Calculus In Parabola

Calculus Level 5

Let S S be the focus of y 2 = 4 x . y^2 = 4x. Point P P is moving on this curve such that its abscissa is increasing at a rate of 4 4 units per second. Then the rate of increase of the length of projection of S P SP on x + y = 1 x + y =1 when P P is at ( 4 , 4 ) (4,4) is γ \gamma .

Find 1 0 3 γ 10^3\gamma to 1 decimal place.


The answer is -1414.2.

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2 solutions

Dhruva Patil
Feb 3, 2015

T h e l i n e x + y = 1 p a s s e s t h r o u g h B = ( 0 , 1 ) a n d A = ( 1 , 0 ) L e t t h e v e c t o r u b e t h e v e c t o r a l o n g t h a t l i n e ( B A ) . u = A B = < 0 1 , 1 0 > u = < 1 , 1 > = i ^ + j ^ U n i t v e c t o r a l o n g u i s u ^ = u u u ^ = < 1 2 , 1 2 > = 1 2 ( i ^ + j ^ ) U s i n g t h e e q u a t i o n y 2 = 4 x , w e g e t t h e f o c u s t o b e ( 1 , 0 ) a n d t h e d i r e c t r i x t o b e x = 1. D i s a n y p o i n t o n t h e p a r a b o l a w i t h c o o r d i n a t e s ( x , y ) L e t v b e t h e v e c t o r a l o n g B D v = D B v = < x 1 , y > = ( x 1 ) i ^ + y j ^ T h e l e n g t h o f t h e p r o j e c t i o n o f D B a l o n g l i n e y + x = 1 i s g i v e n b y L = v . u ^ = ( x 1 ) × ( 1 2 ) + ( y ) × ( 1 2 ) L = 1 2 ( 1 x + y ) U s i n g t h e p a r a b o l a e q u a t i o n L = 1 2 ( 1 x + 4 x ) D i f f e r e n t i a t i n g i t w i t h r e s p e c t t o s o m e v a r i a b l e t d L d t = 1 2 ( d x d t + 2 2 x d x d t ) A t p o i n t ( 4 , 4 ) d L d t = 1 2 ( d x d t + 1 4 d x d t ) d L d t = 1 2 2 d x d t G i v e n d x d t = 4 d L d t = 2 10 3 × ( 2 ) = 1414.2 The\quad line\quad x+y=1\quad passes\quad through\quad B=(0,1)\quad and\quad A=(1,0)\\ Let\quad the\quad vector\quad \vec { u } \quad be\quad the\quad vector\quad along\quad that\quad line\quad (\overrightarrow { BA } ).\\ \vec { u } =\overrightarrow { A } -\overrightarrow { B } =\left< 0-1,1-0 \right> \\ \vec { u } =\left< -1,1 \right> =-\hat { i } +\hat { j } \\ Unit\quad vector\quad along\vec { u } \quad is\\ \hat { u } =\frac { \vec { u } }{ \left| \vec { u } \right| } \\ \hat { u } =\left< \frac { -1 }{ \sqrt { 2 } } ,\frac { 1 }{ \sqrt { 2 } } \right> =\frac { 1 }{ \sqrt { 2 } } (-\hat { i } +\hat { j } )\\ \\ Using\quad the\quad equation\quad { y }^{ 2 }=4x,\quad we\quad get\quad the\quad focus\quad to\quad be\quad (1,0)\quad \\ and\quad the\quad directrix\quad to\quad be\quad x=-1.\\ D\quad is\quad any\quad point\quad on\quad the\quad parabola\quad with\quad coordinates\quad (x,y)\\ Let\quad \vec { v } \quad be\quad the\quad vector\quad along\quad BD\\ \vec { v } =\overrightarrow { D } -\overrightarrow { B } \\ \vec { v } =\left< x-1,y \right> =(x-1)\hat { i } +y\hat { j } \\ The\quad length\quad of\quad the\quad projection\quad of\quad DB\quad \\ along\quad line\quad y+x=1\quad is\quad given\quad by\\ L=\vec { v } .\hat { u } =(x-1)\times (\frac { -1 }{ \sqrt { 2 } } )+(y)\times (\frac { 1 }{ \sqrt { 2 } } )\\ L=\frac { 1 }{ \sqrt { 2 } } (1-x+y)\\ Using\quad the\quad parabola\quad equation\\ L=\frac { 1 }{ \sqrt { 2 } } (1-x+\sqrt { 4x } )\\ Differentiating\quad it\quad with\quad respect\quad to\quad some\quad variable\quad t\\ \frac { dL }{ dt } =\frac { 1 }{ \sqrt { 2 } } (-\frac { dx }{ dt } +\frac { 2 }{ 2\sqrt { x } } \frac { dx }{ dt } )\\ At\quad point\quad (4,4)\\ \frac { dL }{ dt } =\frac { 1 }{ \sqrt { 2 } } (-\frac { dx }{ dt } +\frac { 1 }{ \sqrt { 4 } } \frac { dx }{ dt } )\\ \frac { dL }{ dt } =\frac { -1 }{ 2\sqrt { 2 } } \frac { dx }{ dt } \\ Given\quad \frac { dx }{ dt } =4\\ \frac { dL }{ dt } =-\sqrt { 2 } \\ { 10 }^{ 3 }\times (-\sqrt { 2 } )=\boxed { \boxed { -1414.2 } } \\ \\

Lu Chee Ket
Feb 13, 2015

kindly provide image for second solution asap.

Goku Han - 4 years, 9 months ago

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