Question #270540

the characteristic of the equation PQ=x and the integral surface which passes through the curve y=1/2 z=x are =?


1
Expert's answer
2021-11-24T14:56:16-0500
f(x,y,z,p,q)=pqx=0f(x, y, z, p, q)=pq-x=0fx=1,fy=0,fz=0,fp=q,fq=pf_x=-1, f_y=0, f_z=0, f_p=q, f_q=p




dxfp=dyfq=dzpfp+qfq=dp(fx+pfz)=dq(fy+qfz)\dfrac{dx}{f_p}=\dfrac{dy}{f_q}=\dfrac{dz}{pf_p+qf_q}=\dfrac{dp}{-(f_x+pf_z)}=\dfrac{dq}{-(f_y+qf_z)}



dxq=dyp=dz2pq=dp1=dq0\dfrac{dx}{q}=\dfrac{dy}{p}=\dfrac{dz}{2pq}=\dfrac{dp}{1}=\dfrac{dq}{0}


dx/q=dpdx/q=dp

dq=0    q=cdq=0\implies q=c

p=x/c+c1p=x/c+c_1

z=x2/(2c)+c1x+c3z=x^2/(2c)+c_1x+c_3


dy=pdpdy=pdp

p2=2y+c2p^2=2y+c_2

(x/c+c1)2=2y+c2(x/c+c_1)^2=2y+c_2


F(zx2/(2c)c1x,(x/c+c1)22y)=0F(z-x^2/(2c)-c_1x,(x/c+c_1)^2-2y)=0


we have y=1/2 and z=x, so


F(x(1c1)x2/(2c),(x/c+c1)21)=0F(x(1-c_1)-x^2/(2c),(x/c+c_1)^2-1)=0


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS