nunostreet Posted June 9, 2015 Report Share Posted June 9, 2015 So, my teacher gave me this past paper with a problem that i'm having difficulties to solve. So, it goes like this: Consider the curve with equation (x^2+y^2)^2 = 4xy^2(a) Use implicit differentiation to find an expression for dy/dx. (b) Find the equation of the normal to the curve at the point (1, 1). Really need some help, ASAP. Thanks! Reply Link to post Share on other sites More sharing options...
Vioh Posted June 10, 2015 Report Share Posted June 10, 2015 (edited) So, my teacher gave me this past paper with a problem that i'm having difficulties to solve. So, it goes like this: Consider the curve with equation (x^2+y^2)^2 = 4xy^2(a) Use implicit differentiation to find an expression for dy/dx. (b) Find the equation of the normal to the curve at the point (1, 1). Really need some help, ASAP. Thanks! haha, I recognized this question right at the moment when I read your post, because it's from my exam session. More specifically it is question 10 from paper 2 HL Math TZ2 (may 2014). If you haven't read the markscheme, then here it is: And if you have any problems understanding any of those steps, tell me, and I'll help you. Edited June 10, 2015 by Vioh 2 Reply Link to post Share on other sites More sharing options...
nunostreet Posted June 10, 2015 Author Report Share Posted June 10, 2015 You saved my day! Yeah I had no idea the year it belonged to, but now I know thank you! Reply Link to post Share on other sites More sharing options...
Awesomeness Posted June 10, 2015 Report Share Posted June 10, 2015 Hi Could anyone tell me where can I get the Nov/Dec 2014 past papers For Maths, Physics and Chemistry HL and also Bio Reply Link to post Share on other sites More sharing options...
Emmi Posted June 10, 2015 Report Share Posted June 10, 2015 Hi Could anyone tell me where can I get the Nov/Dec 2014 past papers For Maths, Physics and Chemistry HL and also Bio From your teacher, from the IB store, or on the Internet. We do not distribute past papers here because they are copyrighted property, please don't ask people for them. Reply Link to post Share on other sites More sharing options...
thetamaverick Posted June 15, 2015 Report Share Posted June 15, 2015 you can literally just google it. its also the first 2 results ... Hi Could anyone tell me where can I get the Nov/Dec 2014 past papers For Maths, Physics and Chemistry HL and also Bio Reply Link to post Share on other sites More sharing options...
Tsepiso Posted July 29, 2015 Report Share Posted July 29, 2015 So in all cases the gradient of a normal is equal to -1/(dy/dx)? Reply Link to post Share on other sites More sharing options...
kevG Posted July 29, 2015 Report Share Posted July 29, 2015 dy/dx = gradientwhen l1 is perpendicular to l2, then the l1 gradient = -1/gradient of l2 Reply Link to post Share on other sites More sharing options...
Fiona Posted August 6, 2015 Report Share Posted August 6, 2015 So in all cases the gradient of a normal is equal to -1/(dy/dx)? Yes, that's right. This is because the equation of the normal looks as follows: y - y0 = (-1/(dy/dx)) (x - x0) Hence, when you find the derivative everything will disappear except for (-1/(dy/dx)). Please note that in this case y0 and x0 are simply the points the normal passes through, so they are some numbers. Reply Link to post Share on other sites More sharing options...
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