+0  
 
+14
836
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avatar+98152 

This one should be a little easier than the first Challenge..........!!!

 

Given the hyperbola y = 1/x   with its graph restricted to the first quadrant.......prove that, for any point b on the hyperbola in that quadrant, the triangle with the the vertices (0,0), (b, 1/b), (b, 0 )   always has a constant area........

 

Part 2.........show that for any point on the hyperbola with an x  value of  b >1, that the area bounded by the hyperbola, the x axis, the line x = 1 and the line x = b  equals  ln b

 

Here's a pic to get you started..........

 

 

cool cool cool

 Oct 31, 2015
edited by CPhill  Oct 31, 2015

Best Answer 

 #4
avatar+99324 
+5

For the triangle the area is 

.5 x b x 1/b = 0.5 which is a constant.

 

For the second one Area= int of 1/x  dx = (ln x ) from 1 to b = lnb

  Not much of a challenge CPhill.  Hahaha.

 Oct 31, 2015
 #1
avatar+99324 
+5

That is an hyperbola CPhill. ☺

 Oct 31, 2015
edited by Guest  Oct 31, 2015
 #2
avatar+98152 
0

Oops....thanks, Guest....my bad!!!!!

 

 

 

cool cool cool

 Oct 31, 2015
 #3
avatar+99324 
0

That extra skew line from the origin doesn't seem to make sense ://

 Oct 31, 2015
 #4
avatar+99324 
+5
Best Answer

For the triangle the area is 

.5 x b x 1/b = 0.5 which is a constant.

 

For the second one Area= int of 1/x  dx = (ln x ) from 1 to b = lnb

  Not much of a challenge CPhill.  Hahaha.

Melody Oct 31, 2015
 #5
avatar+98152 
0

Guest....look at the first part of the question, again...it's one side of the triangle whose vertices are (0,0), (b, 1/b) and (b, 0)

 

 

 

 

cool cool cool

 Oct 31, 2015
 #6
avatar+98152 
+5

Good going, Melody.....!!!!   5 points from me!!!!.......yeah....it's not really that difficult.....!!!!!

 

 

 

cool cool cool

 Oct 31, 2015
edited by CPhill  Oct 31, 2015
 #7
avatar+99324 
0

Thanks CPhill

What am I supposed to look at.  I don't understand:/

Sorry, I get it  :))

 Oct 31, 2015
edited by Guest  Oct 31, 2015

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