#ThisWeeksFiddler, 20261009

This week the #puzzle is: How Fast Can You Surf? #trigonometry #maximum #differentiation

A surfer is riding along a crest of a sinusoidal wave given by the function y = 2·sin(x−t), where y is the height above sea level in meters, x is horizontal position in meters, and t is time in seconds. Based on this, the surfer knows that her speed—and the speed of the wave—is 1 meter per second.

But hold on—it appears that another wave is approaching from the opposite direction. This wave has half the amplitude but twice the speed: y = sin(x+2t).

The two waves, the forward and the backward, ”interfere,” meaning they combine. In other words, the height of the water at position x and time t is given by y = 2·sin(x−t) + sin(x+2t). Nevertheless, the surfer stays atop a crest of the combined wave.

As the wave moves along, what are the surfer’s slowest and fastest [horizontal] speeds, in meters per second?

And for extra credit:

The surfer rides the crest of the combined wave, y = 2·sin(x−t) + sin(x+2t), for a very long time.

What is her average [horizontal] speed, in meters per second?

How Fast Can You Surf?

Solution, possibly incorrect:

Desmos Program

Method 1: Code it up in Desmos, just to see what’s happening. Huh! The combined wave (purple) has a funny start and stop pattern, where horizontal movement seems to stop entirely, while the wave grows in size.

Method 2: Math. Everybody seems to agree, when $v$ = speed of wave (or in out case, speed of crest), $\lambda$ = wave length and $f$ = wave frequency:

v=\lambda f

This can’t quite fit our situation, because the speed obviously varies, while the wave length and the wave frequency stay constant.

Let’s just look at a single moment.

t=0, w(x)=2sin(x-t)+sin(x+2t)

w(x)=2sin(x)+sin(x)=3sin(x)

When does this function have a max or min?

w'(x)=3 cos(x),w'(x)=0

3cos(x)=0

cos(x)=0

x=arccos(0)=\frac{\pi}{2}

And this max/min is:

w(\frac{\pi}{2})=3sin(\frac{\pi}{2})=3 \cdot 1 = 3

Yeah, actually a max. So at time 0, the crest (or one crest) is at $(\frac{\pi}{2},3)$ .

Let’s try this with a more general time instead.

t=c,w(x)=2sin(x-t)+sin(x+2t)

Chain rule:

w'(x)=2cos(x-t)\cdot 1+cos(x+2t)\cdot 1

Looking for a max/min:

0=2cos(x-t)\cdot 1+cos(x+2t)\cdot 1

0=2cos(x-t)+cos(x+2t)

Identities:

0=2[cos(x)cos(-t))-sin(x)sin(-t)]+cos(x)cos(2t)-sin(x)sin(2t)

0=2cos(x)cos(-t)-2sin(x)sin(-t)+cos(x)cos(2t)-sin(x)sin(2t)

0=cos(x)[2cos(-t)+cos(2t)]-sin(x)[2sin(-t)+sin(2t)]

cos(x)[2cos(-t)+cos(2t)]=sin(x)[2sin(-t)+sin(2t)]

\frac{2cos(-t)+cos(2t)}{2sin(-t)+sin(2t)}=\frac{sin(x)}{cos(x)}

\arctan(\frac{2cos(-t)+cos(2t)}{2sin(-t)+sin(2t)})=x

I try out this function in Desmos. It actually follows the minimum, but the math is the same.

So far so good! A visual inspection shows these values for the speed of the crest.

v_{min}=0(t=0,\frac{2\pi}{3}),v_{max}=?(t=\frac{\pi}{3})

Doing a test where I look at how much the crest moves over a short amount of time, $\epsilon$ , 0 is confirmed. And the max value seems to be 4!

Method 3: Write a program. This simply supplies Desmos with data to show the path of the crest and a curve for the speed of the crest. As it uses the same math, it spits out the same results.

Result: $0 \le v \le 4$ .

And for extra credit:

This is suspiciously easy.

Average speed = total distance / total time. From reading Desmos again, measuring between 2 maxima:

t=0,x=\frac{\pi}{2}

t=\frac{2\pi}{3},x=\frac{7\pi}{6}

v=\frac{\frac{7\pi}{6}-\frac{\pi}{2}}{\frac{2\pi}{3}}

v=\frac{\frac{7\pi}{6}-\frac{3\pi}{6}}{\frac{2\pi}{3}}

v=\frac{\frac{4\pi}{6}}{\frac{2\pi}{3}}

v=\frac{4\pi}{6} \cdot \frac{3}{2\pi}

v=\frac{2}{2}=1

Result: 1.

Stuff from ommadawn.dk (Lise Andreasen)

#ThisWeeksFiddler, 20261009

Solution, possibly incorrect:

And for extra credit:

Skriv et svar Annuller svar

Solution, possibly incorrect:

And for extra credit:

Del dette:

Skriv et svar Annuller svar