Question #270068

If x = e^ty^2 , t= V cos v , y= v sin v. Find dx/dv at v= 3.142/2

1
Expert's answer
2021-11-23T09:47:05-0500
x=ety2x=e^ty^2

x(v)=evcos(v)v2sin2(v)x(v)=e^{v\cos(v)}v^2\sin^2(v)

dx/dv=evcos(v)(cos(v)vsin(v))dx/dv=e^{v\cos(v)}(\cos(v)-v\sin(v))

+evcos(v)(2vsin(v)+2v2sin(v)cos(v))+e^{v\cos(v)}(2v\sin(v)+2v^2\sin(v)\cos(v))

v=π/2v=\pi/2


dx/dvv=π/2=e0(0π/2)dx/dv|_{v=\pi/2}=e^{0}(0-\pi/2)

+e0(2(π/2)(1)+2(π/2)2(1)(0))=π/2+π=π/2+e^{0}(2(\pi/2)(1)+2(\pi/2)^2(1)(0)) =-\pi/2+\pi=\pi/2


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