Final Exam Review Solution

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

• Notes
• 12

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 .

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.
(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

Subscribe to view the full document.

{[ snackBarMessage ]}

"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

What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern