The gradient of f at point P is ∇f = (3, 4). What is the directional derivative D_u f in the direction u = (3/5, 4/5)?
A5 — f increases at a rate of 5 per unit step
B25 — the dot product of (3, 4) with (3, 4)
C7 — the sum of the gradient components
D1 — dividing by the gradient magnitude
u = (3/5, 4/5) is a unit vector (|u| = √(9/25 + 16/25) = 1). D_u f = ∇f · u = 3(3/5) + 4(4/5) = 9/5 + 16/5 = 5. Since u points in the gradient direction, this equals the maximum rate of change |∇f| = 5. Option B is the most common error: using the non-normalized gradient (3, 4) as the direction vector and computing (3,4)·(3,4) = 25, which conflates direction with gradient magnitude.
Question 2 Multiple Choice
At a point where ∇f = (2, −1), a walker moves in a direction perpendicular to ∇f. What rate of change does she experience?
A0 — perpendicular to the gradient means moving along a level curve
B|∇f| = √5 — she is moving at the steepest rate
C−|∇f| = −√5 — perpendicular means opposite direction
DIt depends on which of the two perpendicular directions she chooses
D_u f = |∇f| cos θ where θ is the angle between u and ∇f. Perpendicular means θ = 90°, cos 90° = 0, so D_u f = 0 regardless of which perpendicular direction is chosen. The level curves of f are exactly the curves perpendicular to ∇f — moving along a level curve means f does not change. Option D is wrong: both perpendicular directions give zero.
Question 3 True / False
The maximum directional derivative of f at a point equals the magnitude of the gradient at that point.
TTrue
FFalse
Answer: True
D_u f = ∇f · u = |∇f| cos θ. This is maximized when cos θ = 1 (u points in the gradient direction), giving max D_u f = |∇f|. The gradient direction is exactly the direction of steepest ascent, and the gradient's magnitude is the rate of that ascent per unit step.
Question 4 True / False
The formula D_u f = ∇f · u gives the correct directional derivative for any nonzero vector u — you just interpret the result as 'rate of change per unit step in direction u'.
TTrue
FFalse
Answer: False
If u is not a unit vector, ∇f · u = |∇f||u| cos θ, which is scaled by |u|. This is not the rate of change per unit distance — it conflates direction and step size. Using a non-unit vector v gives ∇f · v = (D_{v/|v|} f) · |v|, a result that depends on the arbitrary magnitude of v rather than the geometry of f. The unit-vector requirement ensures D_u f is purely about slope, independent of step size.
Question 5 Short Answer
Why must the direction vector u be a unit vector in the directional derivative formula D_u f = ∇f · u, and what goes wrong if you use a non-unit vector?
Think about your answer, then reveal below.
Model answer: A unit vector encodes pure direction without a scale. The directional derivative measures rate of change per unit distance traveled; if u has length 2, the formula gives twice the actual rate of change. Using a non-unit vector v produces ∇f · v = (D_{v/|v|} f) · |v| — a result that depends on the arbitrary magnitude of v rather than the geometry of f. Normalization ensures the result is a property of the direction alone.
The intuition: if you could change the directional derivative by stretching your direction vector, the concept would be meaningless. Normalizing removes the arbitrary scale so that D_u f depends only on which way you are pointing, not how long your direction arrow happens to be.