The base of a circular fence with radius 10 m is given by x = 10 cos(t), y = 10 sin(t). The height of the fence at position (x, y) is given by the function h(x, y) = 5 + 0.05(x2 − y2), so the height varies from 0 m to 10 m. Suppose that 1 L of paint covers 100 m2. Determine how much paint you will need if you paint both sides of the fence. (Round your answer to two decimal places.)

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:

First, we calculate the length of the circular fence i.e. the circumference

So, we have:

Substitute x and y in

Expand

Factor out 100

Substitute

For a sine function

The average value is:

Similarly, the average height in is:

So, the area of both sides of the fence is:

This gives

From the question, we understand that:

The amount of paint for is:

Substitute

Hence, you will need 6.28 L of paint

Read more about areas at:

link

  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.

$4

step-by-step explanation:

$30×5=$150

$150 minus $130=$20

$20÷5=$4.

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