{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Calc 3 Exam 1 Sin

# Calc 3 Exam 1 Sin - k< 1 For k> 1 as k increases the...

This preview shows page 1. Sign up to view the full content.

CALCULUS AND ANALYTIC GEOMETRY III SOLUTIONS TO SAMPLE TEST 1 Please let me know if you find errors or typos. 1. (a) The volume is given by the (absolute value of the) scalar triple product a · ( b × c ). This triple product is equal to the determinant 1 1 0 0 1 2 1 0 1 = 3 . (b) The points will lie in the same plane if and only if the vectors -→ PQ = h 1 , 4 , 5 i , -→ PR = h 2 , - 1 , 1 i and -→ PS = h 4 , 7 , 11 i have triple scalar product equal to zero. The calculation is similar to the that in (a) and the answer is indeed zero. 2. You are supposed to draw pictures, but I will just describe the answers in words. In the xy -plane the trace is a hyperbola. In the yz -plane we have a circle. The intersections with the planes z = k are as follows. For k = 1 we have a pair of lines; for k > 1 and for k < 1 we get hyperbolae having the previous lines as asymptotes, with the branches of the hyperbolae for k > 1 being on the opposite sides of the asymptotic lines from those for
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: k < 1. For k > 1, as k increases, the hyperbolae move further from the asymptotes. 3. The point (1 , 1 , 1) corresponds to t = 1. For a normal vector to the normal plane we can take r (1) = h 1 , 2 , 3 i , so the equation is ( x-1) + 2( y-1) + 3( z-1) = 0. As a normal vector for the osculating plane we can take r (1) × r 00 (1). This is because B ( t ) is perpendicular to both T ( t ) and N ( t ), and the plane of T ( t ) and N ( t ) is the same as the plane of r ( t ) and r 00 ( t ). Thus, the normal vector we want is h 1 , 2 , 3 i × h , 2 , 6 i = h 6 ,-6 , 2 i , so we can write the equation of the osculating plane as 3( x-1)-3( y-1) + ( z-1) = 0. 4. The endpoints correspond to t = 0 and t = 2 π , so the arc length is given by Z 2 π | r ( t ) | dt = Z 2 π p sin 2 t + cos 2 t + 1 dt = 2 π √ 2 ....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online