Fall 2016 Math 19 - HW 5 solutions
(If you find any errors, please let your instructor know ASAP)
Section 2.6, # 10: The acceleration due to gravity, g, varies with height above the surface of the
earth, in a certain way. If you go down below the surface of the earth, g varies in a different way.
It can be shown that g is given by

GM r


for r < R
3
R
g=

 GM for r ≥ R
r2
where R is the radius of the earth, M is the mass of the earth, G is the gravitational constant, and
r is the distance to the center of the earth.
(a) Sketch a graph of g against r.
(b) Is g a continuous function of r? Explain your answer.
(c) Is g a differentiable function of r? Explain your answer.
Solution:
(a) Here is the sketch graph:
g
GM
R2
g=
GM r
R3
O
g=
GM
r2
r
R
(b) Answer: g is a continous function of r
proof : In order to check the continuity of g, we just need to show that g is continuous at r = R.
By definition, the left limit of g at x = R is :
lim− g(r) = lim−
r→R
r→R
GM r
GM R
GM
=
= 2
3
3
R
R
R
On the other hand, the right limit of g at x = R is :
lim+ g(r) = lim+
r→R
r→R
GM
GM
= 2
2
r
R
Thus,
lim− g(r) = lim+ g(r) =
r→R
r→R
GM
R2
So
GM
= g(r)
r→R
R2
As a result, g is continuous at r = R, our proof is completed.
lim g(r) =
(c) Answer: g is NOT a differentiable function of r
proof We will prove this by showing the fact that g is not differentiable at r = R.
The right derivative of g at r = R is
g(R + h) − g(R)
lim+
= lim+
h→0
h→0
h
= lim+
GM
(R+h)2
−
GM
R2
h
GM (R2 −(R+h)2 )
(R+h)2 R2
h
GM (R2 − (R + h)2 )
= lim+
h→0
h(R + h)2 R2
h(−2R − h)
= lim+ GM
h→0
h(R + h)2 R2
(−2R − h)
= lim+ GM
h→0
(R + h)2 R2
2GM
=− 3
R
h→0
However, the left derivative of g at x = R is
lim−
h→0
g(R + h) − g(R)
= lim−
h→0
h
GM (R+h)
R3
−
GM
R2
h
GM (R + h) − GM R
= lim−
h→0
hR3
GM h
= lim−
h→0
hR3
GM
= 3
R
Thus, the left derivative is not equal to the right derivative, which implies that g is not differentiable
at r = R, our proof is completed.
Section 3.1, # 60: Find the equation of the line tangent to f (x) at x = 2, if
f (x) =
x3
4
−
2
3x
Solution: First of all, we should calculate the first derivative of f (x), that is
4
3
f 0 (x) = x2 + 2
2
3x
Secondly, we need to find the slope of the tangent line. Since the line is tangent to f (x) at x = 2,
the slope is equal to
3
4
19
f 0 (2) = 22 +
=
2
2
3·2
3
10
. Thus the equation
Now, notice that the tangent line must pass the point (2, f (2)), which is 2,
3
of the line can be written as
y = f 0 (2)(x − 2) + f (2)
19
10
= (x − 2) +
3
3
28
19
= x−
3
3
The answer is y =
19
28
x−
.
3
3
Section 3.1, # 62 On what intervals is the graph of f (x) = x4 − 4x3 both decreasing and concave
up?
Solution: We first calculate the first and second derivative of f (x):
f 0 (x) = 4x3 − 12x2
f 00 (x) = 12x2 − 24x
We know that a (differentiable) function f (x) is decrasing on an interval I if and only if f 0 (x) < 0
for any x ∈ I, and f (x) is concave up on an interval I if and only if f 00 (x) > 0 for any x ∈ I.
Thus the interval I that we are looking for must satisfy the following two constrictions:
f 0 (x) < 0,
f 00 (x) > 0
for any x ∈ I
Solving the inequality f 0 (x) < 0, we get
f 0 (x) < 0
4x3 − 12x2 < 0
4x2 (x − 3) < 0
x < 3 and x 6= 0
Solving the inequality f 00 (x) > 0, we get
f 00 (x) > 0
12x2 − 24x > 0
12x(x − 2) > 0
x > 2 or x < 0
Hence the intervals on which f (x) is both decreasing and concave up is (−∞, 0) ∪ (2, 3) .
Section 3.1, # 64 If
f (x) = 4x3 + 6x2 − 23x + 7
Find the intervals on which
f 0 (x) ≥ 1
.
Solution: First of all, we should calculate the first derivative of f (x):
f 0 (x) = 12x2 + 12x − 23
Then we solve the inequality f 0 (x) ≥ 1 as following:
f 0 (x) ≥ 1
12x2 + 12x − 23 ≥ 1
12x2 + 12x − 24 ≥ 0
x2 + x − 2 ≥ 0
(x − 1)(x + 2) ≥ 0
x ≥ 1 or x ≤ −2
Note: to arrive at these final inequalities, it’s important to remember that a · b > 0 if a and b are
either both positive OR both negative.
Thus the intervals on on which f 0 (x) ≥ 1 is (−∞, −2] ∪ [1, ∞) .
Section 3.2, # 20 Find the derivative of
f (x) = e1+x
Solution: Recall the following properties of derivative:
Proposition 1: if f is differentiable and c is a constant, then
d
(cf (x)) = cf 0 (x)
dx
Proposition 2:
d x
(e ) = ex
dx
By using the above properties of derivative, we have:
d 1+x
d
(e ) =
(e · ex )
dx
dx
d
= e · (ex )
dx
= e · ex
= e1+x
Therefore the derivative of f (x) = e1+x is e1+x .
(by Proposition 1)
(by Proposition 2)
Section 3.3, # 8 Find the derivative of
y = (t3 − 7t2 + 1)et
Solution:Recall the following property of derivative:
Proposition 3 (Product Rule): If f = f (t) and g = g(t) are differentiable, then
(f g)0 = f 0 g + f g 0
By using the above property of derivative, we have:
d
dy
= ((t3 − 7t2 + 1)et )
dt
dt
d
d
= et (t3 − 7t2 + 1) + (t3 − 7t2 + 1) (et ) (by Proposition 3)
dt
dt
t
2
3
2
t
= e (3t − 14t) + (t − 7t + 1)e
= (t3 − 4t2 − 14t + 1)et
Hence the derivative of y = (t3 − 7t2 + 1)et is (t3 − 4t2 − 14t + 1)et .
Section 3.3, # 20 Find the derivative of
g(t) =
4
√
3+ t
Solution:Recall the following property of derivative:
Proposition 4 (Quotient Rule): If f = f (t) and g = g(t) are differentiable, then
0
f 0g − f g0
f
=
g
g2
By using the above property of derivative, we have:
√
√
(4)0 (3 + t) − 4(3 + t)0
0
√
g (t) =
(3 + t)2
1
0−4· √
2 t
√
=
(3 + t)2
2
√
= − 3/2
t + 6t + 9 t
Hence the derivative of g(t) =
4
2
√ is −
√ .
3+ t
t3/2 + 6t + 9 t
Section 3.3, # 22 Find the derivative of
w=
5 − 3z
5 + 3z
Solution: We use the Quotient Rule to solve this problem:
dw
(5 − 3z)0 (5 + 3z) − (5 − 3z)(5 + 3z)0
=
dz
(5 + 3z)2
−3(5 + 3z) − 3(5 − 3z)
=
(5 + 3z)2
30
=−
(5 + 3z)2
Therefore the derivative of w =
5 − 3z
30
is −
.
5 + 3z
(5 + 3z)2
Section 3.3, # 46: Find the equation of the tangent line at x = 1 to y = f (x), if
f (x) =
3x2
5x2 + 7x
Solution: First of all, we should calculate the first derivative of f (x) as the following:
3x
5x + 7
(3x)0 (5x + 7) − 3x(5x + 7)0
=
(5x + 7)2
3(5x + 7) − 3x · 5
=
(5x + 7)2
21
=
(5x + 7)2
f 0 (x) =
Secondly, we need to find the slope of the tangent line. Since the line is tangent to f (x) at x = 1
the slope is equal to
7
21
=
f 0 (1) =
2
(5 + 7)
48
Now, notice that the tangent line must pass the point (1, f (1)), which is
of the line can be written as
y = f 0 (1)(x − 1) + f (1)
1
7
= (x − 1) +
48
4
5
7
= x+
48
48
The answer is y =
7
5
x+
.
48
48
1
1,
. Thus the equation
4
Section 3.3, # 50 Find the derivative of
f (x)
g(x) + 1
Solution:We use the Quotient Rule to solve this problem
0
f (x)
f 0 (x)(g(x) + 1) − f (x)(g(x) + 1)0
=
g(x) + 1
(g(x) + 1)2
f 0 (x)(g(x) + 1) − f (x)g 0 (x)
=
(g(x) + 1)2
f 0 (x) + f 0 (x)g(x) − f (x)g 0 (x)
=
g 2 (x) + 2g(x) + 1
Therefore the derivative of
f 0 (x) + f 0 (x)g(x) − f (x)g 0 (x)
f (x)
is
.
g(x) + 1
g 2 (x) + 2g(x) + 1
Section 3.4, # 34 Find the derivative of
y=
x2 + 2
3
2
Solution:Recall the following property of derivative:
Proposition 5 (Chain Rule) If f and g are differentiable function, then
d
f (g(x)) = f 0 (g(x)) · g 0 (x)
dx
Let f (x) = x2 and g(x) =
x2 + 2
, then y = f (g(x)), so by the Chain Rule, we have
3
d
dy
=
f (g(x))
dx
dx
= f 0 (g(x)) · g 0 (x)
2
2
0
x +2
x +2
=2
·
3
3
2
x +2
2x
=2
·
3
3
3
4x + 8x
=
9
Therefore the derivative of y =
x2 + 2
3
2
is
4x3 + 8x
.
9
Section 3.4, # 44 Find the derivative of
2
f (x) = e−(x−1)
Solution: Let h(x) = e−x and g(x) = (x − 1)2 , then f (x) = h(g(x)), so by the Chain Rule, we
have
d
d
f (x) =
h(g(x))
dx
dx
= h0 (g(x)) · g 0 (x)
2
= −e−(x−1) · ((x − 1)2 )0
2
= −e−(x−1) (2x − 2)
2
2
Therefore the derivative of f (x) = e−(x−1) is −e−(x−1) (2x − 2) .
Section 3.4, # 48 Find the derivative of
z = (te3t + e5t )9
Solution: Let h(t) = t9 and g(t) = te3t + e5t , then z = h(g(t)), so by the Chain Rule and
Produce Rule, we have
dz
d
= h(g(t))
dt
dt
= h0 (g(t)) · g 0 (t)
= 9(te3t + e5t )8 · (te3t + e5t )0
= 9(te3t + e5t )8 · ((te3t )0 + (e5t )0 )
= 9(te3t + e5t )8 · (1 · e3t + t · 3e3t + 5e5t )
= 9(te3t + e5t )8 ((3t + 1)e3t + 5e5t )
= 9e27 (t + e2t )8 (3t + 1 + 5e2t )
= 9e27 (e2t + t)8 (5e2t + 3t + 1)
Therefore the derivative of z = (te3t + e5t )9 is 9e27 (e2t + t)8 (5e2t + 3t + 1) .
Section 3.4, # 76 Given F (2) = 1, F 0 (2) = 5, F (4) = 3, F 0 (4) = 7 and G(4) = 2, G0 (4) = 6,
G(3) = 4, G0 (3) = 8, find:
(a) H(4) if H(x) = F (G(x))
(b) H 0 (4) if H(x) = F (G(x))
(c) H(4) if H(x) = G(F (x))
(d) H 0 (4) if H(x) = G(F (x))
(e) H 0 (4) if H(x) = F (x)/G(x)
Solution:
(a)
H(4) = F (G(4))
= F (2)
=1
Hence H(4) = 1
(b) By the Chain Rule, we have
H 0 (4) = F 0 (G(4)) · G0 (4)
= F 0 (2) · 6
= 5 · 6 = 30
Hence H 0 (4) = 30
(c)
H(4) = G(F (4))
= G(3)
=4
Hence H(4) = 4
(d) By the Chain Rule, we have
H 0 (4) = G0 (F (4)) · F 0 (4)
= G0 (3) · 7
= 8 · 7 = 56
Hence H 0 (4) = 56
(e) By the Quotient Rule, we have
F 0 (4)G(4) − F (4)G0 (4)
H (4) =
G(4)2
7·2−3·6
=
= −1
22
0
Hence H 0 (4) = −1