/sci/ - Science & Math » Thread #11457196

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plz help me

Anonymous Tue 10 Mar 17:51:26 2020 No.11457196 View Reply Original Report

Quoted By: >>11457207 >>11457260 >>11457264

I'm currently taking trigonometry. My professor decided for some reason to assign a calculus problem that deals with related rates, and optimization. If someone could help explain how to solve it I'd really appreciate it.

Anonymous

Anonymous Tue 10 Mar 2020 17:53:19 No.11457207 Report

Quoted By:

>>11457196
>top speed of 4.5m/sec on land
What kind of hobbling fatass lifeguard is this?

Anonymous

Anonymous Tue 10 Mar 2020 18:06:25 No.11457243 Report

Quoted By: >>11457252

put up an equation for the time the life guard takes to get to the swimmer. it should look something like this:
time total = time moving on land + time moving in water
make the distance a variable and you will get an equation, then look for the minimum. use pythagoras for the distance for water.

Anonymous

Anonymous Tue 10 Mar 2020 18:08:15 No.11457252 Report

Quoted By:

>>11457243
Thank you so much anon. I think I can take it from there.

Anonymous

Anonymous Tue 10 Mar 2020 18:11:45 No.11457260 Report

Quoted By:

>>11457196
the vertex of a curve of the form
$y = bx + \sqrt{c^2 + x^2}$ occurs when the derivative $y'=b + 2x\sqrt{c^2 + x^2}$ equals zero. That's all the calculus that's required for the problem, and it looks like it is indeed necessary to use this fact.

Anonymous

Anonymous Tue 10 Mar 2020 18:13:53 No.11457264 Report

Quoted By: >>11457308

>>11457196
You can use Snell's law.
$ \sin(90° - \theta) = \frac{v_2}{v_1}\\ \theta = 90° - \arcsin\left( \frac{v_2}{v_1} \right) $
The distance he should run on the shore is then just
$ d - \frac{c}{\tan(\theta)} \approx 47.4m $
and the minimum rescue time is
$ \frac{d - \frac{c}{\tan(\theta)}}{v_1} + \frac{\frac{c}{\sin(\theta)}}{v_2} \approx 23.9 s $

At last you see as a sanity check
$\frac{\sqrt{c^2 + d^2}} {v_2} \approx 36.6s \\ \frac{d} {v_1} + \frac{c}{v_2} \approx 24.7 s $
So spending maximum or minimum time in water is worse.

Anonymous

Anonymous Tue 10 Mar 2020 18:25:54 No.11457308 Report

Quoted By:

>>11457264
Thank you! That’s actually what I had done, I just figured I was wrong since 24.7 is so close to 23.9

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