← 2019 Paper 1
UPSC 2019 Maths Optional Paper 1 Q4c-i — Step-by-Step Solution 12 marks · Section A
Partial derivatives · Calculus · asked 9× in 14 yrs · Read the full method →
Question
If u = sin − 1 x 1 / 3 + y 1 / 3 x 1 / 2 + y 1 / 2 u=\sin^{-1}\sqrt{\dfrac{x^{1/3}+y^{1/3}}{x^{1/2}+y^{1/2}}} u = sin − 1 x 1/2 + y 1/2 x 1/3 + y 1/3 sin 2 u \sin^2 u sin 2 u x x x y y y − 1 6 -\dfrac{1}{6} − 6 1 x 2 ∂ 2 u ∂ x 2 + 2 x y ∂ 2 u ∂ x ∂ y + y 2 ∂ 2 u ∂ y 2 = tan u 12 ( 13 12 + tan 2 u 12 ) . x^2\dfrac{\partial^2 u}{\partial x^2}+2xy\dfrac{\partial^2 u}{\partial x\partial y}+y^2\dfrac{\partial^2 u}{\partial y^2}=\dfrac{\tan u}{12}\left(\dfrac{13}{12}+\dfrac{\tan^2 u}{12}\right). x 2 ∂ x 2 ∂ 2 u + 2 x y ∂ x ∂ y ∂ 2 u + y 2 ∂ y 2 ∂ 2 u = 12 tan u ( 12 13 + 12 tan 2 u ) .
Technique
Show sin 2 u \sin^2u sin 2 u u u u − 1 6 -\tfrac16 − 6 1 v x = sin 2 u u x v_x=\sin2u\,u_x v x = sin 2 u u x x u x + y u y = g ( u ) = n 2 tan u xu_x+yu_y=g(u)=\tfrac n2\tan u x u x + y u y = g ( u ) = 2 n tan u x 2 u x x + 2 x y u x y + y 2 u y y = g ( u ) [ g ′ ( u ) − 1 ] x^2u_{xx}+2xyu_{xy}+y^2u_{yy}=g(u)[g'(u)-1] x 2 u xx + 2 x y u x y + y 2 u y y = g ( u ) [ g ′ ( u ) − 1 ]
Solution
Step 1 — sin 2 u \sin^2u sin 2 u − 1 6 -\tfrac16 − 6 1
Let v : = sin 2 u = x 1 / 3 + y 1 / 3 x 1 / 2 + y 1 / 2 v:=\sin^2 u=\dfrac{x^{1/3}+y^{1/3}}{x^{1/2}+y^{1/2}} v := sin 2 u = x 1/2 + y 1/2 x 1/3 + y 1/3 ( x , y ) → ( t x , t y ) (x,y)\to(tx,ty) ( x , y ) → ( t x , t y )
v ( t x , t y ) = ( t x ) 1 / 3 + ( t y ) 1 / 3 ( t x ) 1 / 2 + ( t y ) 1 / 2 = t 1 / 3 ( x 1 / 3 + y 1 / 3 ) t 1 / 2 ( x 1 / 2 + y 1 / 2 ) = t 1 / 3 − 1 / 2 v ( x , y ) = t − 1 / 6 v ( x , y ) . v(tx,ty)=\frac{(tx)^{1/3}+(ty)^{1/3}}{(tx)^{1/2}+(ty)^{1/2}}=\frac{t^{1/3}\big(x^{1/3}+y^{1/3}\big)}{t^{1/2}\big(x^{1/2}+y^{1/2}\big)}=t^{1/3-1/2}\,v(x,y)=t^{-1/6}v(x,y). v ( t x , t y ) = ( t x ) 1/2 + ( t y ) 1/2 ( t x ) 1/3 + ( t y ) 1/3 = t 1/2 ( x 1/2 + y 1/2 ) t 1/3 ( x 1/3 + y 1/3 ) = t 1/3 − 1/2 v ( x , y ) = t − 1/6 v ( x , y ) . Hence v = sin 2 u v=\sin^2 u v = sin 2 u n = − 1 6 n=-\dfrac{1}{6} n = − 6 1 ■ \blacksquare ■
Step 2 — Euler’s theorem (first order)
For a function v v v n n n x v x + y v y = n v x\,v_x+y\,v_y=n\,v x v x + y v y = n v v = sin 2 u v=\sin^2u v = sin 2 u v x = 2 sin u cos u u x = sin 2 u u x v_x=2\sin u\cos u\,u_x=\sin 2u\,u_x v x = 2 sin u cos u u x = sin 2 u u x v y = sin 2 u u y v_y=\sin 2u\,u_y v y = sin 2 u u y
sin 2 u ( x u x + y u y ) = n sin 2 u ⟹ x u x + y u y = n sin 2 u sin 2 u = n 2 tan u . \sin 2u\,(x u_x+y u_y)=n\sin^2u\ \Longrightarrow\ x u_x+y u_y=\frac{n\sin^2 u}{\sin 2u}=\frac{n}{2}\tan u. sin 2 u ( x u x + y u y ) = n sin 2 u ⟹ x u x + y u y = sin 2 u n sin 2 u = 2 n tan u . With n = − 1 6 n=-\tfrac16 n = − 6 1 g ( u ) : = n 2 tan u = − 1 12 tan u g(u):=\dfrac{n}{2}\tan u=-\dfrac{1}{12}\tan u g ( u ) := 2 n tan u = − 12 1 tan u
x u x + y u y = − 1 12 tan u . (1) x u_x+y u_y=-\tfrac{1}{12}\tan u. \tag{1} x u x + y u y = − 12 1 tan u . ( 1 ) Step 3 — Euler’s theorem (second order)
The standard second-order consequence: if x u x + y u y = g ( u ) x u_x+y u_y=g(u) x u x + y u y = g ( u ) x x x × x \times x × x y y y × y \times y × y
x 2 u x x + 2 x y u x y + y 2 u y y = g ( u ) [ g ′ ( u ) − 1 ] . (2) x^2u_{xx}+2xy\,u_{xy}+y^2u_{yy}=g(u)\big[g'(u)-1\big]. \tag{2} x 2 u xx + 2 x y u x y + y 2 u y y = g ( u ) [ g ′ ( u ) − 1 ] . ( 2 ) (Derivation: ∂ x \partial_x ∂ x u x + x u x x + y u x y = g ′ ( u ) u x u_x+xu_{xx}+yu_{xy}=g'(u)u_x u x + x u xx + y u x y = g ′ ( u ) u x x x x ∂ y \partial_y ∂ y x u x y + u y + y u y y = g ′ ( u ) u y xu_{xy}+u_y+yu_{yy}=g'(u)u_y x u x y + u y + y u y y = g ′ ( u ) u y y y y x 2 u x x + 2 x y u x y + y 2 u y y + ( x u x + y u y ) = g ′ ( u ) ( x u x + y u y ) x^2u_{xx}+2xyu_{xy}+y^2u_{yy}+(xu_x+yu_y)=g'(u)(xu_x+yu_y) x 2 u xx + 2 x y u x y + y 2 u y y + ( x u x + y u y ) = g ′ ( u ) ( x u x + y u y ) = g ( u ) g ′ ( u ) − g ( u ) = g ( u ) [ g ′ ( u ) − 1 ] =g(u)g'(u)-g(u)=g(u)[g'(u)-1] = g ( u ) g ′ ( u ) − g ( u ) = g ( u ) [ g ′ ( u ) − 1 ]
Step 4 — Evaluate with g ( u ) = − 1 12 tan u g(u)=-\tfrac{1}{12}\tan u g ( u ) = − 12 1 tan u
g ′ ( u ) = − 1 12 sec 2 u = − 1 12 ( 1 + tan 2 u ) . g'(u)=-\tfrac{1}{12}\sec^2u=-\tfrac{1}{12}(1+\tan^2u). g ′ ( u ) = − 12 1 sec 2 u = − 12 1 ( 1 + tan 2 u ) . g ′ ( u ) − 1 = − 1 12 − 1 12 tan 2 u − 1 = − 13 12 − tan 2 u 12 = − ( 13 12 + tan 2 u 12 ) . g'(u)-1=-\tfrac{1}{12}-\tfrac{1}{12}\tan^2u-1=-\tfrac{13}{12}-\tfrac{\tan^2u}{12}=-\Big(\tfrac{13}{12}+\tfrac{\tan^2u}{12}\Big). g ′ ( u ) − 1 = − 12 1 − 12 1 tan 2 u − 1 = − 12 13 − 12 t a n 2 u = − ( 12 13 + 12 t a n 2 u ) . Therefore
x 2 u x x + 2 x y u x y + y 2 u y y = g ( u ) [ g ′ ( u ) − 1 ] = ( − 1 12 tan u ) ( − ( 13 12 + tan 2 u 12 ) ) , x^2u_{xx}+2xyu_{xy}+y^2u_{yy}=g(u)[g'(u)-1]=\Big(-\tfrac{1}{12}\tan u\Big)\Big(-\big(\tfrac{13}{12}+\tfrac{\tan^2u}{12}\big)\Big), x 2 u xx + 2 x y u x y + y 2 u y y = g ( u ) [ g ′ ( u ) − 1 ] = ( − 12 1 tan u ) ( − ( 12 13 + 12 t a n 2 u ) ) , Answer
x 2 ∂ 2 u ∂ x 2 + 2 x y ∂ 2 u ∂ x ∂ y + y 2 ∂ 2 u ∂ y 2 = tan u 12 ( 13 12 + tan 2 u 12 ) . \boxed{\;x^2\frac{\partial^2u}{\partial x^2}+2xy\frac{\partial^2u}{\partial x\partial y}+y^2\frac{\partial^2u}{\partial y^2}=\frac{\tan u}{12}\left(\frac{13}{12}+\frac{\tan^2u}{12}\right).\;} x 2 ∂ x 2 ∂ 2 u + 2 x y ∂ x ∂ y ∂ 2 u + y 2 ∂ y 2 ∂ 2 u = 12 tan u ( 12 13 + 12 tan 2 u ) .