Find ∂w/∂u?

Let
w = x^2 + y^2 + z^2,
x = uv, y = u cos(v), z = u sin(v).
Use the chain rule to find ∂w/∂u when (u, v) = (9, 0).

Let
w = x^2 + y^2 + z^2,
x = uv, y = u cos(v), z = u sin(v).
Use the chain rule to find ∂w/∂u when (u, v) = (9, 0).

1 Answer
Write your answer here...
Start with a one sentence answer
Then teach the underlying concepts
Don't copy without citing sources
preview
?

Answer

Write a one sentence answer...

Answer:

Explanation

Explain in detail...

Explanation:

I want someone to double check my answer

Describe your changes (optional) 200

1
Feb 19, 2018

#w = x^2 + y^2 + z^2#

#w=(uv)^2 + (ucosv)^2 + (usinv)^2#

#color(white)(d)#

#=>(dw)/(du) = 2uv * (u'v+uv') + 2(ucosv)* (u'cosv+ucosv') +2(usinv)* (u'sinv+usinv')#

#color(white)(d)#

#=>(dw)/(du) = 2uv * (v+u(dv)/(du)) + 2(ucosv)* (cosv-usinv(dv)/(du)) +2(usinv)* (sinv+ucosv(dv)/(du))#

#color(white)(d)#

#=>(dw)/(du) = 2uv^2+ 2u^2(dv)/(du)+ 2ucos^2v-2u^2cosvsinv(dv)/(du) +2usin^2v+2u^2sinvcosv(dv)/(du)#

#color(white)(d)#

#=>(dw)/(du) = 2uv^2+ 2u^2(dv)/(du)+ 2u(cos^2v+sin^2v)#

#color(white)(d)#

#=>(dw)/(du) = 2uv^2+ 2u^2(dv)/(du)+ 2u#

Was this helpful? Let the contributor know!
1500
Impact of this question
32 views around the world
You can reuse this answer
Creative Commons License