+0  
 
0
214
1
avatar

Please solve this for me..

Guest Dec 31, 2014

Best Answer 

 #1
avatar+85644 
+10

Here's (b)

x^2 + xy - y^2  = 1    using implicit differentiation, we have

2x + y + xy' - 2yy'  = 0

xy' - 2yy'  = -2x - y       multiply through by - 1 on both sides

2yy' - xy'  = 2x + y

y'(2y - x) = 2x + y

y' =  [2x + y ] /  [2y - x ]      and the slope of the tangent line at (2.3) is given by

y' = [2(2) + 3] / [2(3) - 2 ]   =   [7] / [ 6 - 2]  =  7 / 4  

Here's a graph of the tangent line to the "rotated" hyperbola at the point (2,3)....https://www.desmos.com/calculator/yuxxmj7ud9

Note, that the slope of the tangent line at this point = (7/4).....just as we expected...!!!

 

CPhill  Dec 31, 2014
Sort: 

1+0 Answers

 #1
avatar+85644 
+10
Best Answer

Here's (b)

x^2 + xy - y^2  = 1    using implicit differentiation, we have

2x + y + xy' - 2yy'  = 0

xy' - 2yy'  = -2x - y       multiply through by - 1 on both sides

2yy' - xy'  = 2x + y

y'(2y - x) = 2x + y

y' =  [2x + y ] /  [2y - x ]      and the slope of the tangent line at (2.3) is given by

y' = [2(2) + 3] / [2(3) - 2 ]   =   [7] / [ 6 - 2]  =  7 / 4  

Here's a graph of the tangent line to the "rotated" hyperbola at the point (2,3)....https://www.desmos.com/calculator/yuxxmj7ud9

Note, that the slope of the tangent line at this point = (7/4).....just as we expected...!!!

 

CPhill  Dec 31, 2014

7 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details