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the climbing frame name: amazon the pta are investigating some new play…

Question

the climbing frame
name: amazon

the pta are investigating some new playground equipment. they are thinking of buying a climbing frame like this.

  • this shape is made out of two identical parabolic metal frames with nets between them.
  • the maximum height is 3.5 metres.
  • the width of one metal frame is 5.44m.
  • the net is attached at from the bottom of the frame and one rope goes up to the vertex of the parabolic frame.
  • the net has 18 wires this length.
  • what is the equation of the parabola?
  • investigate the total length of the wire needed for the 18 wires in the net
  • safety regulations state that the angle of elevation of the wire from the base of the frame to the highest point cannot be more than \\(45^\circ\\). investigate whether the climbing frames net will meet this requirement?
  • the climbing frame must be surrounded by a safe fall zone.
  • the fall zone must have an area of \\(44.2\text{ m}^2\\).
  • the fall zone will be covered in bark to a depth of 15cm. the bark costs \\$195 a \\(\text{m}^3\\) (this price includes gst).
  • the bark in the fall zone must be kept in by an edging.
  • the edging comes in straight lengths. the lengths are 150mm high and come in 2200mm lengths.
  • each length costs \\$67.50 including gst.
  • investigate possible shapes for the fall zone and the dimensions of this fall zone.
  • what volume of bark will be required to cover the fall zone?
  • as a school, the pta do not need to pay gst. how much will the pta have to pay for the bark to fill the fall zone and the edging for the fall zone?

state and discuss any assumptions that you have made while calculating your answers.

Explanation:

Find the parabolic equation

We model the parabolic frame on a Cartesian plane.
Let the vertex be on the y-axis at \((0, 3.5)\).
Since the total width is \(5.44\text{ m}\), the x-intercepts are at \((-2.72, 0)\) and \((2.72, 0)\).
Using the vertex form \(y = ax^2 + c\):

$$ y = ax^2 + 3.5 $$

Substitute the point \((2.72, 0)\) to find \(a\):

$$ 0 = a(2.72)^2 + 3.5 \implies a = -\frac{3.5}{7.3984} \approx -0.473 $$

Thus, the equation of the parabola is:

$$ y = -0.473x^2 + 3.5 $$

Calculate wire length

The net has 18 wires of "this length", which go from the bottom of the frame up to the vertex.
Assuming the wires connect the base boundary to the vertex \((0, 3.5)\):
The distance from a base corner \((2.72, 0)\) to the vertex \((0, 3.5)\) along a straight line is:

$$ L = \sqrt{2.72^2 + 3.5^2} = \sqrt{7.3984 + 12.25} = \sqrt{19.6484} \approx 4.43\text{ m} $$

For 18 wires of this length:

$$ \text{Total length} = 18 \times 4.4327 \approx 79.79\text{ m} $$

Check angle of elevation

The angle of elevation \(\theta\) from the base corner to the vertex is:

$$ \tan(\theta) = \frac{\text{Height}}{\text{Half-width}} = \frac{3.5}{2.72} \approx 1.2868 $$
$$ \theta = \arctan(1.2868) \approx 52.1^\circ $$

Since \(52.1^\circ > 45^\circ\), the climbing frame's net does not meet the safety requirement.

Analyze fall zone dimensions

The fall zone must have an area of \(44.2\text{ m}^2\).
If we assume a rectangular fall zone surrounding the climbing frame:
Let the frame width be \(5.44\text{ m}\) and depth be \(d\).
If the fall zone extends \(1.5\text{ m}\) around the frame:

$$ \text{Width} = 5.44 + 2(1.5) = 8.44\text{ m} $$
$$ \text{Depth} = d + 2(1.5) $$

For an area of \(44.2\text{ m}^2\):

$$ \text{Length} \times \text{Width} = 44.2 \implies \text{Length} = \frac{44.2}{8.44} \approx 5.24\text{ m} $$

Thus, a rectangular zone of \(8.44\text{ m} \times 5.24\text{ m}\) gives an area of \(44.2\text{ m}^2\).
The perimeter of this rectangular zone is:

$$ P = 2(8.44 + 5.24) = 27.36\text{ m} $$

Calculate bark volume

The depth of the bark is \(15\text{ cm} = 0.15\text{ m}\).
The area of the fall zone is \(44.2\text{ m}^2\).

$$ V = \text{Area} \times \text{depth} = 44.2 \times 0.15 = 6.63\text{ m}^3 $$

Calculate cost excluding GST

GST in New Zealand is \(15\%\). To exclude GST, we divide the retail price by \(1.15\).

  1. Bark Cost:

Retail price is \(\$195\text{ per m}^3\) (including GST).
Price excluding GST:

$$ \text{Price}_{\text{excl}} = \frac{195}{1.15} \approx \$169.57\text{ per m}^3 $$

Total bark cost:

$$ \text{Cost}_{\text{bark}} = 6.63 \times 169.565 \approx \$1,124.22 $$
  1. Edging Cost:

Perimeter is \(27.36\text{ m} = 27360\text{ mm}\).
Each edging length is \(2200\text{ mm}\).
Number of lengths needed:

$$ N = \frac{27360}{2200} \approx 12.44 \implies 13\text{ lengths} $$

Retail price per length is \(\$67.50\) (including GST).
Price excluding GST:

$$ \text{Price}_{\text{excl}} = \frac{67.50}{1.15} \approx \$58.70 $$

Total edging cost:

$$ \text{Cost}_{\text{edging}} = 13 \times 58.696 \approx \$763.05 $$
  1. Total Cost:

\[
\text{Total Cost} = 1,124.22…

Answer:

Question 1

The equation of the parabola is:

$$y = -0.473x^2 + 3.5$$

Question 2

The total length of wire needed for the 18 wires is approximately:

$$79.79\text{ m}$$

Question 3

The angle of elevation is:

$$\theta \approx 52.1^\circ$$

Since \(52.1^\circ > 45^\circ\), the net does not meet the safety requirement.

Question 4

A possible rectangular fall zone has dimensions of:

$$8.44\text{ m} \times 5.24\text{ m}$$

with a perimeter of \(27.36\text{ m}\).

Question 5

The volume of bark required is:

$$6.63\text{ m}^3$$

Question 6

The total cost for the PTA excluding GST is:

$$\$1,887.27$$

(comprising \(\$1,124.22\) for bark and \(\$763.05\) for 13 lengths of edging)