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 one year ago
In the assifnments from this course on the 1 problem set, in exercise 3 I simply dont understand what I'm doing wrong.
y = 1000  x^2
y' = 2x
I need to find the line that is tangent to y and that also passes through (0,1100).
I get yyo=y'(xo)(xxo)
But i dont know (xo,yo)
Please help
Original question:
3. On the planet Quirk, a cell phone tower is a 100foot pole on top of a green
mound 1000 feet tall whose outline is described by the parabolic equation y = 1000 − x^2
. An ant climbs up the mound starting from ground level (y = 0). At what height y does the ant begin to
see the tower?
 one year ago
In the assifnments from this course on the 1 problem set, in exercise 3 I simply dont understand what I'm doing wrong. y = 1000  x^2 y' = 2x I need to find the line that is tangent to y and that also passes through (0,1100). I get yyo=y'(xo)(xxo) But i dont know (xo,yo) Please help Original question: 3. On the planet Quirk, a cell phone tower is a 100foot pole on top of a green mound 1000 feet tall whose outline is described by the parabolic equation y = 1000 − x^2 . An ant climbs up the mound starting from ground level (y = 0). At what height y does the ant begin to see the tower?

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Waynex
 one year ago
Best ResponseYou've already chosen the best response.1There is a nice explanation of the method to solve this at this link: http://math.stackexchange.com/a/137068 The post I linked to has the solution, but it might be helpful to read a few other posts there. Even with that information, the solution might be difficult. I'll help you get there, if needed. By the way, thanks for asking this question. I stumbled on this type of problem a few months back, and just couldn't crack it. That forum post I linked to helped a lot. And I did crack it. When you asked the question, that gave me an opportunity to see if I still remembered it. I vaguely did, but I had to refer back to that forum post lol.

matheusoliveira
 one year ago
Best ResponseYou've already chosen the best response.0Thanks a lot for the post. I just craked it. I'm really happy. It was really just a system because the tangent point belongs to both the tangent line and the curve :D thx a lot man!!

Waynex
 one year ago
Best ResponseYou've already chosen the best response.1You're welcome. Very true, I haven't tried to solve one that way. It did occur to me that setting it up as a system of equations might be easier than parameterizing a point on the curve.
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