Multivariable-Calculus Β· Unit 6 Β· Video 5 Β· Interactive Practice

Kepler's Second Law Says the Force Is Central

IKey Formulas

FormulaNameWhat it says
r⃗×v⃗=constant vector\vec{r} \times \vec{v} = \text{constant vector}Kepler's second lawConstant length (equal areas) and constant direction (one plane)
ddt(u⃗×w⃗)=du⃗dt×w⃗+u⃗×dw⃗dt\frac{d}{dt}(\vec{u} \times \vec{w}) = \frac{d\vec{u}}{dt} \times \vec{w} + \vec{u} \times \frac{d\vec{w}}{dt}Product rule for ×\timesLeft factor stays left, right factor stays right
ddt(r⃗×v⃗)=v⃗×v⃗+r⃗×a⃗=r⃗×a⃗\frac{d}{dt}(\vec{r} \times \vec{v}) = \vec{v} \times \vec{v} + \vec{r} \times \vec{a} = \vec{r} \times \vec{a}Derivative of r⃗×v⃗\vec{r} \times \vec{v}v⃗×v⃗=0⃗\vec{v} \times \vec{v} = \vec{0}
rβƒ—Γ—aβƒ—=0βƒ—β€…β€ŠβŸΊβ€…β€Šaβƒ—βˆ₯rβƒ—\vec{r} \times \vec{a} = \vec{0} \iff \vec{a} \parallel \vec{r}Central forceThe force acts along the line through the origin

Key Insight: Kepler's second law β€…β€ŠβŸΊβ€…β€Šrβƒ—Γ—vβƒ—=constantβ€…β€ŠβŸΊβ€…β€Šrβƒ—Γ—aβƒ—=0βƒ—β€…β€ŠβŸΊβ€…β€Šaβƒ—βˆ₯rβƒ—\iff \vec{r} \times \vec{v} = \text{constant} \iff \vec{r} \times \vec{a} = \vec{0} \iff \vec{a} \parallel \vec{r}. Every step runs both ways, and nothing decides whether the force points toward the origin or away from it.

IIThe Product Rule Predicts the Change in r Γ— v

Along rβƒ—(t)=⟨t,Β t2+1⟩\vec{r}(t) = \langle t,\ t^2 + 1 \rangle, only the order-preserving product rule gives the true slope of rβƒ—Γ—vβƒ—\vec{r} \times \vec{v}.

IIIWhen Does r Γ— a Vanish?

∣rβƒ—Γ—aβƒ—βˆ£|\vec{r} \times \vec{a}| is the parallelogram's area, so rβƒ—Γ—aβƒ—=0βƒ—\vec{r} \times \vec{a} = \vec{0} exactly when aβƒ—\vec{a} lies along rβƒ—\vec{r}.

IVEqual Areas Need a Force Aimed at the Sun

A planet sweeps equal areas about the sun in equal times exactly when the force on it points at the sun.

πŸ’‘ Reverse the force so that it pushes away from CC and the areas about CC are still equal: Kepler's second law cannot tell attraction from repulsion.

VQuiz Questions

Problem 1 Β· Product Rule for a Cross Product

Given: uβƒ—(t)=⟨t,Β 1,Β 0⟩\vec{u}(t) = \langle t,\ 1,\ 0 \rangle and wβƒ—(t)=⟨1,Β t2,Β 0⟩\vec{w}(t) = \langle 1,\ t^2,\ 0 \rangle. Find ddt(uβƒ—Γ—wβƒ—)\frac{d}{dt}(\vec{u} \times \vec{w}) at t=1t = 1.

βœ… Correct! duβƒ—dtΓ—wβƒ—=⟨0,0,t2⟩\frac{d\vec{u}}{dt} \times \vec{w} = \langle 0, 0, t^2 \rangle and uβƒ—Γ—dwβƒ—dt=⟨0,0,2t2⟩\vec{u} \times \frac{d\vec{w}}{dt} = \langle 0, 0, 2t^2 \rangle, which add to ⟨0,0,3⟩\langle 0, 0, 3 \rangle at t=1t = 1.
❌ Check the order. You wrote the second term as dwβƒ—dtΓ—uβƒ—\frac{d\vec{w}}{dt} \times \vec{u}, which equals βˆ’β€‰uβƒ—Γ—dwβƒ—dt-\,\vec{u} \times \frac{d\vec{w}}{dt} and flips its sign. Keep uβƒ—\vec{u} on the left.
❌ Only one term. The product rule has two terms: differentiate uβƒ—\vec{u} with wβƒ—\vec{w} held, then wβƒ—\vec{w} with uβƒ—\vec{u} held, and add.
❌ Not quite. Compute duβƒ—dtΓ—wβƒ—+uβƒ—Γ—dwβƒ—dt\frac{d\vec{u}}{dt} \times \vec{w} + \vec{u} \times \frac{d\vec{w}}{dt} with the factors in their original order.
Show solution

The derivatives are duβƒ—dt=⟨1,0,0⟩\frac{d\vec{u}}{dt} = \langle 1, 0, 0 \rangle and dwβƒ—dt=⟨0,2t,0⟩\frac{d\vec{w}}{dt} = \langle 0, 2t, 0 \rangle. Keeping each factor on its side:

duβƒ—dtΓ—wβƒ—=⟨1,0,0βŸ©Γ—βŸ¨1,t2,0⟩=⟨0,Β 0,Β t2⟩\frac{d\vec{u}}{dt} \times \vec{w} = \langle 1, 0, 0 \rangle \times \langle 1, t^2, 0 \rangle = \langle 0,\ 0,\ t^2 \rangle uβƒ—Γ—dwβƒ—dt=⟨t,1,0βŸ©Γ—βŸ¨0,2t,0⟩=⟨0,Β 0,Β 2t2⟩\vec{u} \times \frac{d\vec{w}}{dt} = \langle t, 1, 0 \rangle \times \langle 0, 2t, 0 \rangle = \langle 0,\ 0,\ 2t^2 \rangle ddt(uβƒ—Γ—wβƒ—)=⟨0,Β 0,Β 3t2⟩⟹⟨0,Β 0,Β 3⟩ atΒ t=1\frac{d}{dt}(\vec{u} \times \vec{w}) = \langle 0,\ 0,\ 3t^2 \rangle \quad\Longrightarrow\quad \langle 0,\ 0,\ 3 \rangle \text{ at } t = 1

Check: directly, uβƒ—Γ—wβƒ—=⟨0,0,t3βˆ’1⟩\vec{u} \times \vec{w} = \langle 0, 0, t^3 - 1 \rangle, whose derivative is ⟨0,0,3t2⟩\langle 0, 0, 3t^2 \rangle. The swapped rule would give t2βˆ’2t2=βˆ’t2t^2 - 2t^2 = -t^2, i.e. ⟨0,0,βˆ’1⟩\langle 0, 0, -1 \rangle: wrong sign on the second term.

Problem 2 Β· The Derivative of r Γ— v

Given: a particle moves along rβƒ—(t)=⟨1,Β t,Β t2⟩\vec{r}(t) = \langle 1,\ t,\ t^2 \rangle. Find ddt(rβƒ—Γ—vβƒ—)\frac{d}{dt}(\vec{r} \times \vec{v}) at t=1t = 1.

βœ… Correct! The derivative is vβƒ—Γ—vβƒ—+rβƒ—Γ—aβƒ—=rβƒ—Γ—aβƒ—=⟨2t,βˆ’2,0⟩\vec{v} \times \vec{v} + \vec{r} \times \vec{a} = \vec{r} \times \vec{a} = \langle 2t, -2, 0 \rangle, so ⟨2,βˆ’2,0⟩\langle 2, -2, 0 \rangle at t=1t = 1.
❌ Check the order. That is aβƒ—Γ—rβƒ—=βˆ’β€‰rβƒ—Γ—aβƒ—\vec{a} \times \vec{r} = -\,\vec{r} \times \vec{a}. The product rule keeps rβƒ—\vec{r} on the left: the surviving term is rβƒ—Γ—aβƒ—\vec{r} \times \vec{a}.
❌ That is rβƒ—Γ—vβƒ—\vec{r} \times \vec{v} itself. At t=1t = 1, rβƒ—Γ—vβƒ—=⟨1,βˆ’2,1⟩\vec{r} \times \vec{v} = \langle 1, -2, 1 \rangle; the question asks for its rate of change.
❌ Not every motion obeys Kepler's law. The derivative is zero only for a central force; here aβƒ—=⟨0,0,2⟩\vec{a} = \langle 0, 0, 2 \rangle is not parallel to rβƒ—\vec{r}.
❌ Not quite. Use ddt(rβƒ—Γ—vβƒ—)=rβƒ—Γ—aβƒ—\frac{d}{dt}(\vec{r} \times \vec{v}) = \vec{r} \times \vec{a}.
Show solution

Differentiate: vβƒ—=⟨0,1,2t⟩\vec{v} = \langle 0, 1, 2t \rangle and aβƒ—=⟨0,0,2⟩\vec{a} = \langle 0, 0, 2 \rangle. By the product rule with the order kept,

ddt(rβƒ—Γ—vβƒ—)=vβƒ—Γ—vβƒ—+rβƒ—Γ—aβƒ—=0βƒ—+rβƒ—Γ—aβƒ—.\frac{d}{dt}(\vec{r} \times \vec{v}) = \vec{v} \times \vec{v} + \vec{r} \times \vec{a} = \vec{0} + \vec{r} \times \vec{a}. rβƒ—Γ—aβƒ—=⟨1,t,t2βŸ©Γ—βŸ¨0,0,2⟩=⟨tβ‹…2βˆ’t2β‹…0,Β Β t2β‹…0βˆ’1β‹…2,Β Β 1β‹…0βˆ’tβ‹…0⟩=⟨2t,Β βˆ’2,Β 0⟩\vec{r} \times \vec{a} = \langle 1, t, t^2 \rangle \times \langle 0, 0, 2 \rangle = \langle t \cdot 2 - t^2 \cdot 0,\ \ t^2 \cdot 0 - 1 \cdot 2,\ \ 1 \cdot 0 - t \cdot 0 \rangle = \langle 2t,\ -2,\ 0 \rangle

At t=1t = 1 this is ⟨2,βˆ’2,0⟩\langle 2, -2, 0 \rangle. Check: rβƒ—Γ—vβƒ—=⟨t2,βˆ’2t,1⟩\vec{r} \times \vec{v} = \langle t^2, -2t, 1 \rangle directly, and its derivative is ⟨2t,βˆ’2,0⟩\langle 2t, -2, 0 \rangle. Because it is not zero, this force is not central.

Problem 3 Β· Equal Areas With No Force at All

Given: a particle drifts in a straight line, rβƒ—(t)=⟨3t,Β 2,Β 0⟩\vec{r}(t) = \langle 3t,\ 2,\ 0 \rangle, with no force acting. Find rβƒ—Γ—vβƒ—\vec{r} \times \vec{v} and the area swept out by the segment from the origin to the particle between t=0t = 0 and t=4t = 4.

What is r⃗×v⃗\vec{r} \times \vec{v}?

What area is swept out from t=0t = 0 to t=4t = 4?

βœ… Correct! rβƒ—Γ—vβƒ—\vec{r} \times \vec{v} is constant because aβƒ—=0βƒ—\vec{a} = \vec{0} is trivially parallel to rβƒ—\vec{r}: a free particle sweeps area at the constant rate 12β‹…6=3\tfrac{1}{2} \cdot 6 = 3.
❌ Check the order. That is vβƒ—Γ—rβƒ—\vec{v} \times \vec{r}. Here rβƒ—Γ—vβƒ—\vec{r} \times \vec{v} has kk-component (3t)(0)βˆ’(2)(3)=βˆ’6(3t)(0) - (2)(3) = -6.
❌ aβƒ—=0βƒ—\vec{a} = \vec{0} does not make rβƒ—Γ—vβƒ—\vec{r} \times \vec{v} zero. It makes its derivative rβƒ—Γ—aβƒ—\vec{r} \times \vec{a} zero; rβƒ—\vec{r} and vβƒ—\vec{v} themselves are not parallel.
❌ Check rβƒ—Γ—vβƒ—\vec{r} \times \vec{v}. With vβƒ—=⟨3,0,0⟩\vec{v} = \langle 3, 0, 0 \rangle, only the kk-component survives: xvyβˆ’yvxx v_y - y v_x.
❌ Missing the 12\tfrac{1}{2}. The area rate is 12∣rβƒ—Γ—vβƒ—βˆ£\tfrac{1}{2}|\vec{r} \times \vec{v}|, not ∣rβƒ—Γ—vβƒ—βˆ£|\vec{r} \times \vec{v}|.
❌ Check the area. Multiply the constant area rate 12∣rβƒ—Γ—vβƒ—βˆ£\tfrac{1}{2}|\vec{r} \times \vec{v}| by the elapsed time 44.
Show solution

Step 1: vβƒ—=⟨3,0,0⟩\vec{v} = \langle 3, 0, 0 \rangle, so

rβƒ—Γ—vβƒ—=⟨3t,2,0βŸ©Γ—βŸ¨3,0,0⟩=⟨0,Β 0,Β (3t)(0)βˆ’(2)(3)⟩=⟨0,Β 0,Β βˆ’6⟩.\vec{r} \times \vec{v} = \langle 3t, 2, 0 \rangle \times \langle 3, 0, 0 \rangle = \langle 0,\ 0,\ (3t)(0) - (2)(3) \rangle = \langle 0,\ 0,\ -6 \rangle.

It is constant, as it must be: a⃗=0⃗\vec{a} = \vec{0} gives ddt(r⃗×v⃗)=r⃗×a⃗=0⃗\frac{d}{dt}(\vec{r} \times \vec{v}) = \vec{r} \times \vec{a} = \vec{0}. (The minus sign only says the sweep is clockwise.)

Step 2: the area rate is 12∣rβƒ—Γ—vβƒ—βˆ£=3\tfrac{1}{2}|\vec{r} \times \vec{v}| = 3, so over 44 time units

area=3β‹…4=12.\text{area} = 3 \cdot 4 = 12.

Check: the swept region is the triangle with vertices (0,0)(0, 0), (0,2)(0, 2), (12,2)(12, 2), whose area is 12β‹…12β‹…2=12\tfrac{1}{2} \cdot 12 \cdot 2 = 12.

Problem 4 Β· Aimed at an Axis, Not the Origin

Given: a particle moves on the helix rβƒ—(t)=⟨cos⁑2t,Β sin⁑2t,Β t⟩\vec{r}(t) = \langle \cos 2t,\ \sin 2t,\ t \rangle. Its acceleration aβƒ—=βŸ¨βˆ’4cos⁑2t,Β βˆ’4sin⁑2t,Β 0⟩\vec{a} = \langle -4\cos 2t,\ -4\sin 2t,\ 0 \rangle always points straight at the zz-axis. Find rβƒ—Γ—aβƒ—\vec{r} \times \vec{a} at t=Ο€4t = \frac{\pi}{4} and decide whether the force is central.

What is r⃗×a⃗\vec{r} \times \vec{a} at t=π4t = \frac{\pi}{4}?

Is the force central (with respect to the origin)?

βœ… Correct! "Central" means along the line through the origin. The helix never lies in one plane, so rβƒ—Γ—vβƒ—\vec{r} \times \vec{v} keeps turning and Kepler's second law fails.
❌ The axis is not the origin. aβƒ—\vec{a} points at (0,0,t)(0, 0, t), but rβƒ—\vec{r} runs from (0,0,0)(0, 0, 0); the two are parallel only when t=0t = 0.
❌ Check the order. That is aβƒ—Γ—rβƒ—=βˆ’β€‰rβƒ—Γ—aβƒ—\vec{a} \times \vec{r} = -\,\vec{r} \times \vec{a}.
❌ Check the cross product. At t=Ο€4t = \frac{\pi}{4}, rβƒ—=⟨0,1,Ο€4⟩\vec{r} = \langle 0, 1, \frac{\pi}{4} \rangle and aβƒ—=⟨0,βˆ’4,0⟩\vec{a} = \langle 0, -4, 0 \rangle.
❌ Not quite. Central means aβƒ—βˆ₯rβƒ—\vec{a} \parallel \vec{r}, i.e. rβƒ—Γ—aβƒ—=0βƒ—\vec{r} \times \vec{a} = \vec{0}; the size of aβƒ—\vec{a} and attraction versus repulsion are irrelevant.
Show solution

At t=Ο€4t = \frac{\pi}{4}: cos⁑2t=0\cos 2t = 0 and sin⁑2t=1\sin 2t = 1, so rβƒ—=⟨0,1,Ο€4⟩\vec{r} = \langle 0, 1, \frac{\pi}{4} \rangle and aβƒ—=⟨0,βˆ’4,0⟩\vec{a} = \langle 0, -4, 0 \rangle.

rβƒ—Γ—aβƒ—=⟨(1)(0)βˆ’Ο€4(βˆ’4),Β Β Ο€4(0)βˆ’(0)(0),Β Β (0)(βˆ’4)βˆ’(1)(0)⟩=βŸ¨Ο€,Β 0,Β 0⟩\vec{r} \times \vec{a} = \left\langle (1)(0) - \tfrac{\pi}{4}(-4),\ \ \tfrac{\pi}{4}(0) - (0)(0),\ \ (0)(-4) - (1)(0) \right\rangle = \langle \pi,\ 0,\ 0 \rangle

In general rβƒ—Γ—aβƒ—=⟨4tsin⁑2t,Β βˆ’4tcos⁑2t,Β 0⟩\vec{r} \times \vec{a} = \langle 4t \sin 2t,\ -4t \cos 2t,\ 0 \rangle, nonzero for every tβ‰ 0t \neq 0. So ddt(rβƒ—Γ—vβƒ—)β‰ 0βƒ—\frac{d}{dt}(\vec{r} \times \vec{v}) \neq \vec{0}: the direction of rβƒ—Γ—vβƒ—\vec{r} \times \vec{v} changes, matching the fact that a helix does not lie in a plane through the origin.

The force points at the zz-axis, not at the origin, so it is not central. Constant ∣aβƒ—βˆ£|\vec{a}| is irrelevant, and a central force may point either toward or away from the origin.

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