MATH 151 Midterm: MATH 151 TAMU Y2013 2013a Exam 2b Solutions
Spring 2013 Math 151
Exam II Version B Solutions
1. Ev=s′= 2t−1 = 0 when t=
1
2. Therefore the total distance trav-
eled is
f1
2−f(0)
+
f(2) −f1
2
=
−1
4−0
+
2−−1
4
=21
2ft .
2. BL(x) = f(a) + f′(a)(x−a).f′(x) = −1
x2,
so f(2) = 1
2and f′(2) = −1
4. Therefore,
L(x) = 1
2−1
4(x−2).f(2.1) ≈L(2.1) =
1
2−1
4(2.1−2) = 1
2−1
40 =19
40 .
3. Cdy
dx =dy/dt
dx/dt =2t−2
3t2−12t. To find t, set
t3−6t2+ 9 = 4 and t2−2t+ 1 = 0.t= 1
is the only solution to both equations. Then
dy
dx =2−2
3−12 = 0, which means the tangent
line is horizontal.
4. Ady
dx =dy/dt
dx/dt =πcos(π t)
6t2. To find t,
set 2t3= 2 and sin(π t) = 0.t= 1 is
the only solution to both equations. Then
dy
dx =πcos(π)
6=−π
6.
5. CU′(x) = f(x)g′(x)−g(x)f′(x)
(f(x))2,
so U′(1) = f(1)g′(1) −g(1)f′(1)
(f(1))2=
(−2)(2) −(1)(3)
22=−7
4.
6. DV′(x) = g(x)f′(x)+f(x)g′(x), so V′(1) =
g(1)f′(1) + f(1)g′(1) = (1)(3) + (−2)(2) =
−1.
7. DS′(x) = 4(f(x))3(f′(x)), so S′(1) =
4(f(1))3(f′(1)) = 4(−2)3(3) = −96.
8. AFrom the figure below, d2=p2+ 1, with
dp
dt =1
2, p = 3,and dd
dt what we are solving
for. Then 2ddd
dt = 2pdp
dt . When p= 3, d =
√10, so dd
dt =2(3)(1
2)
2√10 =3
2√10
m
s.
9. EDifferentiate implicitly: 3y2dy
dx −2x=
−7dy
dx,dy
dx(3y2+ 7) = 2x,dy
dx =2x
3y2+7.
10. Df(4)(x) = f(x) = sin x, so the derivatives
repeat every four. Since 2013 ÷4has a re-
mainder of 1, f(2013)(x) = f′(x) = cos x.
11. Br′(t) = (2t+ 3)i+2
t3j.r′(1) = 5i+ 2j
is a tangent vector, so a tangent vector of
unit length is ˆ
r(1) = 1
√52+ 22(5i+ 2j) =
5
√29 i+2
√29 j.
12. EH′(x) = f′x
21
2,H′′ (x) =
f′′ x
21
22
, etc. Following this pat-
tern, H(16)(x) = f(16) x
21
216
. Then
H(16)(0) = f(16)(0) 1
216
= 2 1
216
=
1
215 .
13. Alim
t→∞ e−2t= 0,so
lim
t→∞
500
5 + 95e−2t=500
5 + 0 =100.
14. Blim
x→0
sin(5x)
sin(7x)= lim
x→0
sin(5x)
1·1
sin(7x)
= lim
x→0
sin(5x)
5x ·7x
sin(7x)·5x
7x →5
7.
15. AApply the Chain Rule: d
dx(sin(cos x)) =
cos(cos x)·(−sin x).
16. .
(a) f′(x) = (x2+ 1)(1) −(x)(2x)
(x2+ 1)2, so
m=f′(3) = (9 + 1)(1) −(3)(6)
(9 + 1)2=
−8
100 =−2
25. When x= 3, y =
3
32+ 1 =3
10. Therefore, an equa-
tion of the line tangent to the curve
at x= 3 is y−3
10 =−2
25 (x−3), or
y=−2
25 x+27
50 .
1
Document Summary
Exam ii version b solutions: e v = s = 2t 1 = 0 when t = 2(cid:19)(cid:12)(cid:12)(cid:12)(cid:12) f (2) f(cid:18) 1 ft . = 2: b l(x) = f (a) + f (a)(x a). f (x) = . 1 (x 2). f (2. 1) l(2. 1) = 2 dy dx (2. 1 2) = dy/dt dx/dt: c. To nd t, set t3 6t2 + 9 = 4 and t2 2t + 1 = 0. t = 1 is the only solution to both equations. 3t2 12t: a dy dx dy/dt dx/dt. To nd t: a set 2t3 = 2 and sin( t) = 0. t = 1 is the only solution to both equations. 6 lim t : b f (x)g (x) g(x)f (x) (f (x))2 f (1)g (1) g(1)f (1) (f (1))2, c u (x) so u (1) = ( 2)(2) (1)(3)
