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Consider the parametric curve: x=1+9cost,y=7+9sint,π/2≤t≤3π/2 x=1+9cos⁡t,y=7+9sin⁡t,π/2≤t≤3π/2 The cartesian equation of the curve has the form (x−h)2+(y−k)2=R2(x−h)2+(y−k)2=R2

a. Find h,k and Rb. The initial point has coordinates:?c. The terminal point has coordinates:? The curve is traced? clockwise or counter?

Respuesta :

rodolforodriguezr

Answer:

a)

h = 1

k = 7

R = 9

b)  Initial point: (1,16)

c) Terminal point : (1 -2)

The path is traced counterclockwise

Step-by-step explanation:

We have the parametric curve

x(t) = 1+9cos(t)

y(t) = 7+9sin(t)  

with π/2 ≤ t ≤ 3π/2

a)

We can see that

x-1 = 9cos(t)

y-7 = 9sin(t)

hence

[tex]\bf (x-1)^2+(y-7)^2=9^2cos^2(t)+9^2sin^2(t)=9^2(cos^2(t)+sin^2(t))=9^2[/tex]

so

h = 1

k = 7

R = 9

We notice that the curve is part of a circumference with radio 9 and center (1,7)

b)

The initial point is obtained when t = π/2

x(π/2) = 1+9cos(π/2) = 1

y(π/2) = 7+9sin(π/2) = 7+9 = 16

The initial point is then (1,16)

c)

The terminal point is obtained when t = 3π/2

x(3π/2) = 1+9cos(3π/2) = 1

y(3π/2) = 7+9sin(3π/2) = 7-9 = -2

The final point is (1,-2)

The path is the part of the circumference with center  (1,7) and radio 9 traced counterclockwise from (1,16) to (1,-2)

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