Final Exam Review Solution

Final Exam Review Solution - Name Math 247 Honors Calculus...

This preview shows page 1 - 4 out of 12 pages.

Name: Math 247– Honors Calculus III Fall 2012 FINAL REVIEW 12/4/12 1. Find the angle between the vectors h 1 , 1 , 2 i and h 4 , 1 , - 2 i . Solution: Compute | ~v | = 6, | ~w | = 21, and ~v · ~w = 1. Since ~v · ~w = | ~v || ~w | cos θ , we obtain θ = cos - 1 ( 1 3 14 ). 2. Suppose ~ r (3) = h 1 , 2 , 1 i ~ r 0 (3) = h- 2 , 1 , 0 i ~s (3) = h- 1 , 2 , 3 i ~s 0 (3) = h 2 , - 2 , 5 i Compute ( ~ r · ~s ) 0 (3) and ( ~ r × ~s ) 0 (3). Solution: ( ~ r · ~s ) 0 (3) = ~ r 0 (3) · ~s (3) + ~ r (3) · ~s 0 (3) = h- 2 , 1 , 0 i · h- 1 , 2 , 3 i + h 1 , 2 , 1 i · h 2 , - 2 , 5 i = - 1 + 3 = 2 ( ~ r × ~s ) 0 (3) = ~ r 0 (3) × ~s (3) + ~ r (3) × ~s 0 (3) = h- 2 , 1 , 0 i × h- 1 , 2 , 3 i + h 1 , 2 , 1 i × h 2 , - 2 , 5 i = h 15 , 3 , - 7 i 3. Find a vector normal to the plane containing the points (3 , 3 , 1), (4 , - 1 , 3), and (2 , 0 , 0). Solution: First find two vectors in the plane: ~v = h 1 , - 4 , 2 i and ~w = h 1 , 3 , 1 i . Then a normal vector is given by ~ N = ~v × ~w = h- 10 , 1 , 7 i 4. Find an equation of the plane containing the point (2 , 1 , 0) and the line ~ r ( t ) = h- 1 + t, 3 - t, 2 + t i . Solution: We have a point P = (2 , 1 , 0), so we need a normal vector. To get this, we need two vectors in the plane. Since the line is in the plane, its direction vector ~v = h 1 , - 1 , 1 i is in the plane. We can also find a point on the line, e.g. Q = ~ r (0) = ( - 1 , 3 , 2), and let ~w = Q - P = h- 3 , 2 , 2 i . Then set ~ N = ~w × ~v = h 4 , 5 , 1 i . So our plane has equation 4 x + 5 y + z = 13. 5. Find an equation of the plane containing the lines ~ r ( t ) = h 8 , 3 , 4 i + t h 1 , - 1 , 0 i and ~s ( t ) = h 5 , 0 , 0 i + t h 0 , 3 , 2 i . Solution: To find a point in the plane, we find where the two lines meet. So we set h 8 + t, 3 - t, 4 i = h 5 , 3 s, 2 s i , and find they meet at P = (5 , 6 , 4). The direction vectors of each line give us vectors in the plane ~v = h 1 , - 1 , 0 i and ~w = h 0 , 3 , 2 i , so we can get a normal vector ~ N = ~w × ~v = h 2 , 2 , - 3 i . So our plane has equation 2 x + 2 y - 3 z = 10. 6. Find the unit tangent and the unit normal vectors to the curve ~ r ( t ) = h t 3 , 3 t, t i at the point h 1 , 3 , 1 i .
Image of page 1

Subscribe to view the full document.

Solution: ~ T ( t ) = ~ r 0 ( t ) | ~ r 0 ( t ) | = h 3 t 2 , 3 , 0 i 3 t 4 + 1 = h t 2 t 4 + 1 , 1 t 4 + 1 , 0 i ~ T (1) = h 1 2 , 1 2 , 0 i ~ N ( t ) = ~ T 0 ( t ) | ~ T 0 ( t ) | = h 2 t ( t 4 +1) 3 2 , - 2 t 3 ( t 4 +1) 3 2 , 0 i 2 t t 4 +1 ( t 4 +1) 3 2 = h 1 t 4 + 1 , - t 2 t 4 + 1 , 0 i ~ N (1) = h 1 2 , - 1 2 , 0 i 7. Find a unit vector that is tangent to the curve ~ r ( t ) = h sin t, cos t, t i at the poitn (1 , 0 , π 2 ). Solution: ~ T ( π 2 ) = ~ r 0 ( π 2 ) | ~ r 0 ( π 2 ) | = h 0 , - 1 , 1 i |h 0 , - 1 , 1 i| = h 0 , - 1 2 , 1 2 i 8. Find the length of the curve ~ r ( t ) = h 4 sin t, 4 cos t, 3 t i between (0 , 4 , 0) to (0 , - 4 , 3 π ). Solution: Z π 0 | ~ r 0 ( t ) | dt = Z π 0 5 dt = 5 π 9. Find the length of the curve ~ r ( t ) = h t 3 3 , 2 t, t 2 i from (0 , 0 , 0) to ( 8 3 , 4 , 4). Solution: Z 2 0 | < t 2 , 2 , 2 t > | dt = Z 2 0 | t 2 + 2 | dt = Z 2 0 t 2 + 2 dt 8 3 + 4 = 20 3 10. Determine each of the following limits: (a) lim ( x,y ) (0 , 0) y 4 x 4 +5 y 4 (b) lim ( x,y ) (0 , 0) x 2 y x 2 + y 2 (c) lim ( x,y ) (0 , 0) x 2 - y 2 x 2 + y 2 (d) lim ( x,y ) (0 , 0) yx 2 x 4 + y 2 (e) lim ( x,y ) (0 , 0) x 2 y 2 x 4 +3 y 4 (f) lim ( x,y ) (0 , 0) x 2 y 2 x 2 + y 2 Solution: (a) Along y = mx , we have lim x 0 ( mx ) 4 x 4 +5( mx ) 4 = m 4 1+5 m 4 , DNE (b) Along y = mx , we have lim x 0 x 2 ( mx ) x 2 +( mx ) 2 = lim x 0 m 1+ m 2 x = 0. Indeed, by the squeeze theorem, we have lim ( x,y ) (0 , 0) | x 2 y x 2 + y 2 | ≤ lim ( x, y ) (0 , 0) | y | = 0, so the limit is 0.
Image of page 2
(c) Along y = mx , we have lim x 0 x 2 - ( mx ) 2 x 2 +( mx ) 2 = 1 - m 2 1+ m 2 , DNE (d) Along y = mx , we have lim x 0 ( mx ) x 2 x 4 +( mx ) 2 = lim x 0 mx x 2 + m 2 = 0. However, along y = x 2 , we hae lim x 0 ( x 2 ) x 2 x 4 +( x 2 ) 2 = 1 2 . DNE (e) Along y = mx , we have lim x 0 x 2 ( mx
Image of page 3

Subscribe to view the full document.

Image of page 4

{[ snackBarMessage ]}

Get FREE access by uploading your study materials

Upload your study materials now and get free access to over 25 million documents.

Upload now for FREE access Or pay now for instant access
Christopher Reinemann
"Before using Course Hero my grade was at 78%. By the end of the semester my grade was at 90%. I could not have done it without all the class material I found."
— Christopher R., University of Rhode Island '15, Course Hero Intern

Ask a question for free

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern