Multivariable-Calculus · Unit 11 · Video 1 · Interactive Practice

The Gradient Vector: Why It Is Perpendicular to Level Sets

IKey Formulas

FormulaNameWhat it says
∇w=⟨wx, wy, wz⟩\nabla w = \langle w_x,\ w_y,\ w_z \rangleGradientThe partial derivatives stacked into one vector, one vector at each point
dwdt=∇w⋅dr⃗dt\dfrac{dw}{dt} = \nabla w \cdot \dfrac{d\vec r}{dt}Chain rule, vector formThe rate of change of ww along a path is the gradient dotted with the velocity
∇w⊥{w=c}\nabla w \perp \{w = c\}Main theoremThe gradient is perpendicular to the level curve or level surface through the point, pointing toward higher ww
w=a1x+a2y+a3z  ⇒  ∇w=⟨a1,a2,a3⟩w = a_1 x + a_2 y + a_3 z \;\Rightarrow\; \nabla w = \langle a_1, a_2, a_3 \rangleLinear checkThe coefficients are the normal vector of every level plane

Key Insight: On a level surface ww stays equal to cc, so for any curve on it dwdt=∇w⋅v⃗=0\frac{dw}{dt} = \nabla w \cdot \vec v = 0: every tangent velocity v⃗\vec v is perpendicular to ∇w\nabla w.

IIPerpendicular at Every Point

Wherever ∇w≠0⃗\nabla w \neq \vec 0, the gradient meets the level curve through the point at a right angle.

IIIThe Proof: Moving Along a Level Curve

Along any path dwdt=∇w⋅v⃗\frac{dw}{dt} = \nabla w \cdot \vec v; on a level curve that rate is zero, so v⃗⊥∇w\vec v \perp \nabla w.

💡 In three variables the same argument runs over every curve on the level surface through the point; their velocities fill the tangent plane, so ∇w\nabla w is perpendicular to the whole tangent plane.

IVReading the Gradient off a Contour Plot

For w=13x3−x+y2w = \tfrac{1}{3}x^3 - x + y^2, the contours alone fix the direction of ∇w\nabla w: normal to the curve, toward higher values.

VQuiz Questions

Problem 1 · Compute a Gradient

Given: w=x2y+yz3w = x^2 y + y z^3 — find ∇w\nabla w at the point (1,2,−1)(1, 2, -1).

✅ Correct! wx=2xy=4w_x = 2xy = 4, wy=x2+z3=1−1=0w_y = x^2 + z^3 = 1 - 1 = 0, wz=3yz2=6w_z = 3yz^2 = 6.
❌ Check the sign of z3z^3. With z=−1z = -1, z3=−1z^3 = -1, so wy=x2+z3=1+(−1)=0w_y = x^2 + z^3 = 1 + (-1) = 0, not 22.
❌ Check z2z^2. wz=3yz2w_z = 3yz^2 and (−1)2=+1(-1)^2 = +1, so wz=3⋅2⋅1=6w_z = 3 \cdot 2 \cdot 1 = 6.
❌ Not quite. Differentiate term by term, holding the other two variables fixed: wx=2xyw_x = 2xy, wy=x2+z3w_y = x^2 + z^3, wz=3yz2w_z = 3yz^2; then substitute (1,2,−1)(1, 2, -1).
Show solution

The gradient stacks the three partial derivatives:

wx=2xy,wy=x2+z3,wz=3yz2.w_x = 2xy, \qquad w_y = x^2 + z^3, \qquad w_z = 3yz^2.

At (x,y,z)=(1,2,−1)(x, y, z) = (1, 2, -1):

wx=2(1)(2)=4,wy=12+(−1)3=0,wz=3(2)(−1)2=6.w_x = 2(1)(2) = 4, \qquad w_y = 1^2 + (-1)^3 = 0, \qquad w_z = 3(2)(-1)^2 = 6.

So ∇w(1,2,−1)=⟨4,0,6⟩\nabla w(1, 2, -1) = \langle 4, 0, 6 \rangle.

Problem 2 · Normal to an Ellipse

Given: the point (2,1)(2, 1) lies on the ellipse x2+4y2=8x^2 + 4y^2 = 8 — find a vector perpendicular to the ellipse at (2,1)(2, 1).

✅ Correct! The ellipse is the level curve w=8w = 8 of w=x2+4y2w = x^2 + 4y^2, and ∇w=⟨2x,8y⟩=⟨4,8⟩\nabla w = \langle 2x, 8y \rangle = \langle 4, 8 \rangle at (2,1)(2, 1).
❌ That is the position vector. It is normal only for circles centred at the origin; here ⟨2,1⟩⋅⟨−8,4⟩=−12≠0\langle 2, 1 \rangle \cdot \langle -8, 4 \rangle = -12 \neq 0, so it is not perpendicular to the ellipse's tangent direction.
❌ That vector is tangent. ⟨−8,4⟩⋅⟨4,8⟩=0\langle -8, 4 \rangle \cdot \langle 4, 8 \rangle = 0: it runs along the ellipse, perpendicular to the gradient.
❌ Check ∂y(4y2)\partial_y (4y^2). It is 8y8y, not 4y4y, so the second component is 88.
Show solution

Write the ellipse as a level curve: w=x2+4y2w = x^2 + 4y^2 and w(2,1)=4+4=8w(2, 1) = 4 + 4 = 8.

∇w=⟨2x, 8y⟩,∇w(2,1)=⟨4, 8⟩.\nabla w = \langle 2x,\ 8y \rangle, \qquad \nabla w(2, 1) = \langle 4,\ 8 \rangle.

By the theorem, ⟨4,8⟩\langle 4, 8 \rangle is perpendicular to the ellipse at (2,1)(2, 1). Check: implicit differentiation gives 2x+8y y′=02x + 8y\,y' = 0, so y′=−x4y=−12y' = -\frac{x}{4y} = -\frac{1}{2}, a tangent direction ⟨2,−1⟩\langle 2, -1 \rangle, and ⟨4,8⟩⋅⟨2,−1⟩=8−8=0\langle 4, 8 \rangle \cdot \langle 2, -1 \rangle = 8 - 8 = 0.

The position vector ⟨2,1⟩\langle 2, 1 \rangle is not normal here: the ellipse is not a circle about the origin.

Problem 3 · Rate Along a Circle

Given: w=x2−y2w = x^2 - y^2 and the path r⃗(t)=⟨5cos⁡t, 5sin⁡t⟩\vec r(t) = \langle \sqrt5\cos t,\ \sqrt5\sin t \rangle, at the moment it passes through (−1,2)(-1, 2) — find the velocity there and dwdt=∇w⋅v⃗\frac{dw}{dt} = \nabla w \cdot \vec v.

What is the velocity v⃗\vec v at (−1,2)(-1, 2)?

What is dw/dtdw/dt there?

✅ Correct! ∇w(−1,2)=⟨−2,−4⟩\nabla w(-1, 2) = \langle -2, -4 \rangle and ⟨−2,−4⟩⋅⟨−2,−1⟩=4+4=8\langle -2, -4 \rangle \cdot \langle -2, -1 \rangle = 4 + 4 = 8.
❌ Check the derivative of cos⁡t\cos t. ddtcos⁡t=−sin⁡t\frac{d}{dt}\cos t = -\sin t, so x′(t)=−5sin⁡tx'(t) = -\sqrt5\sin t; at this point 5sin⁡t=2\sqrt5 \sin t = 2, giving x′=−2x' = -2.
❌ That is the position. The velocity is r⃗ ′(t)=⟨−5sin⁡t,5cos⁡t⟩\vec r\,'(t) = \langle -\sqrt5\sin t, \sqrt5\cos t \rangle, tangent to the circle, not along the radius.
❌ That is ∇w\nabla w at the point. The velocity comes from the path alone: r⃗ ′(t)=⟨−5sin⁡t,5cos⁡t⟩\vec r\,'(t) = \langle -\sqrt5\sin t, \sqrt5\cos t \rangle.
❌ The circle is not a level curve of ww. w=x2−y2w = x^2 - y^2 changes around the circle (it equals 5cos⁡2t5\cos 2t), so ∇w⋅v⃗\nabla w \cdot \vec v need not vanish.
❌ That dots ∇w\nabla w with the position. The chain rule uses the velocity: ∇w⋅v⃗\nabla w \cdot \vec v, with v⃗=⟨−2,−1⟩\vec v = \langle -2, -1 \rangle.
❌ Check the velocity. Differentiate each component of r⃗(t)\vec r(t), then use 5cos⁡t=−1\sqrt5\cos t = -1 and 5sin⁡t=2\sqrt5\sin t = 2.
❌ Check the dot product. ∇w=⟨2x,−2y⟩=⟨−2,−4⟩\nabla w = \langle 2x, -2y \rangle = \langle -2, -4 \rangle; multiply component by component with v⃗=⟨−2,−1⟩\vec v = \langle -2, -1 \rangle and add.
Show solution

Step 1: the velocity. At (−1,2)(-1, 2) we have 5cos⁡t=−1\sqrt5\cos t = -1 and 5sin⁡t=2\sqrt5\sin t = 2, so

v⃗=r⃗ ′(t)=⟨−5sin⁡t, 5cos⁡t⟩=⟨−2, −1⟩.\vec v = \vec r\,'(t) = \langle -\sqrt5\sin t,\ \sqrt5\cos t \rangle = \langle -2,\ -1 \rangle.

Step 2: the gradient. ∇w=⟨2x,−2y⟩\nabla w = \langle 2x, -2y \rangle, so ∇w(−1,2)=⟨−2,−4⟩\nabla w(-1, 2) = \langle -2, -4 \rangle.

Step 3: the chain rule.

dwdt=∇w⋅v⃗=(−2)(−2)+(−4)(−1)=4+4=8.\frac{dw}{dt} = \nabla w \cdot \vec v = (-2)(-2) + (-4)(-1) = 4 + 4 = 8.

Check: on the path w=5cos⁡2t−5sin⁡2t=5cos⁡2tw = 5\cos^2 t - 5\sin^2 t = 5\cos 2t, so dwdt=−10sin⁡2t=−20sin⁡tcos⁡t=−20⋅25⋅−15=8\frac{dw}{dt} = -10\sin 2t = -20\sin t\cos t = -20 \cdot \tfrac{2}{\sqrt5} \cdot \tfrac{-1}{\sqrt5} = 8.

The rate is not zero because the circle crosses the level curves of x2−y2x^2 - y^2; only a path that stays on one level curve has ∇w⋅v⃗=0\nabla w \cdot \vec v = 0.

Problem 4 · A Curve on a Level Surface

Given: at a point PP on the level surface w=5w = 5, ∇w(P)=⟨3,−1,2⟩\nabla w(P) = \langle 3, -1, 2 \rangle. A curve lying in the surface passes through PP with velocity v⃗=⟨1,a,−1⟩\vec v = \langle 1, a, -1 \rangle — find aa.

✅ Correct! ww is constant along the curve, so 0=∇w⋅v⃗=3−a−2=1−a0 = \nabla w \cdot \vec v = 3 - a - 2 = 1 - a, giving a=1a = 1.
❌ Check the middle product. (−1)(a)=−a(-1)(a) = -a, so the equation is 3−a−2=03 - a - 2 = 0, not 3+a−2=03 + a - 2 = 0.
❌ Check the last product. (2)(−1)=−2(2)(-1) = -2, so 3−a−2=03 - a - 2 = 0; dropping that sign gives 3−a+2=03 - a + 2 = 0.
❌ The velocity is constrained. The curve stays on w=5w = 5, so dwdt=∇w⋅v⃗=0\frac{dw}{dt} = \nabla w \cdot \vec v = 0; that one equation fixes aa.
Show solution

Along a curve r⃗(t)\vec r(t) in the level surface, w(r⃗(t))=5w(\vec r(t)) = 5 for all tt, so by the chain rule

0=dwdt=∇w⋅v⃗=⟨3,−1,2⟩⋅⟨1,a,−1⟩=3−a−2=1−a.0 = \frac{dw}{dt} = \nabla w \cdot \vec v = \langle 3, -1, 2 \rangle \cdot \langle 1, a, -1 \rangle = 3 - a - 2 = 1 - a.

Hence a=1a = 1, and v⃗=⟨1,1,−1⟩\vec v = \langle 1, 1, -1 \rangle lies in the tangent plane at PP: this is exactly the step of the proof that makes ∇w\nabla w perpendicular to the level surface.

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