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University of the People *
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Course
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Subject
Mathematics
Date
Nov 24, 2024
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Uploaded by AngelRoseBenedict
Written Assignment Unit 6
University of the People
MATH 1211: Calculus
Dr. Sanjay Yadav
December 27, 2023
1. First, make the approximation that the radar station is at ground level. Thus, we have a right triangle. If y is the radar station, then y is constant at 6 km.
Using the pythagorean theorem,
s
2
=
x
2
+
y
2
Now we take the derivative of the Pythagorean relationship with respect to time t using the chain rule:
2
s∙
d s
dt
=
2
x∙
d x
d t
Since y is constant, we do not have a 2y(dy/dt). Thus, we cancel out the constant
2 in the equation above:
s∙
d s
dt
=
x∙
d x
d t
The problem states that s is decreasing at a rate of 400 km/hr when s=10 km. The problem is asking for the horizontal speed dx/dt with respect to time. Therefore, we solve for the derivative equation for dx/dt:
s∙
d s
dt
x
=
d x
d t
But since x = (s
2
- y
2
)
1/2
s = 10 km
y = 6 km
ds/dt = 400 km/hr
10
∙
400
¿¿
Thus,
10
∙
400
¿¿
4000
¿¿
4000
8
=
d x
dt
d x
d t
=
500
Therefore, the horizontal speed of the plane as it approaches towards the radar station at a decreasing rate of 400 km/hr is 500 km/hr
.
2. Suppose that s is the distance from the boat to the pulley and x is the distance to the dock. Therefore,
s
2
=
s
2
+
x
2
Solving for dx/dt when ds/dt = 3 and x = 8:
2
s∙
ds
dx
=
2
x ∙
dx
dt
If x = 8,
s
=
√
6
2
+
8
2
s
=
10
Plugging the values in the equation 2
s∙
ds
dx
=
2
x ∙
dx
dt
:
2
(
10
)
∙
3
=
2
(
8
)
∙
dx
dt
dx
dt
=
60
16
Therefore, the boat is approaching the dock at 3.75 m/s
.
3. Let a = 120 ft be the height of the kite above the ground and c =130 ft the length of the string of the kite and b be the distance of the girl from the kite. Let db/dt = 6 ft/s be the rate of the kite flying away fro the girl.
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Applying the pythagorean relationship to the scenario:
a
2
+
b
2
=
c
2
120
2
+
b
2
=
130
2
b
2
=
130
2
−
120
2
b
=
√
2500
b
=
50
Now, we differentiate the equation with respect to time t on both sides of the pythagorean equation:
2
a∙
d a
dt
+
2
b∙
d b
d t
=
2
c∙
dc
dt
Here, we plug in the values in the variables of the equation above:
2
(
120
)
∙
0
+
2
(
50
)
∙
6
=
2
(
130
)
∙
d c
dt
0
+
600
=
260
∙
d c
dt
600
/
260
=
dc
dt
600
/
260
=
dc
dt
2.3077
=
dc
dt
Therefore, the approximate rate of the string as the girl is releasing the string is 2.3077 ft/sec.
4. Find thelinearization of f
(
x
)=
sinx
at x
=
π
/
2
.
L
(
x
)=
f
(
x
)+
f '
(
x
)(
x −a
)
f
(
π
/
2
)=
sin
(
π
/
2
)=
1
f '
(
π
/
2
)=
cos
(
π
/
2
)=
0
L
(
x
)=
1
+
0
(
x −
(
π
/
2
))
L
(
x
)=
1
+
0
L
(
x
)=
1
5. Use linear approximation to estimate e
−
0.01
at x=0.
If we consider the function f
(
x
)=
e
x
. Then f
(
0
)=
e
0
=
1
and f '
(
0
)=
e
0
=
1.
Then
L
(
x
)=
1
+
1
(
x−
0
)
L
(
x
)=
x
+
1
L
(
−
0.01
)=
−
0.01
+
1
L
(
−
0.01
)=
−
0.01
+
1
L
(
−
0.01
)=
0.99
6. Find the minimum and maximum value of the functions:
A. f
(
x
)=
x
3
−
3
x
+
2
Differentiating the function:
f '
(
x
)=
3
x
2
−
3
Substituting f’(x)=0 in the equation to find the critical points:
0
=
3
x
2
−
3
3
x
2
=
3
Dividing both side of the equation by 3:
x
2
=
1
Squaring both side:
√
x
2
=
√
1
x
=
±
1
Therefore, the critical values are -1 and 1.
Now we find the second derivative:
f ' '
(
x
)=
6
x
Substituting the critical values in the equation above to determine if the derivative is less than or greater than 0.
f ' '
(
1
)=
6
(
1
)=
6
f
(
1
)=(
1
)
3
−
3
(
1
)+
2
f
(
1
)=
0
f ' '
(
−
1
)=
6
(
−
1
)=
−
6
f
(
−
1
)=(
−
1
)
3
−
3
(
−
1
)+
2
f
(
−
1
)=
−
1
+
3
+
2
=
4
Thus, the minimum value of the function is (1, 0) and the maximum value is at (-
1, 4).
B. f
(
x
)=
x
4
−
8
x
+
3
Differentiating the function: f '
(
x
)=
4
x
3
−
8
Finding the critical points if f’(x)=0:
0
=
4
x
3
−
8
−
4
x
3
=
8
Dividing both side by -4:
x
3
=
−
2
3
√
x
=
3
√
2
x
=
±
1.26
Critical values are -1.26 and 1.26
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Second derivative: f ' '
(
x
)=
12
x
2
Third derivative: f ' ' '
(
x
)=
24
x
Substituting the critical values in the 3
rd
derivative:
f ' ' '
(
1.26
)=
24
(
1.26
)=
30.24
f
(
1.26
)=(
1.26
)
4
−
8
(
1.26
)+
3
f
(
1.26
)=
2.52
−
10.08
+
3
f
(
1.26
)=
−
4.46
f ' ' '
(
−
1.26
)=
24
(
−
1.26
)=
−
30.24
f
(
−
1.26
)=(
−
1.26
)
4
−
8
(
−
1.26
)+
3
f
(
−
1.26
)=
2.52
+
10.08
+
3
f
(
−
1.26
)=
15.6
Therefore, the minimum value of the function is (1.26, -4.46) and the maximum value is (-1.26, 15.6).
7. First, we find f’(x) for f
(
x
)=
x
+
ln
(
x
2
−
1
)
.
f '
(
x
)=
1
+
2
x
x
2
−
1
Now setting f’(x) to 0:
0
=
1
+
2
x
x
2
−
1
x
2
−
1
+
2
x
=
0
x
2
+
2
x −
1
=
0
Solving for x using quadratic formula to find critical points:
x
=
−b±
√
b
2
−
4
ac
2
a
x
=
−
2
±
√
2
2
−
4
(
2
)(
−
1
)
2
(
1
)
x
=
−
2
±
√
4
−
8
2
x
=
−
2
±
√
4
+
8
2
x
=
−
2
±
√
12
2
x
=
−
1
±
6
√
4
x
=
±
7
√
4
Critical points are 7
√
4
and -7
√
4
.
8. To show that the equation x
3
+
x−
1
=
0
has one root exactly, we use the Rolle’s Theorem, which states that if a real-valued function f(x) is continuous on a closed interval [a, b], differentiable on the open interval [a, b], and f(a) = f(b), then there exists at least one c in the open interval (a, b) such that f '
(
c
)=
0.
First, we verify continuity and differentiability of the function. f
(
x
)=
x
3
+
x −
1
is a polynomial, therefore it is continuous on (
−∞,∞
)
and differentiable on (
−∞,∞
)
.
Then, we find f(a) and f(b). In our equation, a = 1 and b = 1, therefore,
f
(
a
)=
x
3
+
x−
1
f
(
1
)=(
1
)
3
+(
1
)
−
1
f
(
1
)=
1
f
(
b
)=
x
3
+
x−
1
f
(
1
)=(
1
)
3
+(
1
)
−
1
f
(
1
)=
1
Therefore, f(a) = f(b).
Now, using Rolle’s Theorem:
f '
(
x
)=
x
3
+
x −
1
f '
(
x
)=
3
x
2
+
1
0
=
3
x
2
+
1
−
3
x
2
=
1
x
2
=
−
1
/
3
x
=
±
√
1
/
3
Critical points are √
1
/
3
and −
√
1
/
3
.
Using the Mean Value Theorem:
f '
(
c
)=
f
(
b
)
−f
(
a
)
b−a
f '
(
c
)=
1
−
1
1
−
1
=
0
Therefore, there is exactly one root for the equation x
3
+
x−
1
, since f(a) = f(b), and f '
(
c
)=
0.
9. If f(1) = 10 and f '
(
x
)
≥
2
for
1
≤ x≤
4
, how small can f(4) possibly be?
Using the Mean Value Theorem to solve for f(4). Given that f '
(
x
)
≥
2
for
1
≤ x≤
4
, we set a = 1 and b = 4. Thus,
f '
(
c
)=
f
(
b
)
−f
(
a
)
b−a
f '
(
c
)
≥
f
(
4
)
−f
(
1
)
4
−
1
f
(
4
)
−f
(
1
)
≤
2
∙
(
4
−
1
)
f
(
4
)
−f
(
1
)
≤
2
∙
(
4
−
1
)
f
(
4
)
−
10
≤
6
f
(
4
)
≤
16
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Therefore, the smallest possible value for f(4) is 16
.
10. From the given graph, the function is concaving down between the intervals [-2.5, 0]
and [0, 5.8]
. Conversely, the function is concaving up between the interval [-1.8, 3.5].
Reference
All formulas and theorems are derived from: Herman, E. & Strang, G. (2020).
Calculus volume 1
. OpenStacks. Rice University.
https://openstax.org