{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

midterm2_solution_problems1thru3

# midterm2_solution_problems1thru3 - 1 Let C be a curve...

This preview shows pages 1–2. Sign up to view the full content.

1 Let C be a curve defined by the position function r ( t ) = < sin 2 t, t, cos 2 t > . (a) Calculate the equation of the Normal plane at the point (0 , π, 1). It’s easier than you think. Solution : The Normal plane is spanned by N and B , so it is perpendicular to T . The point (0 , π, 1) corresponds t = π . r ( t ) = < 2 cos 2 t, 1 , - 2 sin 2 t > T = r ( t ) | r ( t ) | = 1 5 < 2 cos 2 t, 1 , - 2 sin 2 t > T ( π ) = < 2 5 , 1 5 , 0 > Then the equation of the Normal plane at (0 , π, 1) is 2 5 ( x - 0) + 1 5 ( y - π ) + 0( z - 1) = 0 which can be simplified as 2 x + y - π = 0 (b) Calculate the equation of the Osculating plane at the point (0 , π, 1). Solution : The Normal plane is spanned by T and N , so it is perpendicular to B . T = 1 5 < - 4 sin 2 t, 0 , - 4 cos 2 t > N = T | T | = < - sin 2 t, 0 , - cos 2 t > N ( π ) = < 0 , 0 , - 1 > B = T × N = < - 1 5 , 2 5 , 0 > Then the equation of the Osculating plane at (0 , π, 1) is - 1 5 ( x - 0) + 2 5 ( y - π ) + 0( z - 1) = 0 which can be simplified as - x + 2 y - 2 π = 0 (c) Is the Osculating plane perpendicular to the Normal plane? Explain.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 2

midterm2_solution_problems1thru3 - 1 Let C be a curve...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online