1/ The circle x^2 + y^2= 4 in R^2 is not a function. For example, if x=0 , y has

Question

1/ The circle x^2 + y^2= 4 in R^2 is not a function. For example, if x=0 , y has two possible values: + and - 2.but the vector equation for this circle r= 2cos (t)i + 2sin(t) j is a vector function. Explain how the vector equation is a function when the original curve is not.

2/ Based on the question 1 keeps revisiting the same points on the circle as we increase the parameter t. Explain why this does not violate the statement that the vector equation is a vector function.

3/ vector function: r(t)= sqrt[4-2t(i)] + cos(t) i - (1/t+1) k

a/ what is the domain of this function?

b/ compute vector r' (0)=

c/ How is the vector r'(0) oriented geometrically with the curve at the point where t=0?

d/ Are the vector r(0) and vector r'(0) orthogonal to each other ?

e/ Is the curve given by vector r(t) smooth over its entire domain? Explain

f/ Find the indefinite integral vector r(t)*dt =

g/ Your answer should have a constant vector C attatched to it . Why ?

Please explain for me the answers details. Thanks a lot !

Explanation / Answer

F

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