Explanation Given information: r ( t ) = t i + t 2 j + t 3 k (1) Formula used: Write the general expression to calculate the velocity. v = d r d t (2) Here, d r d t is the rate of change in displacement with respect to time. Calculation: Substitute 1 for t in equation (1). r ( 1 ) = ( 1 ) i + ( 1 ) 2 j + ( 1 ) 3 k = i + j + k For r at t = 1 , the point P is ( 1 , 1 , 1 ) . Substitute equation (1) in equation (2). v = d d t ( t i + t 2 j + t 3 k ) = ( 1 ) i + ( 2 t ) j + ( 3 t 2 ) k v = 1 i + 2 t j + 3 t 2 k (3) Take magnitude on both sides of the above equation. | v | = 1 2 + ( 2 t ) 2 + ( 3 t 2 ) 2 | v | = 1 + 4 t 2 + 9 t 4 (4) | v ( 1 ) | = 1 + 4 ( 1 ) 2 + 9 ( 1 ) 4 = 1 + 4 + 9 = 14 The tangential vector T is calculated as follows. T = v | v | (5) Substitute equation (3) and (4) in equation (5). T = 1 i + 2 t j + 3 t 2 k 1 + 4 t 2 + 9 t 4 (6) At t = 1 , the equation (6) becomes, T ( 1 ) = 1 i + 2 ( 1 ) j + 3 ( 1 ) 2 k 1 + 4 ( 1 ) 2 + 9 ( 1 ) 4 = 1 i + 2 j + 3 k 1 + 4 + 9 = 1 i + 2 j + 3 k 14 T ( 1 ) = 1 14 i + 2 14 j + 3 14 k (7) The equation (7) is normal to the normal plane. Therefore, it can be expressed as, 1 14 ( x − 1 ) + 2 14 ( y − 1 ) + 3 14 ( y − 1 ) = 0 On solving the above equation, x + 2 y + 3 z = 6 The above equation is an equation of the normal plane

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Author: Hass, Joel., Heil, Christopher , WEIR, Maurice D.

Publisher: Pearson,

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Chapter 13, Problem 26PE

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Find the equations for the osculating, normal, and rectifying planes for given vector at the point (1,1,1).

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Find the equations for the osculating, normal, and rectifying planes for given vector at the point ( 1 , 1 , 1 ) .

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Find the equations for the osculating, normal, and rectifying planes for given vector at the point (1,1,1).

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