Math 1225 Solution Homework 5 Fall 2015
1)
dy
=cotθ (secθ tanθ )+(-csc 2θ ) when y = cot θ secθ
dθ
df(θ ) [( − csc θ cot θ ) cos θ + csc θ ( − sin θ )]sin θ − (csc θ cos θ ) cos θ
=
=
dθ
sin 2 θ
[ −(cot 2 θ + tan θ )]sin θ − (csc θ cos2 θ )
csc θ cos θ
, when f(θ ) =
2
sin θ
sin θ
2)
3) An elastic band is hung on a hook and a mass is hung on the lower
end of the band. When the mass is pulled downward and then released,
it vibrates vertically. The equation of position of the mass at any time t seconds
is given to be s = 2cos2 (t) + 3sin(t/2) centimeters. Find the velocity at t = π .
solution:
velocity =
s/ =
4 cos(t )( − sin(t)) + (3 / 2) cos(t/ 2) =
0, for x = π .
so mass is at rest.
Use the Chain Rule to evaluate the following (Do not back substitute)
5) Find dy/dx if:
y = m2 + m and m =
x . for x = 4
dy dy dm
 1  5
(2m 1) 
= =+
 
=
dx dm dx
2 x  4
6) Find dp/dk if:
p = q– 2q3
and q = r2 and r = k +1. for k = 2
dp dp dq dr
==
(1 − 6q 2 )(2r )(1) =
[1 − 6(81)](2(3))(1) =
−2910
dk dk dr dk
7) Find dy/dx for each of the following (using the inside/outside rule)
a)
y=
3
(5x 3 − x) 2 = (5x 3 − x)2/3 ,
b) y =x3 e3-5x ,
−1
dy 2
= (5x 3 − x) 3 (15 x 2 − 1)
dx 3
dy
=
−5 x 3e3−5 x + 3x 2 e3−5 x
dx
c)
dy
2
=
72 x [6cos(3x )sin(3x )] + 72 x (2 ln(7)) cos2 (3x )
y = 7 2 x cos
(3 x)
dx
d)
y = tan(sin 2 (4 x))
dy
= [sec 2 (sin 2 (4 x ))][8sin(4 x) cos(4 x)]
dx
8)
Find the following limits:
a ) Limit
6sin(3x )
sin(3x )
= 9Limit
= 9(1)
= 9
x →0
2x
3x
b) Limit
sin(3x )sin(2 x )
sin(3x )
sin(2 x )
sin(3x )
sin(2 x )
Limit = 3Limit
Limit = 3(1)(1)
= Limit
= 3
2
0
0
0
0
x
x
x
x
→
→
→
→
2x
2x
3x
2x
x
x →0
x →0
c) Limit
x →0
sin(3x )
sin(3x )
1
 sin(3x ) x 
= Limit 
3Limit
Limit
=
= 3(1)
= 3

x →0
x →0
x →0  sin x 
sin( x )
3x
 x sin x 


 x 