# soln+of+practice.pdf

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Georgia Institute of Technology

Mathematics

MATH 2401

Jing Hu

Spring

Description

Math 2401 Name (Print):
Spring 2013
Midterm 1 GT ID:
02/19/13
Time Limit: 80 Minutes Section
signature:
Instructions:This exam contains 7 pages (including this cover page) and 6 problems. Check to see
if any pages are missing. Print your name and sign your signature to indicate that you accept the
honor code.
The following rules apply:
▯ Calculator is allowed to use. You may not use
Problem Points Score
your books, notes on this exam.
▯ Box the answer and show the work you did to 1 8
arrive at the answer to receive full credit.
2 9
▯ Organize your work in a reasonably neat and
coherent way, in the space provided. Work scat- 3 9
tered all over the page without a clear ordering
will receive very little credit. 4 8
▯ Mysterious or unsupported answers will not
receive full credit. A correct answer, unsupport- 5 8
ed by calculations, explanation, or algebraic work
will receive no credit; an incorrect answer support- 6 8
ed by substantially correct calculations and expla-
Total: 50
nations might still receive partial credit.
▯ If you need more space, use the back of the pages;
clearly indicate when you have done this. Math 2401 Midterm 1 - Page 2 of 7 02/19/13
1. (8 points) Calculate, simplifying the answer:
R1 ▯t 2
(a) (2 points) 0[e i + sin(▯t)j + t k]dt
solution:
R
1[e▯ti + sin(▯t)j + t k]dt
0 R R R
= 1e▯t dti + 1sin(▯t)dtj + 1e ▯tt dtk
0 0 3
= ▯e ▯t j i + (▯cos▯t)j j +1 t j k
0 ▯ 0 3 0
= e▯1 i + j + k 1
e ▯ 3
d t ▯t ▯t t 1
(b) (3 points) dt[(e i + e j + tk) ▯ (e i + e j + t2k)]
solution:
d[(e i + e▯t j + tk) ▯ (e▯ti + e j + 1 k)]
dt t
= d(1 + 1 + )1
dt t
1
= ▯ t2 .
d 1 1
(c) (3 points) dt[(i + tj + k) ▯ ( t ▯ lntj + k)t
solution:
d[(i + tj + k) ▯ ( i ▯ lntj + k)]
dt 1 t1 t 1
= j ▯ ( t ▯ lntj + k)t+ (i + tj + k) ▯ (▯ )(i + 2j + k)
1 1 t
= ▯ it▯ ▯ k +t0
1 1
= ▯ it▯ ▯ k t Math 2401 Midterm 1 - Page 3 of 7 02/19/13
2. (9 points) Consider the curve r(t) = e sinti + e costj + e k: Note that the point P(0;1;1)
belongs to this curve. Find at P.
(a) (3 points) the unit tangent;
(b) (3 points) the principal normal
(c) (3 points) the curvature;
Solution:
a).
0 t t t t t
r (t) = [epcost + e sint]i + [e cost ▯ e sint]j + e k p
kr (t)k = (e cost + e sint) + (e cost ▯ e sint) + e 2t= 3et
0
T(t) = r0(t) = p1 [(cost + sint)i + (cost ▯ sint)j + k]
kr (t)k 3
At P(0;1;1); e = 1; t = 0:
T(0) = p1 (i + j + k) .
3
b).
0 0
N(t) = T 0t) then N(0) = T 00)
kT (t)k kT (0)k
From part a), T (t) = p1 [(cost ▯ sint)i + (▯cost ▯ sint)j].
3 p
0 1 0 2
At point P, T (0) = p3(i ▯ j),kT (0)k = p 3:
0
N(0) = T0(0) = p1 (i ▯ j)
kT (0)k 2
c).
0
▯(t) = kT (t)k
ds=dt
p p p
kT (0)k p2= 3

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