help forming geometrical intuition for why dy/dt is as follows:

KrabLord

KrabLord

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gilbert strang, chapter 6.4.png
why is it the case that dy/dt is as stated? I cannot get myself to see why that is the case.
 
I would choose to solve this problem as follows:

The ant's absolute speed is:

[MATH]v=1+y[/MATH]
The ant's speed is it own relative to the band plus the portion of the band it has already traveled times the rate the band is being stretched.

Hence, we can arrange this as a linear first order ODE:

[MATH]\d{x}{t}-\frac{1}{t+2}x=1[/MATH]
Compute integrating factor:

[MATH]\mu(t)=\exp\left(-\int \frac{1}{t+2}\,dt\right)=\frac{1}{t+2}[/MATH]
[MATH]\frac{1}{t+2}\d{x}{t}-\frac{1}{(t+2)^2}x=\frac{1}{t+2}[/MATH]
[MATH]\frac{d}{dt}\left(\frac{1}{t+2}x\right)=\frac{1}{t+2}[/MATH]
Integrate:

[MATH]\frac{1}{t+2}x=\ln(t+2)+C[/MATH]
[MATH]x(t)=(t+2)(\ln(t+2)+C)[/MATH]
[MATH]x(0)=2(\ln(2)+C)=0\implies C=-\ln(2)[/MATH]
And so we find:

[MATH]x(t)=(t+2)(\ln(t+2)-\ln(2))[/MATH]
This does imply:

[MATH]y=\ln(t+2)-\ln(2)\implies y'=\frac{1}{t+2}[/MATH]
But, I don't think I would have thought to begin with that. And then to finish:

[MATH]y=\ln(t+2)-\ln(2)=1\implies \frac{t+2}{2}=e\implies T=2e-2[/MATH]
 
To find part (a) directly, let "s" be the length of the band and "x" be the absolute position of the ant...

Fraction \(\displaystyle y=\frac{x}{s}\)

\(\displaystyle sy=x\)

Differentiate wrt time

\(\displaystyle s\frac{dy}{dt} +\frac{ds}{dt}y = \frac{dx}{dt} \)

Since s=t+2, its derivative is 1

\(\displaystyle (t+2)\frac{dy}{dt} +y = \frac{dx}{dt} \)

Then use the ant's absolute speed, as described in post#2 by @MarkFL

\(\displaystyle (t+2)\frac{dy}{dt} +y = 1+y \)

You can take it from here!
 
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