Answer:
6.28 L
Step-by-step explanation:
The length of the circular fence is ...
C = 2πr = 2π(10 m) = 20π m
The height in terms of t is ...
h(t) = 5 +0.05((10cos(t))^2 -(10sin(t))^2) = 5 +5(1 -2sin(t)^2)
For 0 ≤ t ≤ 2π, the average value of this height function is 5.
The fence has an average height of 5 m and a length of 20π m, so an area of ...
(one side) fence area = (5 m)(20π m) = 100π m²
The area of both sides is double this, or ...
total fence area = 2(100π m²) = 2π(100 m²)
Since 1 liter covers 100 m², we need 2π liters to cover the fence.
The amount of paint needed is 6.28 L.
The area of a shape is the amount of space it covers.
You will need 6.28 L of paint to paint both sides of the fence
The given parameters are:
[tex]\mathbf{r = 10m}[/tex]
[tex]\mathbf{x = 10cos(t)}[/tex]
[tex]\mathbf{y = 10sin(t)}[/tex]
[tex]\mathbf{h(x,y) = 5 + 0.5(x^2 - y^2)}[/tex]
First, we calculate the length of the circular fence i.e. the circumference
[tex]\mathbf{C = 2\pi r}[/tex]
So, we have:
[tex]\mathbf{C = 2\pi \times 10}[/tex]
[tex]\mathbf{C = 20\pi }[/tex]
Substitute x and y in [tex]\mathbf{h(x,y) = 5 + 0.5(x^2 - y^2)}[/tex]
[tex]\mathbf{h(t)= 5 + 0.5((10cos(t))^2 - (10sin(t))^2)}[/tex]
Expand
[tex]\mathbf{h(t)= 5 + 0.5(100cos^2(t) - 100sin^2(t))}[/tex]
Factor out 100
[tex]\mathbf{h(t)= 5 + 50(cos^2(t) - sin^2(t))}[/tex]
Substitute [tex]\mathbf{cos^2(t)= 1 - sin^2(t)}[/tex]
[tex]\mathbf{h(t)= 5 + 50(1 - sin^2(t) - sin^2(t))}[/tex]
[tex]\mathbf{h(t)= 5 + 50(1 - 2sin^2(t))}[/tex]
For a sine function
[tex]\mathbf{h(t)= asin(bt + c) + d}[/tex]
The average value is:
[tex]\mathbf{Average = d}[/tex]
Similarly, the average height in [tex]\mathbf{h(t)= 5 + 50(1 - 2sin^2(t))}[/tex] is:
[tex]\mathbf{h(t) = 5}[/tex]
So, the area of both sides of the fence is:
[tex]\mathbf{Area = 2 \times h \times C}[/tex]
This gives
[tex]\mathbf{Area = 2 \times 5 \times 20\pi}[/tex]
[tex]\mathbf{Area = 200\pi}[/tex]
From the question, we understand that:
[tex]\mathbf{1\ liter = 100m^2}[/tex]
The amount of paint for [tex]\mathbf{200\pi}[/tex] is:
[tex]\mathbf{Paint = \frac{200\pi}{100}L}[/tex]
[tex]\mathbf{Paint = 2\pi \ L}[/tex]
Substitute [tex]\mathbf{\pi = 3.14}[/tex]
[tex]\mathbf{Paint = 2\times 3.14 \ L}[/tex]
[tex]\mathbf{Paint = 6.28 \ L}[/tex]
Hence, you will need 6.28 L of paint
Read more about areas at:
https://brainly.com/question/9541556
