if x=√asin−1t,y=√acos−1t,then show that dydx=−yx.
Given, x=√asin−1t,y=√acos−1t
ie.., x=a12sin−1t and y=a12cos−1t [∵ (ab)c=abc]Differentiating w.r.t. t, we getdxdt=a12sin−1tlog a ddt(12sin−1t) [∵ ddxax=axlog a]=a12sin−1t log a(12√1−t2)=a12sin−1tlog a2√1−t2and dydt=a12cos−1tlog addt(12cos−1t) (Using chain rule)=a12cos−1t log a(−12√1−t2)=−a12cos−1tlog a2√1−t2⇒ dydx=dydtdxdt=−a12cos−1ta12sin−1t=−√acos−1t√asin−1t=−yx
Alternate~method
x=√asin−1tTaking log on both sides log x=log(asin−1t)1/2 ⇒ log x=12log a sin−1⇒ log x=12sin−1t log a⇒ log x=12log a sin−1tDifferentiating w.r.t. t,we get 1x.dxdt=12log a ddt(sin−1 t)=12 log a 1√1−t2⇒ dxdt=x log a2×1√1−t2Again, y=√acos−1tTaking log on both sides, log y=log(acos−1t)1/2⇒ log y=12log acos−1t⇒ log y=12cos−1t log aDifferentiating w.r.t. t,we get1y.dydt=12log a×[−1√1−t2] ⇒ dydt=−y log a2×1√1−t2We know thatdydx=dydt×dtdx⇒dydx=−y log a2×1√1−t2×2×√1−t2x log a⇒ dydx=−yx
Another method
x=√asin−1t ........(i)and y=√acos−1t .......(ii)Multiplying Eqs. (i) and (ii), we get⇒ xy=√asin−1t×√acos−1t⇒ xy=√asin−1t.acos−1t⇒ xy=√asin−1t+cos−1t [∵ sin−1x+cos−1x=π2]⇒ xy=√aπ2Differentiating w.r.t. x,we getx dydx+y=0 ⇒ x dydx=−y ⇒ dydx=−yx (ddx (constant)=0)