Multiple Choice. 1.(6 pts) Find symmetric equations of the line L passing through the point (2, 5, 1) and perpendicular to the plane x + 3y z = 9.

Size: px

Start display at page:

Download "Multiple Choice. 1.(6 pts) Find symmetric equations of the line L passing through the point (2, 5, 1) and perpendicular to the plane x + 3y z = 9."

Beatrice Whitehead
5 years ago
Views:

1 Multiple Choice.(6 pts) Find smmetric equations of the line L passing through the point (, 5, ) and perpendicular to the plane x + 3 z = 9. (a) x = = z (c) (x ) + 3( 3) (z ) = 9 (d) (e) x = 3 5 = z + (b) (x ) = ( 5)( 3) = z + x = 3 5 = z + = 9.(6 pts) The two curves below intersect at the point (,, ) = r () = r (). Find the cosine of the angle of intersection ( r (t) = e 3t i + sin t + π ) j + (t )k r (t) = ti + j + (t )k (a) (b) 3 (c) 5 (d) 5 (e) e e + 3.(6 pts) Find the projection of the vector,, 5 onto the vector,, (a),, 5 (b) 6, 3, (c),, 5 (d) 3, 3, 5 (e),, 5.(6 pts) Find r(x)dx where r(x) = (sec x)i + e x k Recall: sec x dx = tan x + C.

2 (a) (tan x + C )i + C j + (e x + C 3 )k (b) tan x + e x + C (c) (tan x)i + e x k (d) (tan x + C )i + (e x + C )k (e) (tan x + C)i + Cj + (e x + C)k 5.(6 pts) Find the volume of the parallelepiped spanned b the three vectors,,,,, and 3,,. (a) 9 (b) 3 (c) (d) (e) 3 6.(6 pts) Find the area of the triangle formed b the three points (,, ), (,, ) and (3, 3, 3). (a). (b) (c) 3 3 (d) (e) 7.(6 pts) Which of the following is a contour map for f(x, ) = x x +? 3

3 (a) (b) (c) (d) (e)

4 8.(6 pts) A particle is travelling and has position at time t given b r(t) = How far does it travel from time t = to time t = 3. (a) 56 (b) 7 (c) (d) 96 (e) 8 3 t3, t, 3t. 9.(6 pts) Find the radius of the sphere given b the equation x 8x z z + 3 =. (a) 6 (b) (c) (d) 3 (e).(6 pts) Below are five expressions involving two vectors a and b. All of them are alwas equal to either (the scalar) or (the vector) except one. Which one can be nonzero? (a) (a b) (b a) (b) (a b) (b a) (c) (a + b) (a b) (d) b (a a) (e) a (b a) Partial Credit You must show our work on the partial credit problems to receive credit!.( pts.) Consider the curve r(t) = sin(t)i + tj cos(t)k. Give equations for the normal plane and the osculating plane at t =..( pts.) Are the lines 3,, + t,, and 7, 5, 6 + t, 3, 3 parallel, intersecting, or skew? If intersecting, find a point of intersection. 3.( pts.) Suppose the curve C has parametric equations: x(t) = t 3 t, (t) = t, z(t) = t + t 5

5 Find the parametric equation for the tangent line to the above curve C at the point P = (,, ). 6

6 Name: Instructor: ANSWERS Math 55, Practice Exam September 3, The Honor Code is in effect for this examination. All work is to be our own. No calculators. The exam lasts for hour and 5 minutes.. Be sure that our name is on ever page in case pages become detached. Be sure that ou have all pages of the test. Each multiple choice question is 6 points, each partial credit problem is points. You will receive extra points. PLEASE MARK YOUR ANSWERS WITH AN X, not a circle!. ( ) (b) (c) (d) (e). (a) (b) ( ) (d) (e) 3. (a) (b) (c) (d) ( ). ( ) (b) (c) (d) (e) 5. (a) (b) (c) ( ) (e) 6. (a) (b) ( ) (d) (e) 7. (a) (b) (c) ( ) (e) 8. (a) ( ) (c) (d) (e) 9. (a) (b) (c) ( ) (e). (a) ( ) (c) (d) (e) Please do NOT write in this box. Multiple Choice.. 3. Extra Points. Total

7 .The smmetric equations of line are given b (x x )/a = ( )/b = (z z )/c, where (x,, z ) is a point on the line and a, b, c is a direction vector. Since L is perpendicular to the plane x + 3 z = 9, then we can take the normal to the plane as the direction vector, this is,, 3, is a direction vector of L. Therefore, the smmetric equations are x = + 5 = z 3..Note r (t) = ( 3e 3t, cos t + π ), t r (t) =,, t To compute the angle of intersection we find r () = 3,, r () =,, so that cos θ = r () r () r () r () = = ( ),,,, 5 proj,,,, 5 =,, =,, =,,,,,,. ((sec r(x)dx = x)i + e x k ) dx ( ) ( = sec xdx i + ) ( dx j + = (tan x + C ) i + C j + (e x + C 3 ) k ) e x dx k 5.Answer is the absolute value of the triple product 3 = () = = 6.Two vectors which form two sides of the triangle are,, =,,,, and, 3, = 3, 3, 3,,. Hence i j k = 3, ( ), 3 = 3,, 3 3 The area of the parallelogram is 3,, 3 = = 3 and the area of the triangle is half this. 8

8 7.The level curves are given b f(x, ) = we get x x + = c x + cx. = c where c is a constant. Solving for So each level curve for c has a vertical asmptote at x =. Of the five given curves, that onl leaves (a) and (d) as the possible correct answers. Also notice that for each x, there is onl one corresponding value of i.e. is a well defined function of x. So each level curve must pass the vertical test. But clearl there are curves in (a) which fail this. Hence the correct answer is (d). 8. r (t) = t, t, 3. So the distance travelled is given b L(, 3) = r (t) dt = t + t + 9 dt = (t + 3) dt = 7. For the penultimate equalit, notice that t + t + 9 = (t + 3). 9.Completing squares x +8x = (x+) 6, + = (+), z z = (z 5) 5. So x 8x z z + 3 = (x + ) + ( + ) + (z 5) = (x + ) + ( + ) + (z 5) = So radius is = 3..(a) is alwas zero since a b = (b a) and cross product of parallel vectors is zero. (b) is zero if and onl if a b =. So (b) can be non zero. (c) is alwas zero since a b is perpendicular to both a and b and hence perpendicular to a + b. (d) is alwas zero since a a = for an vector a. (e) is alwas zero since a is perpendicular to b a..at t =, r() =,,, r () =,,, r () =,,. The normal vector to the normal plane is given b the velocit vector. So equation of the normal plane is x + =. The osculating plane at t is a plane containing r (t) and r (t). osculating plane at t = is given b i j k r () r () = =, 8, The equation to the osculating plane is thus x 8 =. So a normal to the 9

9 .The are not parallel, since,, is not a multiple of, 3, 3. If the intersect, then one can find a t and s such that 3 + t = 7 + s + t = 5 + 3s + t = 6 + 3s Solving the first two equations we get s =, t =. This clearl satisfies the third equation also. So the lines do intersect. To get the point of intersection we plug in t = in the equation of the first line (or equivalentl s = in the equation of the second line). Doing so we obtain the point of intersection as (5,, ). 3.Let v(t) =,, 3. Then the vector equation for the tangent line to C at P is given b tv() +,, = t, t, 3t +,, = t, t, 3t +. Then the parametric equation for the tangent line to C at P is given b x(t) = t, (t) = t, z(t) = 3t +.

12.1. Cartesian Space

12.1. Cartesian Space In most of your previous math classes, we worked with functions on the xy-plane only meaning we were working only in 2D. Now we will be working in space, or rather 3D. Now we will