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

The slope of the tangent at (x, y) to a curve passing through (1,π4) is given by
yxcos2(yx) , then the equation of the curve is

A
y=tan1(log(ex))
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
y=xtan1(log(xe))
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
y=xtan1(log(ex))
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
None of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C y=xtan1(log(ex))
we have, dydx=yxcos2(yx)
Putting y=vx, so that dydx=v+xdvdx, we get
v+xdvdx=vcos2v
dvcos2v=dxxsec2udu=1xdx
On integration, we get
tanu=logx+logC
tan(yx)=logx+logC
This passes through (1,π4), therefore 1=log⁡C.
so, tan(yx)=log x+1
tan(yx)=logx+1tan(yx)=logx+logey=xtan1(log(ex))

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Methods of Solving First Order, First Degree Differential Equations
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon