I wonder where this came from

Calculus Level 3

Find the arc length of the following curve from $t=1$ to $t=1,000,001$ :

$x=\frac{1}{2}t(\cos(\ln(t))+\sin(\ln(t)))$

$y=\frac{1}{2}t(\cos(\ln(t))-\sin(\ln(t)))$

Bonus: Figure out how I used a tweaked result from my solution to this problem (assuming A is instead not constant) to come up with this question.

The answer is 1000000.

2 solutions

ChengYiin Ong
Jan 18, 2021

The derivatives turn out to be quite nice expressions :) $x'(t)=\cos (\ln (t)) \ , \ y'(t)=-\sin (\ln (t)).$ So, the arc length is given by $\int_{1}^{1000001} \sqrt{[x'(t)]^2+[y'(t)]^2} \, dt=\int_{1}^{1000001} \sqrt{(\cos (\ln (t)))^2+(-\sin (\ln (t)))^2} \, dt=\int_{1}^{1000001} 1 \, dt=\boxed{1000000}$

Veselin Dimov
Jan 19, 2021

I wonder if y(x) expressable I'm terms of normal functions (roots, exponents, trig and log)? I got to the differential equation $\frac{dy}{dx}=\frac{y-x}{y+x}$ and after a lot of wasted time I looked it up in Wolfram alpha where I didn't find what I wanted to.

Interesting supposition. That differential equation doesn't look intimidating, but, of course, looks can be deceiving. Maybe I'll take a look at it later when I finish what I'm doing now.

James Wilson - 4 months, 3 weeks ago

It's almost an exact differential equation but not quite. I was unable to crack it. I honestly doubt it is possible to represent y(x) explicitly using a finite combination of elementary functions.

James Wilson - 4 months, 3 weeks ago

I wonder where this came from

The answer is 1000000.

2 solutions

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