MATH 147
Lab Key #5
3/1/2016
(1) Differentiate each function. Express your answers using positive exponents.
5 √
(a) f ( x ) = x4 + 3x2 − 2 + 12x4 − 8x3 + 7x2 + 1
3
(b) g ( x ) =
4
3
( x − 2)
Solution:
(a) We will use the chain rule to differentiate each function as follows:
5 1/2 d d 4
0
2
4
3
2
+
f (x) =
x + 3x − 2
12x − 8x + 7x + 1
dx
dx
4 d h
i 1
−1/2 d h
i
= 5 x4 + 3x2 − 2
x4 + 3x2 − 2 +
12x4 − 8x3 + 7x2 + 1
12x4 − 8x3 + 7x2 + 1
dx
2
dx
4 3
2
48x − 24x + 14x
= 5 x4 + 3x2 − 2
4x3 + 6x + √
2 12x4 − 8x3 + 7x2 + 1
4 24x3 − 12x2 + 7x
= 5 x4 + 3x2 − 2
4x3 + 6x + √
.
12x4 − 8x3 + 7x2 + 1
−4
(b) First, we will write g ( x ) as g ( x ) = 3 x3 − 2
. Then we will use the
chain rule as follows:
−4 −5 d h
i
d 3
x −2
= 3 · −4 x 3 − 2
·
x3 − 2
g0 ( x ) = 3 ·
dx
dx
=−
12
( x 3 − 2)
5
3x2 = −
36x2
( x 3 − 2)
5
.
1
(2) Find an equation of the tangent line to the curve defined implictly by
x2 + xy + y2 = 3 at (1, 1) . Express your answer in slope-intercept form.
Solution: We need to know the slope of the tangent line and a point on the
tangent line. Since we are given the point (1, 1) , we only need to find the
dy
slope. Hence we need to find dx .
i
d h 2
d
x + xy + y2 =
[3] .
dx
dx
d h 2i
d
d h 2i
x +
y = 0.
[ xy] +
dx
dx
dx
dy
d
dy
+ 2y
2x +
= 0.
[x] · y + x ·
dx
dx
dx
2x + y + x ·
dy
dy
+ 2y
= 0.
dx
dx
Now we can plug in our point to solve for
2 (1) + 1 + 1 ·
dy
dx
at out point.
dy
dy
+2·
= 0.
dx
dx
∴ 3+3
dy
= 0.
dx
dy
= −3.
dx
dy
∴
= −1.
dx
Hence, the slope of the tangent line is m = −1. Now using point-slope form,
we get the equation
y − 1 = −1 · ( x − 1) .
∴3
∴ y − 1 = − x + 1.
∴ y = − x + 2.
2
(3) The radius of a spherical tumor is decreasing at a rate of 2 millimeters
(mm) per week. How fast is the volume of the tumor decreasing when the
radius is 5 mm? Include the appropriate unit for the rate of change.
Solution: First, we need to set up an equation representing the volume of the
tumor. Since the tumor is a sphere, its volume V is given by
V=
4 3
πr ,
3
where r is the radius of the tumor. However, we’re interested in the rate of
change of the volume of the tumor, so we need to find dV
dt .
d 4 3
dV
=
πr .
dt
dt 3
dV
4 d h 3i
= π
r .
dt
3 dt
dV
4
dr
= π · 3r2 · .
dt
3
dt
dV
dr
= 4πr2 · .
dt
dt
Since the rate of decrease of the radius is 2 mm per week,
fore, when r = 5 mm,
dV
mm3
= 4π · 52 · −2
.
dt
week
dr
dt
mm
= −2 week
. There-
dV
mm3
= −200π
.
dt
week
Therefore, the volume
of
the tumor is decreasing at a rate of 200π cubic milmm3
limeters per week week .
∴
3
(4)
√ An object moves along a straight line with displacement function s (t) =
t2 + 1 where s is measured in meters (m) and t is measured in seconds (s).
Find the velocity and acceleration of the object at time t = 1.
Solution: The velocity is given by s0 (1) . Therefore, we need to differentiate
s (t) . For this, we will use the chain rule:
1/2
s ( t ) = t2 + 1
.
−1/2 d h
i
1/2 d 2
12
t +1
·
t2 + 1 .
t +1
=
dt
2
dt
−1/2
1 2
t
∴ s0 (t) =
.
· (2t) =
t +1
1/2
2
( t2 + 1)
∴ s0 (t) =
Hence, the velocity at t = 1 is given by s0 (1) =
1
1/2
( 12 + 1 )
=
acceleration is given by s00 (1) . Hence, we need to differentiate
use the quotient rule:
h
i
d
2 + 1 1/2 − t · d
2 + 1 1/2
t
·
t
t
[
]
dt
dt
s00 (t) =
.
t2 + 1
√1 m/s.
2
s0 (t) . We
The
will
1/2
− t · s0 (t)
.
t2 + 1
1/2
12 + 1
− 1 · s 0 (1)
00
∴ s (1) =
.
12 + 1
√
2√
−1
2 − √1
1
2
2
∴ s00 (1) =
=
= √ .
2
2
2 2
00
∴ s (t) =
Hence, the acceleration at t = 1 is
t2 + 1
1
√
2 2
m/s2 .
4