18.02A PROBLEM SET 9 SOLUTIONS

1. (10 points) The Lasker Rink in the northeast corner of Central Park is a perfect circle having a radius of 50 meters. A skater moves counterclockwise around the edge of the rink at a speed of 5 meters per second. We’ll put a coordinate system on the rink with origin at the center of the rink, the x-axis running east (toward Fifth Avenue) and the y-axis running north (toward Central Park North).

a) Suppose (x, y) is a point at the edge of the rink. Write down a non-zero tangent

vector

T(x, y) = A(x, y)i + B(x, y)j

to the rink at the point (x, y). (That is, write formulas for A(x, y) and B(x, y).)

There are lots of ways to do this. One way is to say that in order to be tangent to a curve f(x, y) = c, a vector has to be perpendicular to the gradient

∇f = \frac{∂f}{∂x} i + \frac{∂f}{∂y} j.

The circle of radius 50 has equation x^{2} + y^{2} = 2500, so we can take f(x, y) = x^{2} + y^{2}. Then the gradient is

∇f = 2xi + 2yj.

A vector perpendicular to the gradient is

T(x, y) = −2yi + 2xj.

(I got that by rotating the gradient through an angle of π/2.)

Another way to proceed is to write down parametric equations x(s) and y(s) for the edge of the rink and then to get a tangent vector by differentiating with respect to the parameter s:

\frac{dx}{ds}i + \frac{dy}{ds} j.

One natural parameter is the angle θ, leading to x(θ) = 50 cos θ, y(θ) = 50 sin θ. This gives the tangent vector

S(x(θ), y(θ)) = −50 sin θi + 50 cos θj = −y(θ)i + x(θ)j.

There are also more exotic choices, like

x(t) = 50(t^{2} − 1)/(t^{2} + 1), y(t) = 50(2t)/(t^{2} + 1).

(This covers all of the circle except the point (50, 0), as t runs over all real numbers). Such choices lead to more complicated formulas for tangent vectors, but they work.

b) Calculate the length of your tangent vector T(x, y). The first one I wrote has length

\sqrt{(−2y)^{2} + (2x)^{2}} = 2\sqrt{x^{2} + y^{2}} = 2 · 50 = 100.

c) The skater’s velocity vector at v(x, y) must be some multiple of your tangent vector:

v(x, y) = c(x, y)T(x, y).

Calculate c(x, y).

Since the tangent vector I wrote points counterclockwise, it corresponds to a velocity of 100 in the counterclockwise direction (with 100 being the length from part (b)). To get the skater’s velocity, I need to reduce that to 5, which I can do by multiplying by

c(x, y) = 1/20.

The result is

v(x, y) = (−y/10)i + (x/10)j.

d) Find a vector field F with the property that the line integral

∫_{C} F·dR

along the skater’s path measures the distance that the skater has travelled to the west. (For example, over a path beginning at the northern extremity of the rink and continuing to the western extremity, the line integral ought to be 50 meters.)

If F(x, y) = −i (a unit vector pointing due west), then F · v is always the westward component of v. It follows that F·dR is the westward motion over the small piece dR of C, so the line integral of F over C gives the total westward motion.

e) Suppose the skater’s path C begins at the point (50\sqrt{3}/2, 25), and continues exactly halfway around the rink to the point (−50\sqrt{3}/2, −25). Evaluate the line integral

∫_{C} F·dR

using your formulas for F and v = dR/dt. (It’s possible to get an answer without writing down any integrals precisely, but part of the problem is to write them down.)

Actually the outline indicated given here is a pretty slow way to compute the line integral; it’s probably easier to ignore the first parts of the problem, simply writing down any old parametrization of C and computing. But I’ll try to do what the problem suggests.

The idea is to use time as the parameter for the curve; that’s what makes dR/dt equal to the velocity. Since the skater is travelling halfway around the rink, a distance of πr = 50π meters, the time required is 10π seconds. The line integral is

∫_{0}^{10π} F(x(t), y(t))·v(x(t), y(t))dt
= ∫_{0}^{10π} −i·((−y(t)/10)i + (x(t)/10)j)dt
= ∫_{0}^{10π} y(t)/10 dt.

To finish the problem this way without thinking, you need a formula for y as a function of t. For motion around a circle of radius r at angular velocity ω, the magic formulas are

x(t) = r cos(ωt + A), y(t) = r sin(ωt + A).

Here A is the angular position at time t = 0; in our case π/6 (which I calculated as the inverse tangent of y/x at time 0; that is, as the inverse tangent of (25)/(50\sqrt{3}/2) = 1/\sqrt{3}). Since the circumference of 100π is covered in time 20π seconds, the angular velocity ω is 2π/(20π) = 1/10 radians per second. Therefore

y(t) = 50 sin(t/10 + π/6),

and

∫_{C} F·dR = ∫_{0}^{10π} 5 sin(t/10 + π/6) dt.
= −50 cos(t/10 + π/6)]_{0}^{10π} = −50(cos(7π/6) − cos(π/6))
= −50(−\sqrt{3}/2 − \sqrt{3}/2) = 50\sqrt{3}.

Remember that this line integral is supposed to measure distance travelled west; so its value should be just the decrease in the x coordinate over C. That’s 50\sqrt{3}/2 − (−50\sqrt{3}/2), which agrees.

2. (8 points) This problem is about whirlpools. A vector field F in the plane (possibly not defined at the origin) is called a whirlpool if its direction at every point is perpendicular to the radius. The most general whirlpool vector field looks like

F(x, y) = −f(r)yi + f(r)xj,

with f a function of the radius r = \sqrt{x^{2} + y^{2}}.

a) How much work is done by the whirlpool force field in moving counterclockwise around the circle of radius r centered at the origin? (The answer depends on f(r).)

A whirlpool vector field always points along the circle, and has magnitude r|f(r)|. It points counterclockwise if f is positive and clockwise if f is negative. The work done by F in moving counterclockwise around the circle of radius r is therefore 2πr (the length of the circle) times rf(r) (the magnitude of the force). This is 2πr^{2}f(r).

b) The coordinates of F are

M(x, y) = −f(r)y, N(x, y) = f(r)x.

Suppose f(r) = 1/r^{2}, so that the whirlpool field has magnitude 1/r. Calculate ∂M/∂y and ∂N/∂x.

We have

M(x, y) = \frac{−y}{x^{2} + y^{2}} , N(x, y) = \frac{x}{x^{2} + y^{2}} .

Therefore

\frac{∂M}{∂y} = \frac{(x^{2} + y^{2})(−1) − (2y)(−y)}{(x^{2} + y^{2})^{2}} = \frac{y^{2} − x^{2}}{(x^{2} + y^{2})^{2}} .

\frac{∂N}{∂x} = \frac{(x^{2} + y^{2})(1) − (2x)(x)}{(x^{2} + y^{2})^{2}} = \frac{y^{2} − x^{2}}{(x^{2} + y^{2})^{2}} .

The thing to notice is that ∂M/∂y = ∂N/∂x.

c) Explain why part (a) shows that the only conservative whirlpool field is the zero field. Explain why part (b) shows that the whirlpool field with f(r) = 1/r^{2} is conservative. [In the original problem, there were misprints here: a reference to “part (c)” and to “f(r) = 1/r.”] Explain this apparent contradiction.

A vector field is conservative exactly when its integral around any closed path is zero. Part (a) shows that the integral of the whirlpool field around the (closed) circle of radius r is 2πr^{2}f(r). This is zero for all r if and only if f is identically zero; that is, if and only if F = 0.

Part (b) shows that the field with f(r) = 1/r^{2} satisfies the derivative condition curl F = 0 to be conservative. Although this condition is necessary for being conservative (every conservative field has to satisfy it), the condition is sufficient only in regions with no holes. The region where F is defined (the whole plane except the origin) has a hole at the origin. So there is no contradiction.

3. (7 points) Problem 21.1(12) on page 757. Since this was a straightforward calculation, I won’t write out a solution.