Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

The maximum distance from origin of a point on the curve $x=a\sin t-b\sin \left ( \frac{at}{b} \right )$

$y=a\cos t-b\cos \left ( \frac{at}{b} \right ),both\; a,b> 0\; is$

Option 1)
$a-b\; \;$

Option 2)
$\; \; a+b\;$

Option 3)
$\; \sqrt{a^{2}+b^{2}}\;$

Option 4)
$\; \; \; \sqrt{a^{2}-b^{2}}$

Answers (1)

best_answer

As we learnt in

Rate Measurement -

Rate of any of variable with respect to time is rate of measurement. Means according to small change in time how much other factors change is rate measurement:

$\Rightarrow \frac{dx}{dt},\:\frac{dy}{dt},\:\frac{dR}{dt},(linear),\:\frac{da}{dt}$

$\Rightarrow \frac{dS}{dt},\:\frac{dA}{dt}(Area)$

$\Rightarrow \frac{dV}{dt}(Volume)$

$\Rightarrow \frac{dV}{V}\times 100(percentage\:change\:in\:volume)$

- wherein

Where dR / dt means Rate of change of radius.

$D^2=\left ( a\sin t-b\sin \frac{at}{b} \right )^2+\left ( a\:cos\:t- b cos\frac{at}{b}\right )^2$

$D^2= a^{2}\sin ^{2}t+a^{2}\cos ^{2}t+b^{2}\sin ^{2}\frac{at}{b}+b^{2}\cos ^{2}\frac{at}{b}-2ab \sin t\sin \frac{at}{b}-2ab\cos t\cos \frac{at}{b}$

$= a^{2}+b^{2}-2ab\left [ \cos \left ( \frac{at}{b}-1 \right ) \right ]$

$\therefore$ for min $cos \left ( \frac{at}{b}-1 \right )= 1$

$D^{2}=a^{2}+b^{2}-2ab=\left ( a-b \right )^2$

$\therefore D=a-b$

Option 1)

$a-b\; \;$

Correct

Option 2)

$\; \; a+b\;$

Incorrect

Option 3)

$\; \sqrt{a^{2}+b^{2}}\;$

Incorrect

Option 4)

$\; \; \; \sqrt{a^{2}-b^{2}}$

Incorrect

Posted by

prateek

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Advanced 2024: Chemical engineering cut-offs at IITs; know opening, closing ranks

anam-khan

IIT JEE Mains Result 2024 Session 2: How is overall merit list prepared?

anam-khan

JEE Mains Session 2 result 2024 soon; List of top NITs in India

Latest Question