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Flauer [41]
3 years ago
13

The function r(t) traces a circle. Determine the radius, center, and plane containing the circle r(t)=−9i+ (11cos(t))j+ (11sin(t

))k
Mathematics
1 answer:
Feliz [49]3 years ago
6 0

we are given

r(t)=−9i+ (11cos(t))j+ (11sin(t))k

now, we can find x-components , y-component and z-component

x=-9,y=11cos(t),z=11sin(t)

now, we can square and add them

x^2+y^2+z^2=(-9)^2+(11cos(t))^2+(11sin(t))^2

now, we can simplify it

x^2+y^2+z^2=81+121cos^2(t)+121sin^2(t)

x^2+y^2+z^2=81+121(cos^2(t)+sin^2(t))

x^2+y^2+z^2=81+121

x^2+y^2+z^2=202

we can also write as

(x-0)^2+(y-0)^2+(z-0)^2=(\sqrt{202} )^2

now, we can compare it with

(x-0)^2+(y-0)^2+(z-0)^2=(r )^2

Center:

(0,0,0)

radius:

r=\sqrt{202}

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The inverse relation is not a function because of the plus minus. For instance, plugging x = 4 into y = \pm \sqrt{x+5} leads to y = -3 and y = 3 simultaneously. You would have to apply a domain restriction on y = x^2-5 to make it a one-to-one function, to make the inverse a function. One possible domain restriction is x > 0 which would lead to the inverse function y = \sqrt{x+5}

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