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3 years ago

The circumference of a sphere was measured to be 82 cm with a possible error of 0.5 cm.

Mathematics

1 answer:

Nat2105 [25]3 years ago

3 0

Answer:

A) Maximum error = 170.32 cm³

B)Relative error = 0.0575

Step-by-step explanation:

A) Formula for circumference is: C = 2πr

Differentiating with respect to r, we have;

dC/dr = 2π

r is small, so we can write;

ΔC/Δr = 2π

So, Δr = ΔC/2π

We are told that ΔC = 0.5.

Thus; Δr = 0.5/2π = 0.25/π

Now, formula for Volume of a sphere is;

V(r) = (4/3)πr³

Differentiating with respect to r, we have;

dV/dr = 4πr²

Again, r is small, so we can write;

ΔS/Δr = 4πr²

ΔV = 4πr² × Δr

Rewriting, we have;

ΔV = ((2πr)²/π) × Δr

Since C = 2πr, we now have;

ΔV = (C²/π)Δr

ΔV will be maximum when Δr is maximum

Thus, ΔV = (C²/π) × 0.25/π

C = 82 cm

Thus;

ΔV = (82²/π) × 0.25/π

ΔV = 170.32 cm³

B) Formula for relative error = ΔV/V

Relative error = 170.32/((4/3)πr³)

Relative error = 170.32/((4/3)C³/8π³)

Relative errror = 170.32/((4/3)82³/8π³)

Relative error = 170.32/2963.744

Relative error = 0.0575

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3 years ago

A rectangular enclosure is to be created using 82m rope.

atroni [7]

Let
x-----> the length of rectangle
y-----> the width of rectangle

we know that
perimeter of rectangle=2*[x+y]
perimeter of rectangle=82 m
82=2*[x+y]---> divide by 2 both sides---> 41=x+y--> y=41-x---> equation 1

Area of rectangle=x*y
substitute equation 1 in the area formula
Area=x*[41-x]----> 41x-x²

using a graph tool
see the attached figure

the vertex is the point (20.5,420.25)
that means
for x=20.5 m ( length of rectangle)
the area is 420.25 m²

y=420.25/20.5----> 20.5 m

the dimensions are
20.5 m x 20.5 m------> is a square

the answer part 1)
the dimensions of the rectangular with Maximum area is a square with length side 20.5 meters

Part 2)b) Suppose 41 barriers each 2m long, are used instead. Can the same area be enclosed?
divide the length side of the square by 2
so
20.5/2=10.25--------> 10 barriers
the dimensions are 10 barriers x 10 barriers
10 barriers=10*2---> 20 m

the area enclosed with barriers is =20*20----> 400 m²

400 m² < 420.25 m²
so

the answer Part 2) is
the area enclosed by the barriers is less than the area enclosed by the rope

Part 3)How much more area can be enclosed if the rope is used instead of the barriers

area using the rope=420.25 m²
area using the barriers=400 m²

420.25-400=20.25 m²

the answer part 3) is
20.25 m²

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3 years ago

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Answer:

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3 years ago

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