At what points does the curve r(t) = ti + (4t - t^2)k intersect the paraboloid z
ID: 2856150 • Letter: A
Question
At what points does the curve r(t) = ti + (4t - t^2)k intersect the paraboloid z = x^2 + y^2 (If an answer does not exist, enter DNE.) (x, y, z) = (smaller t-value) (x, y, z) = (larger t-value) Find the derivative, r'(t), of the vector function. r(t) = (t sin 3t , t^2, t cos 2t) r'(t) = Find the unit tangent vector T(t) at the point with the given value of the parameter t. r(t) = (2te^-t, 4 arctan t, 4e^t), t = 0 T(0) = Find the unit tangent vector T(t) at the point with the given value of the parameter t. r( t) = cos ti + 8tj + 3 si n 2tk, t = 0 T(0) =Explanation / Answer
7)r(t)=ti +(4t-t2)k
x=t,y=0,z=(4t-t2)
z=x2+y2
4t-t2=t2+02
2t2-4t=0
2t(t-2)=0
t=0,t=2
answer is (x,y,z)=(0,0,0)
(x,y,z)=(2,0, 4(2) -22) =(2,0,4)
8)r(t)=<tsin3t ,t2 ,tcos2t>
product rule:(uv)'=u'v +uv'
r'(t)=<1sin3t +t(cos3t)*3 ,2t ,1cos2t+ t(-sin2t)*2>
r'(t)=<sin3t +3tcos3t ,2t ,cos2t -2tsin2t>
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