Question

the tangent and normal to the curve y= sin x at the point P where x= π/3 cut the x-axis at A and B respectively
(a). show that AB = (5√3)/4.
(b). find the area of the triangle PAB.

Answer 1

the tangent to the curve is what..cos(pi/3) right?

Answer 2

the normal is 90 from that....

Answer 3

i dont quite understand the cuts the x axis at A and B

Answer 4

maybe x and y axis at A and B?

Answer 5

no

Answer 6

they mean the tangent cuts at a, and the normal cuts at b

Answer 7

let me post the digaram

Answer 8

its very simple

find the eqns of the lines , find the intercepts of those lines , then use distance formula

Answer 9

lol.... yeah, if you post it that would be more helpful :)

Answer 10

simple yes, once you understand the question ;)

Answer 11

just distance formula and area of a triangle = (1/2) bh , thats all there is , you dont even need a picture for this, should be able to draw one yourself, or visualise in your mind pretty easily

Answer 12

well the tangent is simply its derivative; cos(pi/3)

Answer 13

the equation of the line then becomes:

y-(sin(pi/3)) = cos(pi/3)(x-(pi/3))

Answer 14

cos(pi/3) = 1/2 right?

Answer 15

the normal will have a slope of -2 then

Answer 16

y-(sin(pi/3)) = -2(x-(pi/3)) for the normal equation

Answer 17

Guys i still cant understand it , can sun1 explain it to me step by step ? plz

Answer 18

they give you a point (pi/3, sin(pi/3)); or rather (pi/3, sqrt(3)/2) right?

Answer 19

first take the derivative of sine, which is cosine. then plug in \[\frac{\pi}{3}\] into the derivative to get the slope.

\[cos(\frac{\pi}{3})=\frac{1}{2}\] so that is the slope of the tangent line

Answer 20

at this level you should have some redimentary skills at a graph id assume :)

Answer 21

use the poit-slope form of a line to get the euation for the tangent and the norml

Answer 22

amistre is in right direction,find the eqn of tangent and normal to the curve, find the x intercepts of both to find out the coordiantes of the two points, point P is given, it forms the triangle ABP

Answer 23

someones been rearragnging my keyboard lol

Answer 24

the equation of the line use the point slope formula. the point is
\[(\frac{\pi}{3},\frac{\sqrt{3}}{2})\] and the slope is \[\frac{1}{2}\] so the equation is
\[y-\frac{\sqrt{3}}{2}=\frac{1}{2}(x-\frac{\pi}{3})\]

Answer 25

this crosses the x axis where \[y=0\] so set \[y=0\] and solve to get
\[x=\frac{\pi}{3}-\sqrt{3}\] if i did the algebra right

Answer 26

repeat the process for the line with slope -2 through \[(\frac{\pi}{3},\frac{\sqrt{3}}{2})\]

Answer 27

x/2 - pi/6 = -sqrt.3/2

x = -sqrt3 + pi/3 its good ;)

Answer 28

amistre did i screw this up? it looks good to me

Answer 29

thnx

Answer 30

the normal just uses -2 for th eslope then

Answer 31

\[y-\frac{\sqrt{3}}{2}=-2(x-\frac{\pi}{3})\]
is the normal line yes?

Answer 32

yes

Answer 33

let \[y=0\] get \[x=\sqrt{3}+\frac{\pi}{3}\]

Answer 34

at least that is what i got

Answer 35

/ -2 to begin with to ease the process

Answer 36

sqrt.3/4 + pi/3 i think

Answer 37

\[-\frac{\sqrt{3}}{2}=-2(x-\frac{\pi}{3})\]
\[\sqrt{3}=x-\frac{\pi}{3}\]
\[x=\sqrt{3}+\frac{\pi}{3}\]

Answer 38

whoooooooooooooooooops

Answer 39

lorda mercy

Answer 40

\[\frac{\sqrt{3}}{4}+\frac{\pi}{3}\]

Answer 41

another dumb guy award for me.

Answer 42

thnx

Answer 43

lol.... thats what all my medals are for ;)

Answer 44

ok fine. now that we have the x-intercepts, after some elementary algebra mistakes on my part, what are we supposed to do with them?

Answer 45

throw tham at moving trains :)

Answer 46

u both deserve for it :)

Answer 47

i forgot what we were supposed to do with these. multiply? add? find the length?

Answer 48

i mean for medals ;)..not throwing at the moving train :P

Answer 49

the distance between them is just large - small

Answer 50

now the final task, area of the triangle

Answer 51

since they are on the number line; distance between them is large - small

Answer 52

distance between 6 and -3 is 6--3 = 9 ;)

Answer 53

height = sqrt3/2 ; distance = l-s
--------------------------- = area
2

Answer 54

meant to find an area

lol but you guys have done it all over the place and its highly likely that the OP has just giving up on trying to follow what you guys have wrote

Answer 55

the journey taken is more important the the final destination....... grasshopper

Answer 56

sats been consistently and systematically finding the points... its good

Answer 57

ooh some arithmetic i can do!
\[\frac{\sqrt{3}}{4}+\frac{\pi}{3}-(\frac{\pi}{3}-\sqrt{3})=\frac{\sqrt{3}}{4}+\sqrt{3}=\frac{5\sqrt{3}}{4}\]

Answer 58

wow just what it said it was!

Answer 59

very east to set out working out to make the solution of these questions look really easy instead of making it look so complicated with 1000s of posts etc.

Answer 60

true but very hard to type all in in one box. specially if typesetting.

Answer 61

depends on if you wanna read a novel or have a conversation type environment; there isnt only one way to do things :)

Answer 62

hi yo man

Answer 63

this type of environment is suited best for interaction between questioneer and answeer

Answer 64

if they really want a static read thru answer; theres always the internet lol

Answer 65

Asadkarim7 i will be happy to work all this out on one paper and send it to you if this is confusing. the idea is this: find the slope of the tangent line, then find its equation. find the slope of the normal line, and its equation. then in each equation set y = 0 and solve for x. that is the whole idea. once you have both x intercepts subtract to get the distance. the rest is mechanics, but i will be happy to write it out in all its gory details without my algebra errors if you like

Answer 66

asad is intelligent :)

Answer 67

i am sure he is but i will be happy to do it to practice my latex

Answer 68

make a pdf

Answer 69

:) latex.... i perfer ascii lol

Answer 70

thanks :P amistre64

Answer 71

luddite

Answer 72

^ "its very simple

find the eqns of the lines , find the intercepts of those lines , then use distance formula"

esessentially what I wrote like a whole hour ago

Answer 73

ohhh..... so everything after that is useless and should be erased becasue its just utter garbage? .... comes across a bit pompuos dontcha think ;)

Answer 74

in fact the "i still dont understand can you explain it to me step by step" kinda leads me to believe otherwise ... lol

Answer 75

also, geez, way to spam the question guys, nw my computer is going extremely slow when I view this particular question because you guys spammed it so much :|

Answer 76

youre prolly using internet explorer; i had to down firefox to avoid that

Answer 77

the equation editor stuff tends to put a strain on the processor; which is why i tend to avoid it

Answer 78

so asad; how can we help you with this problem?

Answer 79

here it is

Answer 80

let me know if you have any questions

Answer 81

you forgot to put in the area lol

Answer 82

oh heavens. fine.

Answer 83

and here my fine useless garbage contribution ;)

Answer 84

sattelite73 u made the equation y = sqrt 3/ 2
it should have been y- sqrt 3/2

Answer 85

\[\frac{1}{2}bh=\frac{1}{2}\times \frac{5\sqrt{3}}{4}\times \frac{\sqrt{3}}{2}=\frac{15}{16}\]

Answer 86

its his latex.... makes him go a little insane at times and starts putting = signs for - signs :)

Answer 87

everyone is a critic. ok that was a typo and look, it is gone!

Answer 88

that ones just blank.....

Answer 89

u r genius sattelite 73

Answer 90

what lovely font! what nice equations!

Answer 91

how didi u make pdf file

Answer 92

eventually i will learn this latex ... eventually.

Answer 93

he starts out with some fine kindling, then adds a bit of tinder to feed the flame..

Answer 94

i am using texmaker as an editor. just converts for you if you ask, and then can ask for external viewer. really quite friendly. i tried to use this on micordemon windows whatever and it took me maybe a day and a half to download what i needed. miktex and all that garbage. never worked right. left the darkside for linux a few months ago, to maybe 2 minutes do download and works like a charm.