A calculus problem by Jaypee Dela Peña

Calculus Level 2

What is the normal line to the curve y=3x-x^2 at point (2,0)?

y=2x-3 y=X+2 y=x-2 y=2x+1

Jaypee Dela Peña by Jaypee Dela Peña , 26, Philippines

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2 solutions

Samarth Sangam
Aug 19, 2014

y = 3 x − x 2 d i f f e r e n t i a t e w . r . t . x d y d x = 3 − 2 x s u b s t i t u t e t h e p o i n t ( 2 , 0 ) i n d y d x w e g e t d y d x = − 1 t h e e q u a t i o n o f n o r m a l t o t h e c u r v e 1 i s g i v e n b y y − 0 = − ( x − 2 ) / d y d x t h e r e f o r e t h e e q u a t i o n o f t h e n o r m a l i s y = x − 2 y=3x-{ x }^{ 2 }\\ differentiate\quad w.r.t.\quad x\\ \frac { dy }{ dx } =3-2x\\ substitute\quad the\quad point\quad (2,0)\quad in\quad \frac { dy }{ dx } \\ we\quad get\quad \frac { dy }{ dx } =-1\\ the\quad equation\quad of\quad normal\quad to\quad the\quad curve\quad 1\quad is\quad given\quad by\\ y-0=-(x-2)/\frac { dy }{ dx } \\ therefore\quad the\quad equation\quad of\quad the\quad normal\quad is\quad y=x-2

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y = 3 x − x 2 d i f f e r e n t i a t e w . r . t . x d y d x = 3 − 2 x s u b s t i t u t e t h e p o i n t ( 2 , 0 ) i n d y d x w e g e t d y d x = − 1 t h e e q u a t i o n o f n o r m a l t o t h e c u r v e 1 i s g i v e n b y y − 0 = − ( x − 2 ) / d y d x t h e r e f o r e t h e e q u a t i o n o f t h e n o r m a l i s y = x − 2 y=3x-{ x }^{ 2 }\ differentiate\quad w.r.t.\quad x\ \frac { dy }{ dx } =3-2x\ substitute\quad the\quad point\quad (2,0)\quad in\quad \frac { dy }{ dx } \ we\quad get\quad \frac { dy }{ dx } =-1\ the\quad equation\quad of\quad normal\quad to\quad the\quad curve\quad 1\quad is\quad given\quad by\ y-0=-(x-2)/\frac { dy }{ dx } \ therefore\quad the\quad equation\quad of\quad the\quad normal\quad is\quad y=x-2 sorry for the format in solution

samarth sangam - 6 years, 9 months ago
Hasan Kassim
Aug 3, 2014

only one of the given lines passes the point (2,0), which is y=x-2 .

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