Due: 02/25/2016 in class
Instructor: Janina Letz
MATH 1210, Spring 2016
Lab #6 Solution
Name:
Question:
1
2
3
4
5
Total
Points:
8
10
12
10
10
50
Score:
Instructions: Please show all of your work as partial credit will be given where appropriate, and there may
be no credit given for problems where there is no work shown. All answers should be completely simplified,
unless otherwise stated. No calculators or electronics of any kind are allowed.
1. (Linear approximation) Find the linear approximation to the given functions at the specified points.
With the help of a graphing calculator, draw a sketch of the function and its linear approximation in
the indicated interval.
(a) (4 points) f (x) = x2 at a = 2, for x ∈ [0, 3]
Solution: f 0 (x) = 2x, and L(x) = 4+4(x−
2)
(b) (4 points) f (x) = x + sin(2x) at a =
π
, for x ∈ [0, π]
2
Solution: f 0 (x) = 1 + 2 cos(2x), L(x) =
π
π
− (x − )
2
2
1
MATH 1210, Spring 2016
Lab #6 Solution
2. (Related Rates I)
To celebrate the Utes’ victory on the football game last Saturday, your friends decide to have a party
and you are invited to help out with the balloons. Assume that the balloon is cylindrical, and is
inflated at a fixed rate of k liters per minute. Let function R(t) be the radius of the balloon and L(t)
be the length of the balloon at a given time t. Then the relationship between the radius of the balloon,
R(t), and its length, L(t) , is given by R(t) = q · L(t) for some constant q. Given that the volume of the
balloon at time t is described by V (t), where V (t) = π(R(t))2 L(t), answer the following questions:
(a) (4 points) At what rate is the length of the balloon changing? Express
L(t).
Solution:
dL
dt
depending on k, q and
dR
dt
depending on k, q and
dV
dL
= πq 2 3L(t)2
dt
dt
dL
k
=
dt
3πq 2 L(t)2
V (t) = πq 2 L(t)3
k=
(b) (4 points) At what rate is the radius of the balloon changing? Express
R(t).
Solution:
R(t) = qL(t)
dR
dL
kq
kq
=q
=
=
dt
dt
3πq 2 L(t)2
3πR(t)2
(c) (2 points) Use the result from part (b) to explain that why the radius is changing faster when it is
smaller.
Solution: The rate of change of the radius is proportional to the reciprocal of the square of
the radius. So it is small, if the radius is big, and big, if the radius is small.
2
Due: 02/25/2016 in class
Instructor: Janina Letz
MATH 1210, Spring 2016
Lab #6 Solution
Name:
3. (Approximation) Find an approximation for the following numerical values using differentials.
Compare with the exact calculator values after you are done.
√
(a) (4 points) 4.15
√
√
1
x, x = 4, dx = 0.15, dy = √ dx ' 0.0375. Hence 4.15 ' 4 + 0.0375 =
2 x
2.0375. Actual value 2.0371 . . . .
Solution: y =
(b) (4 points)
√
3
√
125.5
Solution: y =
√
3
x, x = 125, dx = 0.5, dy =
0.0067 = 5.0067. Actual value 5.00665 . . . .
(c) (4 points)
√
√
√
1
√
dx ' 0.0067. Hence 3 125.5 ' 3 125 +
3
2
3 x
15.99
√
√
1
x, x = 16, dx = −0.01, dy = √ dx = −0.00125. Hence 15.99 ' 16 −
2 x
0.00125 = 3.99875. Actual value 3.998749 . . .
Solution: y =
√
3
MATH 1210, Spring 2016
Lab #6 Solution
4. (10 points) (Related Rates II) A 20-foot ladder is leaning against a building. If the bottom of the
ladder is sliding along the level pavement away from the building at 1 foot per second, how fast is the
top of the ladder moving down when the foot of the ladder is 5 feet from the wall?
Solution: The ladder, the pavement and the building form a right triangle,
so that if the bottom
√
of the ladder is x feet from the wall then the top of the ladder is h = 202 − x2 feet from the floor.
Taking derivatives with respect to t (measured in seconds), we get
dx
1
dh
· (−2x) .
= √
dt
dt
2 202 − x2
For x = 5 and
dh
5
dx
= 1 we get
= −√
≈ −0.258 feet per second.
dt
dt
375
5. (10 points) (Related Rates III) A light in a lighthouse 2 kilometer offshore from a straight line is
rotating at 2 revolutions per minute. How fast is the beam moving along the shoreline, when it passes
the point 2 kilometer from the point opposite the lighthouse.
lighthouse
α
d=2km
Solution:
shoreline
s=2km
The light is rotating at 2 revolutions per minute, so
π
4.
= 4π
1
min .
ds
1
dα
=d
2
dt
cos(α) dt
s = d tan(α)
For s = 2 km it is tan(α) = 1, so α =
dα
dt
So
ds
1
1
km
= 2 km 1 4π
= 16π
dt
min
min
2
4
It is tan(α) = ds . So