wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

if x=asin1t,y=acos1t,then show that dydx=yx.

Open in App
Solution

Given, x=asin1t,y=acos1t

ie.., x=a12sin1t and y=a12cos1t [ (ab)c=abc]Differentiating w.r.t. t, we getdxdt=a12sin1tlog a ddt(12sin1t) [ ddxax=axlog a]=a12sin1t log a(121t2)=a12sin1tlog a21t2and dydt=a12cos1tlog addt(12cos1t) (Using chain rule)=a12cos1t log a(121t2)=a12cos1tlog a21t2 dydx=dydtdxdt=a12cos1ta12sin1t=acos1tasin1t=yx

Alternate~method

x=asin1tTaking log on both sides log x=log(asin1t)1/2 log x=12log a sin1 log x=12sin1t log a log x=12log a sin1tDifferentiating w.r.t. t,we get 1x.dxdt=12log a ddt(sin1 t)=12 log a 11t2 dxdt=x log a2×11t2Again, y=acos1tTaking log on both sides, log y=log(acos1t)1/2 log y=12log acos1t log y=12cos1t log aDifferentiating w.r.t. t,we get1y.dydt=12log a×[11t2] dydt=y log a2×11t2We know thatdydx=dydt×dtdxdydx=y log a2×11t2×2×1t2x log a dydx=yx

Another method

x=asin1t ........(i)and y=acos1t .......(ii)Multiplying Eqs. (i) and (ii), we get xy=asin1t×acos1t xy=asin1t.acos1t xy=asin1t+cos1t [ sin1x+cos1x=π2] xy=aπ2Differentiating w.r.t. x,we getx dydx+y=0 x dydx=y dydx=yx (ddx (constant)=0)


flag
Suggest Corrections
thumbs-up
17
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Property 2
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon