71how do you guarantee that the numerator is increasing in cosθ ?
What if it doesn't? Sorry But I couldn't get it.
find Min. and Max. value of r in 5r2 + 12r.cos(a) + 7 = 0 (where r ≥0)
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9 Answers
1For convenience, I'm writing the equation as 5r2 + 12r.cosθ + 7 = 0.
By using quadratic formula,
r =-(12cosθ) ± √144cos2θ - 4.5.72x5
=-12cosθ ± 2√36cos2θ - 3510
=-6cosθ ± √36cos2θ - 355
Now since the maximum value of cosθ is (+1),
putting cosθ=+1, we get
r = -6(1) ± √36.(1)2 - 355 = - 75 or -1.
Since minimum value of cosθ is (-1), putting cosθ = -1, we get,
r = -6(-1) ± √36(-1)2 - 355 = 75 or 1
But since r≥0,
Maximum value of r is 75..
Minimum value of r is 1.. [Ans]
11@ Abhisekh, how do you guarantee that the numerator is increasing in cosθ ?
Set -6cos\alpha \pm \sqrt{36cos^2\alpha-35}=k
Squaring and bringing the terms to 1 side,we get k^2+12kcos\alpha+35=0\Rightarrow -1\le \frac{-(k^2+35)}{12k}\le 1.
That helps, I guess.
71
23he means, wats d guarantee that numerator also increases as cosθ increases
235r2 +12rcosθ + 7 = 0 ......(1)
differentiate wrt θ
→ drdθ = 12rsinθ
10r+12cosθ
for r max or min , dr/dθ = 0
hence r= 0 or sinθ = 0
but r=0 is not permissible bcz of eqn 1 ,
so sinθ= 0
so cosθ = ±1, put this in eqn 1
5r2 - 12r + 7 = 0 , i.e r = 1 or 7/5
5r2 + 12r + 7 = 0, i.r r = -1 or - 7/5
hence rmin = 1 , rmax = 7/5 as r≥ 0
1@ qwerty:
Can u please specify what was wrong with my method?
I never said that numerator would increase as cosθ increases....in fact, the maximum and minimum values are coming only for the minimum value of cosθ
i.e. -1....
(and the answers also match.)
:-|
23how did u came to kno that the value of r that u obtained by putting minimum value of cosθ is the minimum/maximum possible value of r ?
1I went on with the assumption that the extreme values of r would come with the extreme values of cosθ , irrespective of whether the value of cosθ is its maximum or minimum....
But anyways, MISTAKE ACCEPTED !
Thanks everyone for clearing my misconception ! :)
