11
rkrish
·2009-03-14 07:04:18
p = (x2 - x1)2 + (2√x2 - x1 - 2)2
p' = -2(x2 - x1) - 2(2√x2 - x1 - 2) = 0 [ p' is ∂p/∂x1 ]
→ (x1 - x2) = (2√x2 - x1 - 2)
→ x1 = √x2 + x2/2 - 1
p = (x2/2 - √x2 + 1)2 + (√x2 - x2/2 - 1)2
= 2(x2/2 - √x2 + 1)2
Let √x2 = x
p = (x2 - 2x + 2)2/2
p'=0
→ (x2 - 2x + 2)(x - 1) = 0
→ x = 1
So, pmin = 1/2
1
greatvishal swami
·2009-03-14 04:55:24
bahut badiya rajat i used the same method [4][4][4][4]
21
tapanmast Vora
·2009-03-14 04:56:04
OH O k k..... read the latter partr of ur point later!!
But hey how do u know exactly these sub will work?
21
tapanmast Vora
·2009-03-14 04:57:51
followin is jus an eg.
If P = (2x1 - x + 2)2 + ( x2 - 3x1 + 5)2
then wat wud hav bn the curves?
1
rajat sen
·2009-03-14 04:58:34
\texttt{The terms in the second bracket are } 2.\sqrt{x_{2}} \texttt{ and } x_1+2 \texttt{ respectively thats why !}
21
tapanmast Vora
·2009-03-14 04:59:59
yah but y do u consider jus da 2nd bracket?
mayb my que in POST # 25 will make my dbt more clear
21
tapanmast Vora
·2009-03-14 06:33:08
@Nishant Sir, Sir I m not gettin the ans wid partial differentiation approach [2] can u help me proceed
62
Lokesh Verma
·2009-03-14 06:36:31
@tapan..
if you keep x1 constant..
then it will be a quadratic in x2
you can find the x2 for minimum value in terms of x1
substitute that in original form
now minimise this for the single varaible x1
21
tapanmast Vora
·2009-03-14 06:46:10
Ya sir, i know how to partial differentiate,
"@Nishant Sir, Sir I m not gettin the ans wid partial differentiation approach can u help me proceed"
wait 2 min, I'll post my workin
21
tapanmast Vora
·2009-03-14 06:56:27
Does the min value of x in abov graf give the necessary value of P?
then how cum is it not 0.5
21
tapanmast Vora
·2009-03-14 04:53:24
Oh k! yah i got how to find the min. dis b/w these 2 curves..........
but had i doubt that WHY/HOW does making such assumptions of these 2 specific curves help u get da min val of P?
11
rkrish
·2009-03-14 07:09:31
credit to nishant bhaiyya.....thanx bhaiyya!!![1][1]
62
Lokesh Verma
·2009-03-14 07:21:13
yes tapan.. you are right.. my method gives a very very very very very dirty answer for this one..
rajat's method is very good for this one..
basically that is what should be done..
I dont see another way out!
62
Lokesh Verma
·2009-03-14 07:21:57
oops rkrish solved it :D
wow.. and i am posting that this is getting messy ;) :D
11
rkrish
·2009-03-14 07:24:27
.....BUT bhaiyya ....I worked out acc. to wat you said(partial derivative wrt x1) and got the ans.
Pls tell me if I have made any careless or silly mistake in post#33.
21
tapanmast Vora
·2009-03-14 07:35:33
OH K!!!!
thnx rkrish n SIR ( for a nice methd)
SIR : Can u pl. explain me the rational b/h Rajat's solution! (i.e. can u pl. tell me how can v apply Rajat's methd if the questn was like in POST #25
11
rkrish
·2009-03-14 08:00:44
@tapan...
I think I understood rajat's approach...
Acc. to the ques,
p=(x2 - x1)2 + (2√x2 - (x1+2))2 = d2 (let)
If you take y2 = 2√x2 : C2 (let)
& y1 = (x1+2) : C1 (let)
you get d in the form of :-
d = √(x2-x1)2 + (y2-y1)2
Now, d is minimum distance b/w curves C1 & C2
and pmin = d2
Now see if you can convert post#25 in the form of the "Dist. Formula" by some substitution.....Otherwise......no other way out(follow wat Nishant bhaiyya's approach)!!!!
21
tapanmast Vora
·2009-03-14 08:06:48
THNX rkrish........... thnx a ton buddy!!!
NOW I AM RICH WITH 2 methods of solving this type of sums........
thnx all!!!!!!!!! [1]
1
greatvishal swami
·2009-03-11 23:36:53
ans to a hai but method bhi batao zara mane thode lambe method se kiya tha
62
Lokesh Verma
·2009-03-11 23:10:05
Vishal which si the question?
(x2-x1)^2+(2\sqrt{x2}-x1-2)^2 \\ or\\ (x2-x1)^2+(2\sqrt{x2-x1}-2)^2 \\
1
greatvishal swami
·2009-03-11 23:12:38
upar wala bro root only with x2
21
tapanmast Vora
·2009-03-11 23:32:50
maine ANS nahi dejha hai, but for sure then its gotta be less than 1....
WAS IT an MCQ?
Wat wer da options?