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Find the Line Tangent to the Surface Given by the Function

Question 57

Multiple Choice

Find the line tangent to the surface given by the function at the specified point PP in the direction of the vector u\vec { u } f(x,y) =x3x2y,P=(1,2) ,u=32,12f ( x , y ) = x ^ { 3 } - x ^ { 2 } y , P = ( 1,2 ) , \vec { u } = \left\langle \frac { \sqrt { 3 } } { 2 } , \frac { 1 } { 2 } \right\rangle


A) r(t) =2+12t,2+32t,3+(2+32) t\vec { r } ( t ) = \left\langle 2 + \frac { 1 } { 2 } t , 2 + \frac { \sqrt { 3 } } { 2 } t , 3 + \left( 2 + \frac { \sqrt { 3 } } { 2 } \right) t \right\rangle
B) r(t) =1+32t,2+12t,1+(234) t\vec { r } ( t ) = \left\langle 1 + \frac { \sqrt { 3 } } { 2 } t , 2 + \frac { 1 } { 2 } t , - 1 + ( 2 \sqrt { 3 } - 4 ) _ { t } \right\rangle
C) r(t) =1+32t,2+12t,1(3+12) t\vec { r } ( t ) = \left\langle 1 + \frac { \sqrt { 3 } } { 2 } t , 2 + \frac { 1 } { 2 } t , - 1 - \left( \frac { \sqrt { 3 } + 1 } { 2 } \right) t \right\rangle
D) r(t) =1+32t,2+12t,1t\vec { r } ( t ) = \left\langle 1 + \frac { \sqrt { 3 } } { 2 } t , 2 + \frac { 1 } { 2 } t , - 1 - t \right\rangle

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