MEMORIAL UNIVERSITY OF NEWFOUNDLAND
DEPARTMENT OF MATHEMATICS AND STATISTICS
MATH 1000
Assignment 5
SOLUTIONS
[4]
1. (a) We use the Quotient Rule, followed by the Product Rule:
g 0 (t) =
(t2 et )0 (t2 + et ) − (t2 + et )0 (t2 et )
(t2 + et )2
[(t2 )0 et + (et )0 t2 ](t2 + et ) − (2t + et )(t2 et )
=
(t2 + et )2
(2tet + t2 et )(t2 + et ) − (2t + et )(t2 et )
=
(t2 + et )2
=
[4]
t4 et + 2te2t
.
(t2 + et )2
(b) We use the Product Rule twice:
y 0 = 2x [sin(x) cos(x)]0 + (2x )0 sin(x) cos(x)
= 2x [[sin(x)]0 cos(x) + [cos(x)]0 sin(x)] + 2x ln(2) sin(x) cos(x)
= 2x [[cos(x)] cos(x) + [− sin(x)] sin(x)] + 2x ln(2) sin(x) cos(x)
= 2x cos2 (x) − 2x sin2 (x) + 2x ln(2) sin(x) cos(x).
[4]
(c) We use the Product Rule, followed by the Chain Rule:
f 0 (x) = (x3 )0 (2x9 − 5)6 + [(2x9 − 5)6 ]0 x3
= 3x2 (2x9 − 5)6 + [6(2x9 − 5)5 (2x9 − 5)0 ]x3
= 3x2 (2x9 − 5)6 + [6(2x9 − 5)5 (18x8 )]x3
= 3x2 (2x9 − 5)6 + 108x11 (2x9 − 5)5 .
[4]
(d) We use the Chain Rule, followed by the Product Rule:
d
dy
= ex csc(x) · [x csc(x)]
dx
dx
d
d
x csc(x)
=e
·
[x] csc(x) + [csc(x)]x
dx
dx
= ex csc(x) · [1 · csc(x) + [− csc(x) cot(x)]x]
= ex csc(x) · [csc(x) − x csc(x) cot(x)]
= ex csc(x) csc(x) − xex csc(x) csc(x) cot(x).
Fall 2016
[4]
(e) We use the Chain Rule twice:
d
d
[f (x)] = 5 sec4 (x3 ) · [sec(x3 )]
dx
dx
d 3
[x ]
dx
= 5 sec4 (x3 ) · sec(x3 ) tan(x3 ) · (3x2 )
= 5 sec4 (x3 ) · sec(x3 ) tan(x3 ) ·
= 15x2 sec5 (x3 ) tan(x3 ).
[4]
(f) We use the Quotient Rule, and then the Chain Rule two separate times:
√
0
f (x) =
√
0
4x2 + 5 [sin(7x) + 3] − [sin(7x) + 3]0 4x2 + 5
[sin(7x) + 3]2
√
1
+ 5)− 2 · (4x2 + 5)0 · [sin(7x) + 3] − [cos(7x) · (7x)0 + 0] 4x2 + 5
=
[sin(7x) + 3]2
√
1
1
(4x2 + 5)− 2 · (8x) · [sin(7x) + 3] − [cos(7x) · 7] 4x2 + 5
2
=
[sin(7x) + 3]2
√
1
4x(4x2 + 5)− 2 [sin(7x) + 3] − 7 cos(7x) 4x2 + 5
=
[sin(7x) + 3]2
1
(4x2
2
=
[4]
4x[sin(7x) + 3] − 7 cos(7x)(4x2 + 5)
√
.
4x2 + 5[sin(7x) + 3]2
(g) We use the Chain Rule, followed by the Quotient Rule:
0
2x − 1
2x − 1
0
2
y = − csc
·
2x + 1
2x + 1
2x − 1
(2x − 1)0 (2x + 1) − (2x + 1)0 (2x − 1)
= − csc2
·
2x + 1
(2x + 1)2
2x − 1
2(2x + 1) − 2(2x − 1)
= − csc2
·
2x + 1
(2x + 1)2
2x − 1
4
= − csc2
·
2x + 1
(2x + 1)2
4
2x − 1
2
=−
csc
.
(2x + 1)2
2x + 1
[5]
2. Using the Quotient Rule, we have
y0 =
[tan(x)]0 cos(x) − [cos(x)]0 tan(x)
[cos(x)]2
=
sec2 (x) cos(x) − [− sin(x)] tan(x)
cos2 (x)
=
sec(x) + sin(x) tan(x)
.
cos2 (x)
Alternatively, we could first rewrite the function as y = tan(x) sec(x) and then use the
Product Rule:
y 0 = [tan(x)]0 sec(x) + [sec(x)]0 tan(x)
= [sec2 (x)] sec(x) + [sec(x) tan(x)] tan(x)
= sec3 (x) + sec(x) tan2 (x).
(It’s not hard to show that these expressions are equivalent. Another way to write this
derivative is as
1 + sin2 (x)
,
y0 =
cos3 (x)
and any of these are acceptable.)
Either way, when x = π3 , y 0 = 14 is the slope of the tangent line. By substituting into the
√
given function, we find that y = 2 3. Thus the equation of the tangent line is
√
π
y − 2 3 = 14 x −
3
√
14π
y = 14x −
+ 2 3.
3
1
. Hence its equation is
The slope of the normal line is − 14
√
1 π
y−2 3=−
x−
14
3
√
1
π
y =− x+
+ 2 3.
14
42
[5]
3. (a) We are told that s(0) = 0, and since s(0) = D, we immediately have D = 0 and therefore
s(t) = At3 + Bt2 + Ct.
We also know that v(0) = 0, and
v(t) = s0 (t) = A(3t2 ) + B(2t) + C(1) = 3At2 + 2Bt + C.
Thus v(0) = C and so C = 0, which means that
s(t) = At3 + Bt2 .
Finally, we are given that s(2) = s(6) = −36. This means that
8A + 4B = −36 and 216A + 36B = −36.
Solving this system of equations gives A = 2 and B = −13.
[2]
(b) We now have
s(t) = 2t3 − 13t2
and v(t) = s0 (t) = 6t2 − 26t.
We want to solve the equation
v(t) = 0
6t2 − 26t = 0
2t(3t − 13) = 0,
so either t = 0 or t = 13
≈ 4.33. Hence the object is again at rest at approximately
3
4.33 seconds. Its position at this time is
2
3
13
2197
13
13
− 13
=−
s
=2
≈ −81.37 metres.
3
3
3
27