If x=sinθ+θcosθ, y=cosθ−θsinθ, then (dydx)θ=π2=

x=sinθ+θcosθ, y=cosθ−θsinθdxdθ=cosθ+cosθ−θsinθ=2cosθ−θsinθdydθ=−sinθ−(sinθ+θcosθ)=−(2sinθ+θcosθ)dydx=dydθdxdθ=−(2sinθ+θcosθ)2cosθ−θsinθ(dydx)θ=π2=−(2sinπ2+π2cosπ2)2cosπ2−π2sinπ2=−2−π2=4π

If x=sinθ+θcosθ, y=cosθ−θsinθ, then (dydx)θ=π2=

x=sinθ+θcosθ, y=cosθ−θsinθdxdθ=cosθ+cosθ−θsinθ=2cosθ−θsinθdydθ=−sinθ−(sinθ+θcosθ)=−(2sinθ+θcosθ)dydx=dydθdxdθ=−(2sinθ+θcosθ)2cosθ−θsinθ(dydx)θ=π2=−(2sinπ2+π2cosπ2)2cosπ2−π2sinπ2=−2−π2=4π

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If $x = \sin θ + θ \cos θ, y = \cos θ - θ \sin θ$ , then ${(\frac{d y}{d x})}_{θ = \frac{π}{2}} =$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{- π}{2}$

$\frac{2}{π}$

$\frac{π}{4}$

$\frac{4}{π}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$x = \sin θ + θ \cos θ, y = \cos θ - θ \sin θ$

$\frac{d x}{d θ} = \cos θ + \cos θ - θ \sin θ = 2 \cos θ - θ \sin θ$

$\frac{d y}{d θ} = - \sin θ - (\sin θ + θ \cos θ) = - (2 \sin θ + θ \cos θ)$

$\frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{- (2 \sin θ + θ \cos θ)}{2 \cos θ - θ \sin θ}$

${(\frac{d y}{d x})}_{θ = \frac{π}{2}} = \frac{- (2 \sin \frac{π}{2} + \frac{π}{2} \cos \frac{π}{2})}{2 \cos \frac{π}{2} - \frac{π}{2} \sin \frac{π}{2}} = \frac{- 2}{- \frac{π}{2}} = \frac{4}{π}$