common mistakes3
  • Giving the vector ∇f\nabla f as the maximum rate. The rate is the number ∣∇f∣|\nabla f|.
  • Using ∇f\nabla f for "decrease most rapidly". That direction is −∇f-\nabla f.
  • On a hill z=h(x,y)z = h(x,y), the direction is the 2D vector ⟨hx,hy⟩\langle h_x, h_y\rangle. It is a way to walk on the map.
3

Find the directions in which a function has zero or a given rate of change

e.g. “Find a direction in which the initial rate of change of the temperature is 0”
  1. 1
    Compute ∇f\nabla f at the point
  2. 2
    Write u=⟨a,b⟩\mathbf u = \langle a, b\rangle with a2+b2=1a^2 + b^2 = 1

    In 3D: ⟨a,b,c⟩\langle a, b, c\rangle with a2+b2+c2=1a^2 + b^2 + c^2 = 1.

  3. 3
    Set ∇f⋅u\nabla f \cdot \mathbf u equal to the target rate
  4. 4
    Solve together with a2+b2=1a^2 + b^2 = 1

    Zero rate in 2D: u=±⟨−fy,fx⟩∣∇f∣\mathbf u = \pm\dfrac{\langle -f_y, f_x\rangle}{|\nabla f|}. It is tangent to the . If ∣target∣>∣∇f∣|\text{target}| > |\nabla f|, no direction works. In 3D, a zero rate allows every direction perpendicular to ∇f\nabla f.

Remember: Zero rate means perpendicular to the gradient.

See it in a problem3 worked examples

Find the unit vectors along which f(x,y)=x2+3xyf(x,y) = x^2 + 3xy has zero rate of change at (1,1)(1, 1). (They are tangent to the level curve there.)

  1. 1
    Compute ∇f\nabla f at the point

    ∇f=⟨2x+3y, 3x⟩\nabla f = \langle 2x + 3y,\ 3x\rangle, so ∇f(1,1)=⟨5,3⟩\nabla f(1, 1) = \langle 5, 3\rangle.

  2. 2
    Write u=⟨a,b⟩\mathbf u = \langle a, b\rangle with a2+b2=1a^2 + b^2 = 1

    u=⟨a,b⟩\mathbf u = \langle a, b\rangle, a2+b2=1a^2 + b^2 = 1.

  3. 3
    Set ∇f⋅u\nabla f \cdot \mathbf u equal to the target rate

    5a+3b=05a + 3b = 0.

  4. 4
    Solve together with a2+b2=1a^2 + b^2 = 1

    ⟨a,b⟩\langle a, b\rangle is a multiple of ⟨3,−5⟩\langle 3, -5\rangle, so u=±134⟨3,−5⟩\mathbf u = \pm\tfrac{1}{\sqrt{34}}\langle 3, -5\rangle.

Answer:

±134⟨3,−5⟩\pm\dfrac{1}{\sqrt{34}}\langle 3, -5\rangle

Common mistakes3
  • Giving one direction when there are two: u\mathbf u and −u-\mathbf u both have zero rate.
  • Missing "no direction works": ∣Duf∣≤∣∇f∣|D_{\mathbf u}f| \le |\nabla f| always.
  • Dropping a2+b2=1a^2 + b^2 = 1. Without it, any multiple seems to work, and the rate comes out wrong.
4

Find the gradient from known directional derivatives

e.g. “Find ∇f(a,b)\nabla f(a,b) given the directional derivatives …”
  1. 1
    Write the unknown gradient: ∇f=⟨p,q⟩\nabla f = \langle p, q\rangle

    Three variables: ⟨p,q,r⟩\langle p, q, r\rangle, and you need three facts.

  2. 2
    Make each given direction a unit vector

    With xx east and yy north: north is ⟨0,1⟩\langle 0, 1\rangle, north-west is 12⟨−1,1⟩\tfrac{1}{\sqrt{2}}\langle -1, 1\rangle.

  3. 3
    Turn each fact into an equation

    A directional derivative gives ∇f⋅u=Duf\nabla f \cdot \mathbf u = D_{\mathbf u}f. "Largest rate MM, in direction u\mathbf u" gives ∇f=M u\nabla f = M\,\mathbf u.

  4. 4
    Solve, then answer what is asked

Remember: Unknown gradient, one equation per fact.

See it in a problem4 worked examples

At a point, Duf=−32D_{\mathbf u}f = -\dfrac{3}{\sqrt{2}} for u=12⟨1,−1⟩\mathbf u = \tfrac{1}{\sqrt{2}}\langle 1, -1\rangle, and Dvf=235D_{\mathbf v}f = \dfrac{23}{5} for v=15⟨4,3⟩\mathbf v = \tfrac{1}{5}\langle 4, 3\rangle. Find ∇f\nabla f there.

  1. 1
    Write the unknown gradient: ∇f=⟨p,q⟩\nabla f = \langle p, q\rangle

    ∇f=⟨p,q⟩\nabla f = \langle p, q\rangle.

  2. 2
    Make each given direction a unit vector

    Both directions are already unit vectors.

  3. 3
    Turn each fact into an equation

    p−q2=−32\dfrac{p - q}{\sqrt{2}} = -\dfrac{3}{\sqrt{2}} and 4p+3q5=235\dfrac{4p + 3q}{5} = \dfrac{23}{5}, so p−q=−3p - q = -3 and 4p+3q=234p + 3q = 23.

  4. 4
    Solve, then answer what is asked

    q=p+3⇒4p+3p+9=23⇒p=2q = p + 3 \Rightarrow 4p + 3p + 9 = 23 \Rightarrow p = 2, q=5q = 5.

Answer:

∇f=⟨2,5⟩\nabla f = \langle 2, 5\rangle

Common mistakes2
  • Using ⟨1,1⟩\langle 1, 1\rangle where the unit vector 12⟨1,1⟩\tfrac{1}{\sqrt{2}}\langle 1, 1\rangle belongs.
  • Compass mix-ups: with xx east and yy north, south is ⟨0,−1⟩\langle 0, -1\rangle and west is ⟨−1,0⟩\langle -1, 0\rangle.
5

Find the rate of change felt by an object moving along a path

e.g. “A bird passes through … traveling towards … with speed 2. At what rate does the air temperature it experiences change?”
  1. 1
    Compute ∇T\nabla T at the current point
  2. 2
    Find the velocity vector v\mathbf v

    A path r(t)\mathbf r(t): v=r′(t)\mathbf v = \mathbf r'(t). A direction and a speed: v=speed×\mathbf v = \text{speed} \times unit direction. Along y=g(x)y = g(x), left to right: direction ⟨1,g′(x)⟩\langle 1, g'(x)\rangle.

  3. 3
    Dot them: dTdt=∇T⋅v\dfrac{dT}{dt} = \nabla T \cdot \mathbf v

    This is the chain rule. Here the speed matters, so do not shrink v\mathbf v to a unit vector.

Remember: Gradient dot velocity.

See it in a problem4 worked examples

The temperature, in degrees, is T=x2y+zT = x^2y + z. A bird passes through (1,2,0)(1, 2, 0) flying toward (3,3,2)(3, 3, 2) at speed 66 m/s. How fast is the temperature it feels changing?

  1. 1
    Compute ∇T\nabla T at the current point

    ∇T=⟨2xy, x2, 1⟩\nabla T = \langle 2xy,\ x^2,\ 1\rangle, so ∇T(1,2,0)=⟨4,1,1⟩\nabla T(1, 2, 0) = \langle 4, 1, 1\rangle.

  2. 2
    Find the velocity vector v\mathbf v

    Direction ⟨2,1,2⟩\langle 2, 1, 2\rangle, length 33: v=6⋅13⟨2,1,2⟩=⟨4,2,4⟩\mathbf v = 6 \cdot \tfrac{1}{3}\langle 2, 1, 2\rangle = \langle 4, 2, 4\rangle.

  3. 3
    Dot them: dTdt=∇T⋅v\dfrac{dT}{dt} = \nabla T \cdot \mathbf v

    dTdt=16+2+4=22\dfrac{dT}{dt} = 16 + 2 + 4 = 22.

Answer:

2222 degrees per second

Common mistakes3
  • Using the unit direction instead of the velocity. The speed multiplies the rate.
  • Flying "in the direction of fastest increase" at speed ss: v=s ∇T∣∇T∣\mathbf v = s\,\dfrac{\nabla T}{|\nabla T|}, so the rate is s ∣∇T∣s\,|\nabla T|.
  • Forgetting units: degrees per second, not just degrees.

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