Applications Of Derivatives Question 290
Question: The maximum distance from the origin of a point on the curve $ x=asint-b $ $ \sin ( \frac{at}{b} ), $ $ y=a,\cos t-b\cos ( \frac{at}{b} ), $ both a, b>0, is
Options:
A) a - b
B) a+b
C) $ \sqrt{a^{2}+b^{2}} $
D) $ \sqrt{a^{2}-b^{2}} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] This distance of the origin form any point (x, y) on the curve is $ \sqrt{x^{2}+y^{2}}=\sqrt{a^{2}+b^{2}-2ab\cos ( t-\frac{at}{b} )} $
$ \le \sqrt{a^{2}+b^{2}+2ab} $
$ [\therefore minimumcos( t-\frac{at}{b} )=-1] $
$ =a+b $
BETA
