ch10 - EASY from 10.1, # 33 in the book Find a vector...

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EASY from 10.1, # 33 in the book Find a vector function that represents the curve of intersection of the two surfaces, for the cone z = sqrt(x^2 +y^2) and the plane z= 1 + y equating z, we get. ..... 1 + y = sqrt(x^2 + y^2) which goes to (1 + y)^2 = x^2 + y^2 y^2 + 2y + 1 = x^2 + y^2 2y + 1 = x^2 2y = x^2 - 1 y = (1/2)x^2 - 1/2 this is also the formula for the projection of the curve of intersection on the xy-plane z = 1 + y = 1 + (1/2)x^2 - 1/2 = (1/2)x^2 + 1/2 parametrized, this is x = t y = (1/2)t^2 - 1/2 z = (1/2)t^2 + 1/2 in vector form this is r (t) = x i + [(1/2)t^2 - 1/2] j + [(1/2)t^2 + 1/2] k
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MEDIUM 10.3 #11 r(t) = <2sin(t), 5t, 2cos(t)> a) Find the unit tangent and unit normal vectors T (t) and N (t). b) Use Formula 9 to find the curvature. Formula 9: = | T ’(t)|/| r ’(t)| a) r’(t) = <2cos(t), 5, -2sin(t)> |r’(t)| = ((2cos(t)) 2 +(5) 2 +(-2sin(t)) 2 ) = 25+4(cos 2 (t)+sin 2 (t))= 29 T (t) = r ’(t)/| r
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ch10 - EASY from 10.1, # 33 in the book Find a vector...

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