Find an expression for a unit vector normal to the surface x = 10 cos (0) sin ($), y = 5 sin (0) sin (4), z = cos (4) for 0 in [0, 2x] and ø in [0, x]. (Enter your solution in the vector form (*,*,*). Use symbolic notation and fractions where needed.) n =

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter3: Functions And Graphs
Section3.2: Graphs Of Equations
Problem 45E
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Find an expression for a unit vector normal to the surface
x = 10 cos (0) sin ($) , y = 5 sin (0) sin (4) , z = cos (4)
for 0 in [0, 27] and o in [0, x].
(Enter your solution in the vector form (*,*,*). Use symbolic notation and fractions where needed.)
n =
Transcribed Image Text:Find an expression for a unit vector normal to the surface x = 10 cos (0) sin ($) , y = 5 sin (0) sin (4) , z = cos (4) for 0 in [0, 27] and o in [0, x]. (Enter your solution in the vector form (*,*,*). Use symbolic notation and fractions where needed.) n =
Expert Solution
Step 1

Given that x=10cosθsinϕy=5sinθsinϕ and z=cosϕ.

We have to find a unit vector normal to the surface.

Let rθ,ϕ=10cosθsinϕ,5sinθsinϕ,cosϕ

Now, rθ=10sinθsinϕ,5cosθsinϕ,0

And rϕ=10cosθcosϕ,5sinθcosϕ,sinϕ.

rθ×rϕ=i^j^k^10sinθsinϕ5cosθsinϕ010cosθcosϕ5sinθcosϕsinϕ=i^5cosθsin2ϕ0j^10sinθsin2ϕ0+k^50sin2θsinϕcosϕ50cos2θsinϕcosϕ=i^5cosθsin2ϕj^10sinθsin2ϕ+k^50sinϕcosϕ=5cosθsin2ϕ,10sinθsin2ϕ,50sinϕcosϕ

Hence, rθ×rϕ=5cosθsin2ϕ,10sinθsin2ϕ,50sinϕcosϕ.

Now,

rθ×rϕ=5cosθsin2ϕ2+10sinθsin2ϕ2+50sinϕcosϕ2=25cos2θsin4ϕ+100sin2θsin4ϕ+2500sin2ϕcos2ϕ=25sin2ϕcos2θsin2ϕ+4sin2θsin2ϕ+100cos2ϕ=5sinϕcos2θsin2ϕ+4sin2θsin2ϕ+100cos2ϕ

Hence, rθ×rϕ=5sinϕcos2θsin2ϕ+4sin2θsin2ϕ+100cos2ϕ.

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