Question

Use chain rule to find dw/dt w=ln(x^2 + y^2 + z^2)^(1/2) , x=sint, y=cost, z=tant

Answer 1

The chain rule states that


`(\mathrm dw)/(\mathrm dt)=(\partial w)/(\partial x)(\mathrm dx)/(\mathrm dt)+(\partial w)/(\partial y)(\mathrm dy)/(\mathrm dt)+(\partial w)/(\partial z)(\mathrm dz)/(\mathrm dt)`

Since
`\ln(x^2+y^2+z^2)^(1/2)=\frac12\ln(x^2+y^2+z^2)`, you have


`(\partial w)/(\partial x)=\frac x{x^2+y^2+z^2}`

`(\partial w)/(\partial x)=\frac y{x^2+y^2+z^2}`

`(\partial w)/(\partial z)=\frac z{x^2+y^2+z^2}`

and


`(\mathrm dx)/(\mathrm dt)=\cos t`

`(\mathrm dy)/(\mathrm dt)=-\sin t`

`(\mathrm dz)/(\mathrm dt)=\sec^2t`

Also, since
`x^2+y^2+z^2=\sin^2t+\cos^2t+\tan^2t=1+\tan^2t=\sec^2t`, the derivative is


`(\mathrm dw)/(\mathrm dt)=(\sin t\cos t)/(\sec^2t)-(\sin t\cos t)/(\sec^2t)+(\tan t\sec^2t)/(\sec^2t)`

`(\mathrm dw)/(\mathrm dt)=\tan t`