Skiing Differential Equation Please help!
Ohh the Tratrix problem :O this looks familiar. hmm
first order equations, but this doesn't translate well in google translate for me
Were you able to figure out part a? Looks like we just use the Pythagorean Theorem, yes?
yes, AB= sqrt(400-y^2) is that right, please shoot me down if not
Ok good stuff.
|dw:1392348132869:dw|Sooooo this is what I'm thinking for part b.
\[\Large\bf\sf \tan(\pi- \theta)\quad=\quad \frac{y}{AB}\]
yaaas, keep going
Tangent is periodic in pi, therefore,\[\Large\bf\sf \tan(\pi-\theta)\quad=\quad \tan(-\theta)\]And since tangent is an odd function, it has this property,\[\Large\bf\sf \tan(-\theta)\quad=\quad -\tan \theta\]
That allows you to simplify the left side down nicely.
\[\Large\bf\sf \tan \theta\quad=\quad -\frac{y}{AB}\]I think that puts us on the right track there.. hmm
and ab is just sqrt(400-y^2), thank you
cool looks good. How bout part c? :)
wouldnt it just be the same thing as part b?
i don't know how to explain that
Ya I think it would be the same. Umm it has something to do with the fact thattttttt... That diagonal is `tangent to the curve. So the slope of that diagonal represents the derivative of the curve. And the slope is given by `rise` / `run`. Which in this case would be `-y` / `AB`. negative y since the y is moving down, yes?
yeah you're right , i thought iwas going to write a novel for this problem
looks like we're done here!
yah wasn't too bad! I thought they were going to make you find the anti-derivative, which is a bit more work :)
thanks for the help :D! it means the world to my education.
one more part to the question d) Solve the equation for y(x)
https://docs.google.com/document/d/1MCidnRDmwE2RxXkZio0ZbCfYEuAqOZxajf6KW7Yrpyo/edit click that please
oo a new part :O
yes a new part :(
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